Chapter 6 Systems of Linear Equations: Exercises

Compiled from the original manuscript of Chapter 6 exercises provided by Professor Ping-Zen Ong: 44 top-level exercises in all, in the same order as in the manuscript. Every exercise comes with a hint and a complete solution that can be expanded; computational exercises are followed by a NumPy verification, while pure proof exercises have no accompanying program.

This part uses 0-based indexing. Inconsistent notation and ambiguities in the statements of the original manuscript are noted in the corresponding exercises, and computed results follow the original data.

import numpy as np
np.set_printoptions(precision=4, suppress=True, linewidth=100)

def print_header(title):
    print("=" * 60)
    print(f"  {title}")
    print("=" * 60)

def print_step(step, desc):
    print(f"\n▶ Step {step}: {desc}")
    print("-" * 40)

def compare_print(label, actual, expected_desc):
    print(f"\n[{label}]")
    print(f"  Computed value: {actual}")
    print(f"  Expected value: {expected_desc}")

def check(label, actual, expected):
    compare_print(label, actual, expected)
    assert np.allclose(actual, expected), label
    print("  check passed")

def rref(A, tol=1e-10):
    """Gauss–Jordan elimination with partial pivoting; only for the small numerical examples of this chapter."""
    R = np.array(A, dtype=float, copy=True)
    row = 0
    for col in range(R.shape[1]):
        if row == R.shape[0]:
            break
        pivot = row + np.argmax(np.abs(R[row:, col]))
        if abs(R[pivot, col]) <= tol:
            continue
        R[[row, pivot]] = R[[pivot, row]]
        R[row] /= R[row, col]
        for i in range(R.shape[0]):
            if i != row:
                R[i] -= R[i, col] * R[row]
        row += 1
    R[np.abs(R) < tol] = 0
    return R

def lu_no_pivot(A):
    """Only for the square matrices in this book whose first n-1 pivots are nonzero."""
    U = np.array(A, dtype=float, copy=True)
    n = len(U)
    L = np.eye(n)
    for j in range(n - 1):
        assert abs(U[j, j]) > 1e-10, "zero pivot: this function does not exchange rows"
        for i in range(j + 1, n):
            L[i, j] = U[i, j] / U[j, j]
            U[i] -= L[i, j] * U[j]
    return L, U

Row Operations, Reduced Forms, and Rank¶

Exercise 1 (Row-Reduced Form)

Let $\mathbf{A}=\begin{pmatrix}1&-2&3&1\\2&-4&8&6\\-1&2&-2&1\end{pmatrix}$ .

Use elementary row operations to find its rref, writing out the steps in detail.
Identify the pivot positions and find $\operatorname{rank}(\mathbf{A})$ .

Hint. First eliminate the entries below the pivot in column 0.

# ============================================================
print_header("Exercise 6-1 | Row-Reduced Form")
# ============================================================

# --- Step 1 ---
print_step(1, "Build the matrix and perform Gauss–Jordan elimination")
A = np.array([[1,-2,3,1],[2,-4,8,6],[-1,2,-2,1]])
check("rref(A)", rref(A), [[1,-2,0,-5],[0,0,1,2],[0,0,0,0]])
# --- Step 2 ---
print_step(2, "Check the rank")
check("rank(A)", np.linalg.matrix_rank(A), 2)

============================================================
  Exercise 6-1 | Row-Reduced Form
============================================================

▶ Step 1: Build the matrix and perform Gauss–Jordan elimination
----------------------------------------

[rref(A)]
  Computed value: [[ 1. -2.  0. -5.]
 [ 0.  0.  1.  2.]
 [ 0.  0.  0.  0.]]
  Expected value: [[1, -2, 0, -5], [0, 0, 1, 2], [0, 0, 0, 0]]
  check passed

▶ Step 2: Check the rank
----------------------------------------

[rank(A)]
  Computed value: 2
  Expected value: 2
  check passed

Exercise 2 (Different Ways of Eliminating)

Let $\mathbf{B}=\begin{pmatrix}2&4&6\\1&3&5\end{pmatrix}$ .

Multiply row 0 by $\frac{1}{2}$ and subtract row 1; call the result $\mathbf{E}_0$ . Is it a ref?
Replace row 1 by “2 times row 1 minus row 0”; call the result $\mathbf{E}_1$ . Is it a ref?
Compare the two refs, then find the rref of each and compare.

Note on the original manuscript. Method one does not say which row the result replaces; under the natural reading $\text{row}_1\leftarrow\frac12\text{row}_0-\text{row}_1$ a ref is obtained, and the statement holds. Method two does not specify which row is replaced either; here we stipulate that it replaces row 1.

Hint. A ref can change with the operations used, but the rref is unique.

# ============================================================
print_header("Exercise 6-2 | Different Ways of Eliminating")
# ============================================================

# --- Step 1 ---
print_step(1, "Compare three readings of the operations")
expected = [[1,0,-1],[0,1,2]]
for B in ([[2,4,6],[0,-1,-2]], [[2,4,6],[0,2,4]], [[0,2,4],[1,3,5]]):
    check("rref", rref(B), expected)

============================================================
  Exercise 6-2 | Different Ways of Eliminating
============================================================

▶ Step 1: Compare three readings of the operations
----------------------------------------

[rref]
  Computed value: [[ 1.  0. -1.]
 [ 0.  1.  2.]]
  Expected value: [[1, 0, -1], [0, 1, 2]]
  check passed

[rref]
  Computed value: [[ 1.  0. -1.]
 [ 0.  1.  2.]]
  Expected value: [[1, 0, -1], [0, 1, 2]]
  check passed

[rref]
  Computed value: [[ 1.  0. -1.]
 [ 0.  1.  2.]]
  Expected value: [[1, 0, -1], [0, 1, 2]]
  check passed

Exercise 3 (Recovering the Matrix)

Suppose $\operatorname{rref}(\mathbf{A})=\begin{pmatrix}1&3&0&2\\0&0&1&-1\\0&0&0&0\end{pmatrix}$ , and columns 0 and 2 are $(1,2,1)^\top$ and $(2,5,3)^\top$ , respectively.

Recover $\mathbf{A}$ .
Find a basis of the row space.
Find a basis of the column space.

Hint. Invertible row operations preserve the linear relations among the columns.

# ============================================================
print_header("Exercise 6-3 | Recovering the Matrix")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the rref of the recovered matrix")
A = np.array([[1,3,2,0],[2,6,5,-1],[1,3,3,-1]])
check("rref(A)", rref(A), [[1,3,0,2],[0,0,1,-1],[0,0,0,0]])
check("column relation", A[:,3], 2*A[:,0]-A[:,2])
check("rank of the two basis columns", np.linalg.matrix_rank(A[:,[0,2]]), 2)

============================================================
  Exercise 6-3 | Recovering the Matrix
============================================================

▶ Step 1: Check the rref of the recovered matrix
----------------------------------------

[rref(A)]
  Computed value: [[ 1.  3.  0.  2.]
 [ 0.  0.  1. -1.]
 [ 0.  0.  0.  0.]]
  Expected value: [[1, 3, 0, 2], [0, 0, 1, -1], [0, 0, 0, 0]]
  check passed

[column relation]
  Computed value: [ 0 -1 -1]
  Expected value: [ 0 -1 -1]
  check passed

[rank of the two basis columns]
  Computed value: 2
  Expected value: 2
  check passed

Exercise 4 (Undoing Row Operations)

Apply the following to a $3\times3$ matrix $\mathbf{M}$ in turn: add -2 times row 0 to row 1, interchange rows 1 and 2, and divide row 2 by 3.

Write down the corresponding elementary matrices $\mathbf{E}_0,\mathbf{E}_1,\mathbf{E}_2$ .
Given that the result is $\boldsymbol{\Gamma}=\begin{pmatrix}3&1&1\\0&2&1\\0&0&1\end{pmatrix}$ , recover $\mathbf{M}$ from $\mathbf{E}_2\mathbf{E}_1\mathbf{E}_0\mathbf{M}=\boldsymbol{\Gamma}$ .

Hint. Undo the operations one by one in reverse order.

# ============================================================
print_header("Exercise 6-4 | Undoing Row Operations")
# ============================================================

# --- Step 1 ---
print_step(1, "Redo the three row operations after recovering M")
E0 = np.array([[1,0,0],[-2,1,0],[0,0,1]])
E1 = np.eye(3)[[0,2,1]]
E2 = np.diag([1,1,1/3])
M = np.array([[3,1,1],[6,2,5],[0,2,1]])
check("E2 E1 E0 M", E2@E1@E0@M, [[3,1,1],[0,2,1],[0,0,1]])

============================================================
  Exercise 6-4 | Undoing Row Operations
============================================================

▶ Step 1: Redo the three row operations after recovering M
----------------------------------------

[E2 E1 E0 M]
  Computed value: [[3. 1. 1.]
 [0. 2. 1.]
 [0. 0. 1.]]
  Expected value: [[3, 1, 1], [0, 2, 1], [0, 0, 1]]
  check passed

Exercise 5 (A Parameter and the Rank)

Let $\mathbf{A}=\begin{pmatrix}1&-2&1\\-3&6&\lambda\end{pmatrix}$ .

Use row operations to find a ref.
When does the column space have dimension 1? What is the rank in the other cases?

Hint. Watch whether row 1 vanishes.

# ============================================================
print_header("Exercise 6-5 | A Parameter and the Rank")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the degenerate value and generic values")
for lam in [-3,-2,0,4]:
    A = [[1,-2,1],[-3,6,lam]]
    check(f"λ={lam}: rank", np.linalg.matrix_rank(A), 1 if lam == -3 else 2)

============================================================
  Exercise 6-5 | A Parameter and the Rank
============================================================

▶ Step 1: Check the degenerate value and generic values
----------------------------------------

[λ=-3: rank]
  Computed value: 1
  Expected value: 1
  check passed

[λ=-2: rank]
  Computed value: 2
  Expected value: 2
  check passed

[λ=0: rank]
  Computed value: 2
  Expected value: 2
  check passed

[λ=4: rank]
  Computed value: 2
  Expected value: 2
  check passed

Exercise 6 (Classifying Reduced Forms)

List all $3\times3$ rref matrices.
If $\boldsymbol{\Gamma}$ is such a matrix of rank $r$ , prove that $\operatorname{Col}(\boldsymbol{\Gamma})=\operatorname{span}(\mathbf{e}_0,\ldots,\mathbf{e}_{r-1})$ ; when $r=0$ , read this as the zero subspace.
Prove that sending an rref matrix to its row space gives a bijection between these matrices and all the subspaces of $\mathbb{R}^3$ .

Hint. Classify by the rank and by the columns containing the pivots, and use the uniqueness of the rref. (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 7 (Row Equivalence)

Let $\mathbf{A},\mathbf{B}$ be $m\times n$ matrices such that $\mathbf{B}$ is obtained from $\mathbf{A}$ by finitely many elementary row operations.

Prove that there exists an invertible $m\times m$ matrix $\mathbf{M}$ with $\mathbf{B}=\mathbf{M}\mathbf{A}$ .
Are the kernels equal? Are the images equal? Give a proof or a counterexample.

Hint. Every elementary matrix is invertible, but left multiplication changes the position of the image within the ambient space. (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 8 (Uniqueness of the Reduced Form)

Prove that the rref of any matrix is unique.

Hint. Use the row space to determine the pivot positions, then compare rows with the same pivot coordinates. (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 9 (Reduced Form with a Parameter)

Let $\mathbf{A}_\lambda=\begin{pmatrix}1&1&\lambda\\1&\lambda&1\\\lambda&1&1\end{pmatrix}$ .

For every $\lambda\in\mathbb{R}$ , use row operations to find the rref.
Classify the rank completely.

Hint. First separate out $\lambda=1$ , then look at the last pivot.

# ============================================================
print_header("Exercise 6-9 | Reduced Form with a Parameter")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the two degenerate values and generic values separately")
for lam in [1,-2,0,2,-3]:
    A = np.array([[1,1,lam],[1,lam,1],[lam,1,1]])
    expected = ([[1,1,1],[0,0,0],[0,0,0]] if lam == 1 else
                [[1,0,-1],[0,1,-1],[0,0,0]] if lam == -2 else np.eye(3))
    check(f"λ={lam}: rref", rref(A), expected)

============================================================
  Exercise 6-9 | Reduced Form with a Parameter
============================================================

▶ Step 1: Check the two degenerate values and generic values separately
----------------------------------------

[λ=1: rref]
  Computed value: [[1. 1. 1.]
 [0. 0. 0.]
 [0. 0. 0.]]
  Expected value: [[1, 1, 1], [0, 0, 0], [0, 0, 0]]
  check passed

[λ=-2: rref]
  Computed value: [[ 1.  0. -1.]
 [ 0.  1. -1.]
 [ 0.  0.  0.]]
  Expected value: [[1, 0, -1], [0, 1, -1], [0, 0, 0]]
  check passed

[λ=0: rref]
  Computed value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
  Expected value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
  check passed

[λ=2: rref]
  Computed value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
  Expected value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
  check passed

[λ=-3: rref]
  Computed value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
  Expected value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
  check passed

Exercise 10 (Rank and Degrees of Freedom)

A linear mapping $T:\mathbb{R}^6\to\mathbb{R}^4$ satisfies $\dim\ker T=3$ , and its representing matrix is the $4\times6$ matrix $\mathbf{A}$ .

How many pivots and free variables does the rref have?
Are there zero rows? How many, and where are they?

Hint. Use the fact that the rank plus the dimension of the kernel equals the dimension of the domain. (This is a derivation/proof exercise; no Python verification is attached.)

Homogeneous Systems of Linear Equations¶

Exercise 11 (Solving Homogeneous Systems)

Use Gaussian elimination to find the complete solution of each of the following three homogeneous systems:

$x+2y-z=0$ , $3x+6y-2z=0$ , $-2x-4y+7z=0$ .
$x-y+2z=0$ , $2x+y-z=0$ , $-x+2y+z=0$ , $x+4y+3z=0$ .
$2z=0$ , $x+2y+2z=0$ , $3x+6y+5z=0$ .

Hint. A homogeneous system always has at least the zero solution; the number of pivots determines the degrees of freedom.

# ============================================================
print_header("Exercise 6-11 | Solving Homogeneous Systems")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the three rrefs and the kernel vectors")
matrices=[[[1,2,-1],[3,6,-2],[-2,-4,7]], [[1,-1,2],[2,1,-1],[-1,2,1],[1,4,3]], [[0,0,2],[1,2,2],[3,6,5]]]
for i,A in enumerate(matrices):
    expected=np.vstack([np.eye(3),np.zeros((1,3))]) if i==1 else [[1,2,0],[0,0,1],[0,0,0]]
    check(f"system {i+1}: rref", rref(A), expected)
    if i != 1:check("kernel vector", np.array(A)@np.array([-2,1,0]), np.zeros(3))

============================================================
  Exercise 6-11 | Solving Homogeneous Systems
============================================================

▶ Step 1: Check the three rrefs and the kernel vectors
----------------------------------------

[system 1: rref]
  Computed value: [[1. 2. 0.]
 [0. 0. 1.]
 [0. 0. 0.]]
  Expected value: [[1, 2, 0], [0, 0, 1], [0, 0, 0]]
  check passed

[kernel vector]
  Computed value: [0 0 0]
  Expected value: [0. 0. 0.]
  check passed

[system 2: rref]
  Computed value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]
 [0. 0. 0.]]
  Expected value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]
 [0. 0. 0.]]
  check passed

[system 3: rref]
  Computed value: [[1. 2. 0.]
 [0. 0. 1.]
 [0. 0. 0.]]
  Expected value: [[1, 2, 0], [0, 0, 1], [0, 0, 0]]
  check passed

[kernel vector]
  Computed value: [0 0 0]
  Expected value: [0. 0. 0.]
  check passed

Exercise 12 (Kernel in Four Unknowns)

Consider $x_0+2x_1-x_2+3x_3=0$ , $2x_0+4x_1-x_2+8x_3=0$ , $-x_0-2x_1+2x_2-x_3=0$ .

Write the system as $\mathbf{A}\mathbf{x}=\mathbf{0}$ .
Find the rref $\boldsymbol{\Gamma}$ of the coefficient matrix.
Identify the pivot variables and the free variables.
Find the general solution, a basis of the kernel, and its dimension.

Hint. Keep $x_1,x_3$ as free parameters.

# ============================================================
print_header("Exercise 6-12 | Kernel in Four Unknowns")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the rref")
A=np.array([[1,2,-1,3],[2,4,-1,8],[-1,-2,2,-1]])
check("rref(A)", rref(A), [[1,2,0,5],[0,0,1,2],[0,0,0,0]])
# --- Step 2 ---
print_step(2, "Check the basis of the kernel")
N=np.array([[-2,-5],[1,0],[0,-2],[0,1]])
check("AN", A@N, np.zeros((3,2)))
check("rank(N)", np.linalg.matrix_rank(N), 2)

============================================================
  Exercise 6-12 | Kernel in Four Unknowns
============================================================

▶ Step 1: Check the rref
----------------------------------------

[rref(A)]
  Computed value: [[1. 2. 0. 5.]
 [0. 0. 1. 2.]
 [0. 0. 0. 0.]]
  Expected value: [[1, 2, 0, 5], [0, 0, 1, 2], [0, 0, 0, 0]]
  check passed

▶ Step 2: Check the basis of the kernel
----------------------------------------

[AN]
  Computed value: [[0 0]
 [0 0]
 [0 0]]
  Expected value: [[0. 0.]
 [0. 0.]
 [0. 0.]]
  check passed

[rank(N)]
  Computed value: 2
  Expected value: 2
  check passed

Exercise 13 (Only the Zero Solution)

Let $\mathbf{B}=\begin{pmatrix}1&1&2\\1&2&5\\2&3&8\end{pmatrix}$ , and consider $\mathbf{B}(x,y,z)^\top=\mathbf{0}$ .

Use row operations to find a ref.
Is there a nonzero solution? Explain.

Hint. Is the last pivot zero?

# ============================================================
print_header("Exercise 6-13 | Only the Zero Solution")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the rank and the reduced form")
B=np.array([[1,1,2],[1,2,5],[2,3,8]])
check("rref(B)", rref(B), np.eye(3))
check("rank(B)", np.linalg.matrix_rank(B), 3)

============================================================
  Exercise 6-13 | Only the Zero Solution
============================================================

▶ Step 1: Check the rank and the reduced form
----------------------------------------

[rref(B)]
  Computed value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
  Expected value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
  check passed

[rank(B)]
  Computed value: 3
  Expected value: 3
  check passed

Exercise 14 (Homogeneous Solutions with a Parameter)

Consider $x+y+\lambda z=0$ , $x+\lambda y+z=0$ , $\lambda x+y+z=0$ , where $\lambda\in\mathbb{R}$ .

Use row operations to determine when there is only the zero solution.
When are there nonzero solutions? Find the dimension of the solution space and the general solution.

Hint. Use the three rrefs of the parameter matrix found earlier.

# ============================================================
print_header("Exercise 6-14 | Homogeneous Solutions with a Parameter")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the kernels and their dimensions in the degenerate cases")
for lam,N in [(1,np.array([[-1,-1],[1,0],[0,1]])),(-2,np.ones((3,1)))]:
    A=np.array([[1,1,lam],[1,lam,1],[lam,1,1]])
    check("AN", A@N, np.zeros((3,N.shape[1])))
    check("nullity(A)", 3-np.linalg.matrix_rank(A), N.shape[1])
check("only the zero solution when λ=0", np.linalg.matrix_rank([[1,1,0],[1,0,1],[0,1,1]]), 3)

============================================================
  Exercise 6-14 | Homogeneous Solutions with a Parameter
============================================================

▶ Step 1: Check the kernels and their dimensions in the degenerate cases
----------------------------------------

[AN]
  Computed value: [[0 0]
 [0 0]
 [0 0]]
  Expected value: [[0. 0.]
 [0. 0.]
 [0. 0.]]
  check passed

[nullity(A)]
  Computed value: 2
  Expected value: 2
  check passed

[AN]
  Computed value: [[0.]
 [0.]
 [0.]]
  Expected value: [[0.]
 [0.]
 [0.]]
  check passed

[nullity(A)]
  Computed value: 1
  Expected value: 1
  check passed

[only the zero solution when λ=0]
  Computed value: 3
  Expected value: 3
  check passed

Exercise 15 (Solving from the Row Space)

The row space of a real matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$ is spanned by $(1,-1,2)$ and $(2,1,1)$ .

Write down a system of equations with the same solutions as $\mathbf{A}\mathbf{x}=\mathbf{0}$ , and explain why.
Find the general solution and determine its geometric shape.

Hint. A vector must be sent to zero by every row vector in the row space.

# ============================================================
print_header("Exercise 6-15 | Solving from the Row Space")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the kernel of the two equivalent equations")
B=np.array([[1,-1,2],[2,1,1]])
check("B(-1,1,1)^T", B@np.array([-1,1,1]), np.zeros(2))
check("rank(B)", np.linalg.matrix_rank(B), 2)

============================================================
  Exercise 6-15 | Solving from the Row Space
============================================================

▶ Step 1: Check the kernel of the two equivalent equations
----------------------------------------

[B(-1,1,1)^T]
  Computed value: [0 0]
  Expected value: [0. 0.]
  check passed

[rank(B)]
  Computed value: 2
  Expected value: 2
  check passed

Exercise 16 (Constraints from a Nonzero Solution)

A real matrix $\mathbf{A}$ is $4\times3$ , and $\mathbf{A}\mathbf{x}=\mathbf{0}$ is known to have a nonzero solution.

What are the largest and the smallest possible ranks?
Must the original system or the ref of its coefficient matrix have a zero row? Explain the difference between the two.

Hint. A nonzero kernel vector gives nullity at least 1, but it does not force the original matrix to have a zero row. (This is a derivation/proof exercise; no Python verification is attached.)

Nonhomogeneous Systems¶

Exercise 17 (Solving Nonhomogeneous Systems)

Use Gaussian elimination to find the complete solution of each of the following three systems; if there are infinitely many solutions, describe the geometric relation to the solution set of the corresponding homogeneous system.

$x+2y-z=1$ , $3x+6y-2z=0$ , $-2x-4y+z=0$ .
$x-y+2z=4$ , $2x+y-z=0$ , $-x+2y+z=-2$ , $x+4y+3z=0$ .
$2z=2$ , $x+2y+2z=1$ , $3x+6y+5z=2$ .

Note on the original manuscript. The missing variable in equation 2 of system 1 has been fixed; but the new coefficient is $z$ , which differs from the $7z$ in system 1 of Exercise 11, so the statement’s “reference” to the coefficients of the homogeneous version no longer matches. With either $z$ or $7z$ there is no solution, which is still consistent with the exercise asking for the complete solution only when a solution exists.

Hint. An augmented matrix with a row $[0 0 0\mid c]$ , $c\ne0$ , means there is no solution.

# ============================================================
print_header("Exercise 6-17 | Solving Nonhomogeneous Systems")
# ============================================================

# --- Step 1 ---
print_step(1, "Check that both the new z and the 7z reading from the homogeneous version have no solution")
for A,b in [(np.array([[1,2,-1],[3,6,-2],[-2,-4,1]]),[1,0,0]),
            (np.array([[1,2,-1],[3,6,-2],[-2,-4,7]]),[1,0,0])]:
    check("rank(A)", np.linalg.matrix_rank(A), 2)
    check("rank([A | b])", np.linalg.matrix_rank(np.column_stack([A,b])), 3)
# --- Step 2 ---
print_step(2, "Verify the unique solution and the affine family of solutions")
A=np.array([[1,-1,2],[2,1,-1],[-1,2,1],[1,4,3]])
check("system 2 substituted", A@np.array([1,-1,1]), [4,0,-2,0])
A=np.array([[0,0,2],[1,2,2],[3,6,5]])
check("system 3 particular solution", A@np.array([-1,0,1]), [2,1,2])
check("system 3 homogeneous direction", A@np.array([-2,1,0]), [0,0,0])

============================================================
  Exercise 6-17 | Solving Nonhomogeneous Systems
============================================================

▶ Step 1: Check that both the new z and the 7z reading from the homogeneous version have no solution
----------------------------------------

[rank(A)]
  Computed value: 2
  Expected value: 2
  check passed

[rank([A | b])]
  Computed value: 3
  Expected value: 3
  check passed

[rank(A)]
  Computed value: 2
  Expected value: 2
  check passed

[rank([A | b])]
  Computed value: 3
  Expected value: 3
  check passed

▶ Step 2: Verify the unique solution and the affine family of solutions
----------------------------------------

[system 2 substituted]
  Computed value: [ 4  0 -2  0]
  Expected value: [4, 0, -2, 0]
  check passed

[system 3 particular solution]
  Computed value: [2 1 2]
  Expected value: [2, 1, 2]
  check passed

[system 3 homogeneous direction]
  Computed value: [0 0 0]
  Expected value: [0, 0, 0]
  check passed

Exercise 18 (Structure of the General Solution of a Nonhomogeneous System)

Consider $x_0+2x_1-x_2+3x_3=2$ , $2x_0+4x_1-x_2+8x_3=9$ , $-x_0-2x_1+2x_2-x_3=3$ .

Write the augmented matrix and find its rref.
Set the free variables to zero to find a particular solution $\mathbf{x}_p$ .
Find the general solution $\mathbf{x}_h$ of the corresponding homogeneous system.
Write the solution as $\mathbf{x}_p+\mathbf{x}_h$ and explain its geometric relation to the kernel.

Hint. Row 2 of the coefficient matrix equals row 1 minus 3 times row 0; the right-hand side also satisfies $3=9-3\cdot2$ .

# ============================================================
print_header("Exercise 6-18 | Structure of the General Solution of a Nonhomogeneous System")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the rref of the augmented matrix and consistency")
A = np.array([[1, 2, -1, 3], [2, 4, -1, 8], [-1, -2, 2, -1]])
b = np.array([2, 9, 3])
augmented = np.column_stack([A, b])
check("rref of the augmented matrix", rref(augmented),
      [[1, 2, 0, 5, 7], [0, 0, 1, 2, 5], [0, 0, 0, 0, 0]])
check("rank(A)", np.linalg.matrix_rank(A), 2)
check("rank([A | b])", np.linalg.matrix_rank(augmented), 2)

# --- Step 2 ---
print_step(2, "Check the particular solution and the homogeneous solution space")
xp = np.array([7, 0, 5, 0])
N = np.array([[-2, -5], [1, 0], [0, -2], [0, 1]])
check("A x_p", A @ xp, b)
check("AN", A @ N, np.zeros((3, 2)))
check("number of kernel basis vectors vs. rank", np.linalg.matrix_rank(N), A.shape[1] - np.linalg.matrix_rank(A))

# --- Step 3 ---
print_step(3, "Substitute the affine general solution x = x_p + x_h")
for s, t in [(0, 0), (1, -2), (-3, 4)]:
    x = xp + N @ np.array([s, t])
    check(f"s={s}, t={t}: general solution", x, [7 - 2*s - 5*t, s, 5 - 2*t, t])
    check("A x", A @ x, b)

============================================================
  Exercise 6-18 | Structure of the General Solution of a Nonhomogeneous System
============================================================

▶ Step 1: Check the rref of the augmented matrix and consistency
----------------------------------------

[rref of the augmented matrix]
  Computed value: [[1. 2. 0. 5. 7.]
 [0. 0. 1. 2. 5.]
 [0. 0. 0. 0. 0.]]
  Expected value: [[1, 2, 0, 5, 7], [0, 0, 1, 2, 5], [0, 0, 0, 0, 0]]
  check passed

[rank(A)]
  Computed value: 2
  Expected value: 2
  check passed

[rank([A | b])]
  Computed value: 2
  Expected value: 2
  check passed

▶ Step 2: Check the particular solution and the homogeneous solution space
----------------------------------------

[A x_p]
  Computed value: [2 9 3]
  Expected value: [2 9 3]
  check passed

[AN]
  Computed value: [[0 0]
 [0 0]
 [0 0]]
  Expected value: [[0. 0.]
 [0. 0.]
 [0. 0.]]
  check passed

[number of kernel basis vectors vs. rank]
  Computed value: 2
  Expected value: 2
  check passed

▶ Step 3: Substitute the affine general solution x = x_p + x_h
----------------------------------------

[s=0, t=0: general solution]
  Computed value: [7 0 5 0]
  Expected value: [7, 0, 5, 0]
  check passed

[A x]
  Computed value: [2 9 3]
  Expected value: [2 9 3]
  check passed

[s=1, t=-2: general solution]
  Computed value: [15  1  9 -2]
  Expected value: [15, 1, 9, -2]
  check passed

[A x]
  Computed value: [2 9 3]
  Expected value: [2 9 3]
  check passed

[s=-3, t=4: general solution]
  Computed value: [-7 -3 -3  4]
  Expected value: [-7, -3, -3, 4]
  check passed

[A x]
  Computed value: [2 9 3]
  Expected value: [2 9 3]
  check passed

Exercise 19 (The Augmented-Rank Criterion)

Consider $\mathbf{A}(x,y,z)^\top=\mathbf{b}$ , where

\mathbf{A}=\begin{pmatrix}1&1&2\\1&2&5\\2&3&7\end{pmatrix}, \mathbf{b}=(2,5,\alpha)^\top.

(28)

Find the rref of the augmented matrix, and determine when there is a solution and what characterizes the case of no solution.
Explain why unequal ranks lead to no solution.
Prove that $\mathbf{b}\in\operatorname{Col}(\mathbf{A})$ when there is a solution.

Hint. Row 2 equals the sum of the first two rows.

# ============================================================
print_header("Exercise 6-19 | The Augmented-Rank Criterion")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the rrefs in the consistent and inconsistent cases")
A=np.array([[1,1,2],[1,2,5],[2,3,7]])
for alpha in [7,0,8]:
    expected=([[1,0,-1,-1],[0,1,3,3],[0,0,0,0]] if alpha==7 else [[1,0,-1,0],[0,1,3,0],[0,0,0,1]])
    check(f"α={alpha}", rref(np.column_stack([A,[2,5,alpha]])), expected)
check("particular solution", A@np.array([-1,3,0]), [2,5,7])
check("homogeneous direction", A@np.array([1,-3,1]), [0,0,0])

============================================================
  Exercise 6-19 | The Augmented-Rank Criterion
============================================================

▶ Step 1: Check the rrefs in the consistent and inconsistent cases
----------------------------------------

[α=7]
  Computed value: [[ 1.  0. -1. -1.]
 [ 0.  1.  3.  3.]
 [ 0.  0.  0.  0.]]
  Expected value: [[1, 0, -1, -1], [0, 1, 3, 3], [0, 0, 0, 0]]
  check passed

[α=0]
  Computed value: [[ 1.  0. -1.  0.]
 [ 0.  1.  3.  0.]
 [ 0.  0.  0.  1.]]
  Expected value: [[1, 0, -1, 0], [0, 1, 3, 0], [0, 0, 0, 1]]
  check passed

[α=8]
  Computed value: [[ 1.  0. -1.  0.]
 [ 0.  1.  3.  0.]
 [ 0.  0.  0.  1.]]
  Expected value: [[1, 0, -1, 0], [0, 1, 3, 0], [0, 0, 0, 1]]
  check passed

[particular solution]
  Computed value: [2 5 7]
  Expected value: [2, 5, 7]
  check passed

[homogeneous direction]
  Computed value: [0 0 0]
  Expected value: [0, 0, 0]
  check passed

Exercise 20 (Nonhomogeneous System with a Parameter)

Consider the nonhomogeneous system with a real parameter $\lambda$ : $x+y+\lambda z=1$ , $x+\lambda y+z=1$ , $\lambda x+y+z=1$ .

When is there exactly one solution? Find it.
When is there no solution?
When are there infinitely many solutions? Find the general solution.

Hint. Check the classification of the rank of the coefficient matrix together with the consistency of the constant vector.

# ============================================================
print_header("Exercise 6-20 | Nonhomogeneous System with a Parameter")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the unique solution for generic parameters")
for lam in [0,2,-3]:
    A=np.array([[1,1,lam],[1,lam,1],[lam,1,1]])
    check(f"λ={lam}: solution", np.linalg.solve(A,np.ones(3)), np.ones(3)/(lam+2))
# --- Step 2 ---
print_step(2, "Check the two special values")
A=np.array([[1,1,-2],[1,-2,1],[-2,1,1]])
check("λ=-2: rank of the augmented matrix", np.linalg.matrix_rank(np.column_stack([A,np.ones(3)])), 3)
A=np.ones((3,3)); N=np.array([[-1,-1],[1,0],[0,1]])
check("λ=1: particular solution", A@np.array([1,0,0]), np.ones(3))
check("λ=1: homogeneous directions", A@N, np.zeros((3,2)))

============================================================
  Exercise 6-20 | Nonhomogeneous System with a Parameter
============================================================

▶ Step 1: Check the unique solution for generic parameters
----------------------------------------

[λ=0: solution]
  Computed value: [0.5 0.5 0.5]
  Expected value: [0.5 0.5 0.5]
  check passed

[λ=2: solution]
  Computed value: [0.25 0.25 0.25]
  Expected value: [0.25 0.25 0.25]
  check passed

[λ=-3: solution]
  Computed value: [-1. -1. -1.]
  Expected value: [-1. -1. -1.]
  check passed

▶ Step 2: Check the two special values
----------------------------------------

[λ=-2: rank of the augmented matrix]
  Computed value: 3
  Expected value: 3
  check passed

[λ=1: particular solution]
  Computed value: [1. 1. 1.]
  Expected value: [1. 1. 1.]
  check passed

[λ=1: homogeneous directions]
  Computed value: [[0. 0.]
 [0. 0.]
 [0. 0.]]
  Expected value: [[0. 0.]
 [0. 0.]
 [0. 0.]]
  check passed

Solution Sets and Affine Translations¶

Exercise 21 (Structure of the General Solution)

Let $\mathbf{A}$ be an $m\times n$ matrix, and suppose $\mathbf{x}_p$ satisfies $\mathbf{A}\mathbf{x}_p=\mathbf{b}$ .

Prove that $\mathbf{x}_p+\mathbf{u}$ is a solution for every $\mathbf{u}\in\ker\mathbf{A}$ .
Prove that every solution $\mathbf{y}$ satisfies $\mathbf{y}-\mathbf{x}_p\in\ker\mathbf{A}$ .
Explain how a particular solution and a basis of the kernel completely determine the general solution.

Hint. Prove the inclusion of the two sets in each direction. (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 22 (True or False: Translations)

For a subspace $W\subseteq\mathbb{R}^n$ , define $W_{\mathbf{a}}=\{\mathbf{w}+\mathbf{a}:\mathbf{w}\in W\}$ . Decide whether each of the following statements is true:

In $\mathbb{R}^2$ , if $W$ is $y=x$ and $\mathbf{a}=(2,0)^\top$ , then $W_{\mathbf{a}}$ is $x-y=2$ .
In $\mathbb{R}^3$ , if $W$ is the line of intersection of $x+y+z=0$ and $x-y-2z=0$ and $\mathbf{a}=(1,2,3)^\top$ , then $W_{\mathbf{a}}=\{(1+t,2-2t,3+t)^\top:t\in\mathbb{R}\}$ .
If $W$ is the plane $x+y+z=0$ and $\mathbf{a}=(1,0,-1)^\top$ , then the translate is still the same plane.
If $W$ is the plane $x+y+z=0$ and $\mathbf{a}=(-3,0,1)^\top$ , then the translate is $\{\mathbf{a}+s(8,-2,-6)^\top+t(-4,1,3)^\top:s,t\in\mathbb{R}\}$ .

Hint. Translation does not change the direction space, and the two directions of a plane must be linearly independent.

# ============================================================
print_header("Exercise 6-22 | True or False: Translations")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the wrong and the correct direction in statement 2")
B=np.array([[1,1,1],[1,-1,-2]])
check("residual of the given direction", B@np.array([1,-2,1]), [0,1])
check("residual of the correct direction", B@np.array([1,-3,2]), [0,0])
# --- Step 2 ---
print_step(2, "Check the remaining statements about translations")
check("statement 1: translation constant", np.array([1,-1])@np.array([2,0]), 2)
check("statement 3: translation constant", np.sum([1,0,-1]), 0)
check("statement 4: rank of the directions", np.linalg.matrix_rank(np.array([[8,-4],[-2,1],[-6,3]])), 1)
check("constant of the correct plane", np.sum([-3,0,1]), -2)

============================================================
  Exercise 6-22 | True or False: Translations
============================================================

▶ Step 1: Check the wrong and the correct direction in statement 2
----------------------------------------

[residual of the given direction]
  Computed value: [0 1]
  Expected value: [0, 1]
  check passed

[residual of the correct direction]
  Computed value: [0 0]
  Expected value: [0, 0]
  check passed

▶ Step 2: Check the remaining statements about translations
----------------------------------------

[statement 1: translation constant]
  Computed value: 2
  Expected value: 2
  check passed

[statement 3: translation constant]
  Computed value: 0
  Expected value: 0
  check passed

[statement 4: rank of the directions]
  Computed value: 1
  Expected value: 1
  check passed

[constant of the correct plane]
  Computed value: -2
  Expected value: -2
  check passed

Exercise 23 (Properties of Translations)

Let $V$ be a linear space, $W\subseteq V$ a subspace, $\mathbf{a},\mathbf{b}\in V$ , and $W_{\mathbf{a}}=\{\mathbf{w}+\mathbf{a}:\mathbf{w}\in W\}$ . Prove:

$\mathbf{a}\in W_{\mathbf{a}}$ .
If $\mathbf{x}\in W_{\mathbf{a}}$ , then $\mathbf{x}-\mathbf{a}\in W$ .
If $\mathbf{x},\mathbf{y}\in W_{\mathbf{a}}$ , then $\mathbf{x}-\mathbf{y}\in W$ .
If $\mathbf{a}\in W$ , then $W_{\mathbf{a}}=W$ .
If $\mathbf{a}\in W_{\mathbf{b}}$ , then $\mathbf{b}\in W_{\mathbf{a}}$ .

Hint. By the definition, write each member of a set as a vector of the subspace plus the translation vector. (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 24 (The Solution Set Is Affine)

Let $\mathbf{A}$ be an $m\times n$ matrix, and let $\mathbf{x}_p$ be a particular solution of $\mathbf{A}\mathbf{x}=\mathbf{b}$ . Prove that the solution set is exactly $\ker(\mathbf{A})_{\mathbf{x}_p}$ .

Hint. Both inclusions for the translated kernel must be proved. (This is a derivation/proof exercise; no Python verification is attached.)

LU and Permutation Matrices¶

Exercise 25 (LU Decompositions of 2×2 Matrices)

Throughout, $\mathbf{L}$ is required to be unit lower triangular and $\mathbf{U}$ upper triangular. For each of the following seven matrices, in order:

\begin{pmatrix}2&1\\5&3\end{pmatrix}, \begin{pmatrix}2&0\\10&5\end{pmatrix}, \begin{pmatrix}0&1\\0&-1\end{pmatrix}, \begin{pmatrix}0&1\\1&0\end{pmatrix}, \begin{pmatrix}0&0\\1&0\end{pmatrix}, \begin{pmatrix}1&0\\1&0\end{pmatrix}, \begin{pmatrix}0&0\\0&1\end{pmatrix}.

(34)

Decide whether an LU decomposition exists; if it exists, find all the decompositions and determine uniqueness; if not, explain why.
Is the $\mathbf{U}$ of each decomposition a ref?

Hint. Compare the four entries of the product.

Solution to Exercise 25

Step 1. Let $\mathbf{L}=\begin{pmatrix}1&0\\\ell&1\end{pmatrix}$ and $\mathbf{U}=\begin{pmatrix}u&v\\0&w\end{pmatrix}$ . For $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , the conditions are $u=a,v=b,\ell a=c,w=d-\ell b$ . If $a\ne0$ , the unique solution is $\ell=c/a$ ; if $a=0,c\ne0$ , there is no solution; if $a=c=0$ , $\ell$ is arbitrary.

Step 2. The complete classification of each matrix is as follows; in the table, $\ell$ determines $\mathbf{L}$ .

Order	$\ell$	$\mathbf{U}$	Uniqueness and ref
1	$5/2$	$\begin{pmatrix}2&1\\0&1/2\end{pmatrix}$	Unique; a ref
2	5	$\begin{pmatrix}2&0\\0&5\end{pmatrix}$	Unique; a ref
3	Arbitrary	$\begin{pmatrix}0&1\\0&-1-\ell\end{pmatrix}$	Infinitely many; a ref only when $\ell=-1$
4	Does not exist	Does not exist	Because $0\ell=1$ is impossible
5	Does not exist	Does not exist	Because $0\ell=1$ is impossible
6	1	$\begin{pmatrix}1&0\\0&0\end{pmatrix}$	Unique; a ref
7	Arbitrary	$\begin{pmatrix}0&0\\0&1\end{pmatrix}$	Infinitely many; none is a ref

Hence $\underline{\text{an upper triangular matrix is not necessarily a ref}}$ ; a ref also requires the pivots to move to the right and the zero rows to be at the bottom.

# ============================================================
print_header("Exercise 6-25 | LU Decompositions of 2×2 Matrices")
# ============================================================

# --- Step 1 ---
print_step(1, "Check every matrix for which a decomposition exists")
cases = [(np.array([[2,1],[5,3]]), [2.5]),
         (np.array([[2,0],[10,5]]), [5]),
         (np.array([[0,1],[0,-1]]), [-1,0,2]),
         (np.array([[1,0],[1,0]]), [1]),
         (np.array([[0,0],[0,1]]), [-2,0,3])]
for A, values in cases:
    for ell in values:
        L = np.array([[1,0],[ell,1]])
        U = np.array([[A[0,0],A[0,1]],[0,A[1,1]-ell*A[0,1]]])
        check(f"ℓ={ell}: LU", L@U, A)

============================================================
  Exercise 6-25 | LU Decompositions of 2×2 Matrices
============================================================

▶ Step 1: Check every matrix for which a decomposition exists
----------------------------------------

[ℓ=2.5: LU]
  Computed value: [[2. 1.]
 [5. 3.]]
  Expected value: [[2 1]
 [5 3]]
  check passed

[ℓ=5: LU]
  Computed value: [[ 2  0]
 [10  5]]
  Expected value: [[ 2  0]
 [10  5]]
  check passed

[ℓ=-1: LU]
  Computed value: [[ 0  1]
 [ 0 -1]]
  Expected value: [[ 0  1]
 [ 0 -1]]
  check passed

[ℓ=0: LU]
  Computed value: [[ 0  1]
 [ 0 -1]]
  Expected value: [[ 0  1]
 [ 0 -1]]
  check passed

[ℓ=2: LU]
  Computed value: [[ 0  1]
 [ 0 -1]]
  Expected value: [[ 0  1]
 [ 0 -1]]
  check passed

[ℓ=1: LU]
  Computed value: [[1 0]
 [1 0]]
  Expected value: [[1 0]
 [1 0]]
  check passed

[ℓ=-2: LU]
  Computed value: [[0 0]
 [0 1]]
  Expected value: [[0 0]
 [0 1]]
  check passed

[ℓ=0: LU]
  Computed value: [[0 0]
 [0 1]]
  Expected value: [[0 0]
 [0 1]]
  check passed

[ℓ=3: LU]
  Computed value: [[0 0]
 [0 1]]
  Expected value: [[0 0]
 [0 1]]
  check passed

Exercise 26 (True or False: LU Decompositions)

For a $2\times2$ square matrix $\mathbf{A}$ , decide whether each of the following statements is true or false, and prove your answer (the lower triangular factor of an LU decomposition must be unit lower triangular):

If $\det\mathbf{A}\ne0$ , there is a unique LU decomposition.
If $\det\mathbf{A}\ne0$ and an LU decomposition exists, the decomposition is unique.
If $\det\mathbf{A}=0$ , an LU decomposition must exist.
If the LU decomposition is unique, then $\det\mathbf{A}\ne0$ .
If there are two or more LU decompositions, then $\det\mathbf{A}=0$ .
If there is no LU decomposition, then $\det\mathbf{A}\ne0$ .

Hint. Use the complete classification of the previous exercise to find counterexamples. (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 27 (Conditions on the Leading Pivots)

If $\mathbf{A}=\begin{pmatrix}a_{00}&a_{01}\\a_{10}&a_{11}\end{pmatrix}$ is invertible and $a_{00}\ne0$ , prove that the LU decomposition is unique.
Explain why the invertibility assumption can be dropped.
If $\mathbf{B}=(b_{ij})_{0\le i,j<3}$ satisfies $b_{00}\ne0$ , $b_{00}b_{11}-b_{10}b_{01}\ne0$ and is invertible, prove that the LU decomposition is unique.

Erratum in the original manuscript. Part 3 mistakenly writes $\mathbf{B}$ as $\mathbf{A}$ ; here it is written uniformly as $\mathbf{B}$ .

Hint. First determine the elimination multipliers, then the upper triangular entries. (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 28 (Uniqueness of the Decomposition)

Let $\mathbf{A}$ be an invertible $n\times n$ matrix, $n\ge1$ .

If the diagonal of the lower triangular factor is not required to consist of 1s, is an existing LU decomposition unique?
Restoring the unit lower triangular requirement, prove that an existing LU decomposition is unique.

Hint. Compare two decompositions to obtain a matrix that is both upper and lower triangular. (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 29 (Elimination for a 3×3 Matrix)

Using the elementary row operations $\mathbf{E}_{ij}(c)$ (adding $c$ times row $j$ to row $i$ ), find the LU decomposition of

\mathbf{A}=\begin{pmatrix}2&1&1\\4&5&2\\2&10&7\end{pmatrix}

(39)

and verify the product.

Hint. Enter the elimination multipliers into a unit lower triangular matrix.

# ============================================================
print_header("Exercise 6-29 | Elimination for a 3×3 Matrix")
# ============================================================

# --- Step 1 ---
print_step(1, "Compute and compare the LU factors")
A = np.array([[2,1,1],[4,5,2],[2,10,7]])
L,U = lu_no_pivot(A)
check("L", L, [[1,0,0],[2,1,0],[1,3,1]])
check("U", U, [[2,1,1],[0,3,0],[0,0,6]])
check("LU", L@U, A)

============================================================
  Exercise 6-29 | Elimination for a 3×3 Matrix
============================================================

▶ Step 1: Compute and compare the LU factors
----------------------------------------

[L]
  Computed value: [[1. 0. 0.]
 [2. 1. 0.]
 [1. 3. 1.]]
  Expected value: [[1, 0, 0], [2, 1, 0], [1, 3, 1]]
  check passed

[U]
  Computed value: [[2. 1. 1.]
 [0. 3. 0.]
 [0. 0. 6.]]
  Expected value: [[2, 1, 1], [0, 3, 0], [0, 0, 6]]
  check passed

[LU]
  Computed value: [[ 2.  1.  1.]
 [ 4.  5.  2.]
 [ 2. 10.  7.]]
  Expected value: [[ 2  1  1]
 [ 4  5  2]
 [ 2 10  7]]
  check passed

Exercise 30 (Decomposing a Rectangular Matrix)

Let $\mathbf{C}=\begin{pmatrix}1&2&0&1\\2&4&1&3\\3&6&1&4\end{pmatrix}$ .

Use row elimination to find a ref $\mathbf{U}$ and a unit lower triangular matrix $\mathbf{L}$ such that $\mathbf{C}=\mathbf{L}\mathbf{U}$ .
Find the rank, and explain the relation between the nonzero rows of $\mathbf{U}$ and $\operatorname{Row}(\mathbf{C})$ .

Hint. Record the elimination multipliers $2,3,1$ .

# ============================================================
print_header("Exercise 6-30 | Decomposing a Rectangular Matrix")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the LU decomposition and the rank")
C = np.array([[1,2,0,1],[2,4,1,3],[3,6,1,4]])
L = np.array([[1,0,0],[2,1,0],[3,1,1]])
U = np.array([[1,2,0,1],[0,0,1,1],[0,0,0,0]])
check("LU", L@U, C)
check("rank(C)", np.linalg.matrix_rank(C), 2)

============================================================
  Exercise 6-30 | Decomposing a Rectangular Matrix
============================================================

▶ Step 1: Check the LU decomposition and the rank
----------------------------------------

[LU]
  Computed value: [[1 2 0 1]
 [2 4 1 3]
 [3 6 1 4]]
  Expected value: [[1 2 0 1]
 [2 4 1 3]
 [3 6 1 4]]
  check passed

[rank(C)]
  Computed value: 2
  Expected value: 2
  check passed

Exercise 31 (Permutation Matrices)

Let $\bar n=\{0,1,\ldots,n-1\}$ and let $\sigma:\bar n\to\bar n$ be a bijection; define $(\mathbf{P}_\sigma)_{ij}=1$ if $j=\sigma(i)$ , and 0 otherwise.

Prove that each row and each column contains exactly one 1, and prove the converse.
Read off the permutations corresponding to the following matrices:
$\mathbf{I}_3, \begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}, \begin{pmatrix}0&1&0&0\\0&0&0&1\\1&0&0&0\\0&0&1&0\end{pmatrix}.$
(43)
For an $n\times m$ matrix $\mathbf{A}$ , describe the effect of $\mathbf{P}_\sigma\mathbf{A}$ .
Prove that $\mathbf{P}_\sigma\mathbf{P}_\tau=\mathbf{P}_{\tau\circ\sigma}$ .
Prove that $\mathbf{P}_\sigma^\top=\mathbf{P}_\sigma^{-1}=\mathbf{P}_{\sigma^{-1}}$ .

Hint. Compute directly from the definition of the entries; do not switch to another convention for permutation matrices.

# ============================================================
print_header("Exercise 6-31 | Permutation Matrices")
# ============================================================

# --- Step 1 ---
print_step(1, "Read off the three permutations")
perms = [[0,1,2],[1,2,0],[1,3,0,2]]
for sigma in perms:
    P = np.eye(len(sigma))[sigma]
    check("column index of the 1 in each row", np.argmax(P,axis=1), sigma)
# --- Step 2 ---
print_step(2, "Check the order of composition with concrete noncommuting permutations")
sigma = np.array([1,3,0,2]); tau = np.array([1,0,2,3])
P = np.eye(4)[sigma]; Q = np.eye(4)[tau]
check("Pσ Pτ", P@Q, np.eye(4)[tau[sigma]])
check("Pσ^T Pσ", P.T@P, np.eye(4))

============================================================
  Exercise 6-31 | Permutation Matrices
============================================================

▶ Step 1: Read off the three permutations
----------------------------------------

[column index of the 1 in each row]
  Computed value: [0 1 2]
  Expected value: [0, 1, 2]
  check passed

[column index of the 1 in each row]
  Computed value: [1 2 0]
  Expected value: [1, 2, 0]
  check passed

[column index of the 1 in each row]
  Computed value: [1 3 0 2]
  Expected value: [1, 3, 0, 2]
  check passed

▶ Step 2: Check the order of composition with concrete noncommuting permutations
----------------------------------------

[Pσ Pτ]
  Computed value: [[1. 0. 0. 0.]
 [0. 0. 0. 1.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]]
  Expected value: [[1. 0. 0. 0.]
 [0. 0. 0. 1.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]]
  check passed

[Pσ^T Pσ]
  Computed value: [[1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]]
  Expected value: [[1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]]
  check passed

Exercise 32 (Decomposition with Row Exchanges)

For each of the following three matrices $\mathbf{A}$ : $$\begin{pmatrix}0&1&2\1&2&3\2&3&4\end{pmatrix},

\begin{pmatrix}1&2&3\\2&4&0\\0&2&1\end{pmatrix}

(44)

\begin{pmatrix}0&1&2\\0&0&1\\1&3&5\end{pmatrix}

(45)

Explain why no LU decomposition is possible without row exchanges.
Find a permutation matrix $\mathbf{P}$ such that $\mathbf{P}\mathbf{A}=\mathbf{L}\mathbf{U}$ , and write $\mathbf{A}=\mathbf{P}^\top\mathbf{L}\mathbf{U}$ .

Hint. When there is still a nonzero entry below a zero pivot, rows must be exchanged.

# ============================================================
print_header("Exercise 6-32 | Decomposition with Row Exchanges")
# ============================================================

# --- Step 1 ---
print_step(1, "Check PA=LU and A=P^T LU separately")
cases = [([[0,1,2],[1,2,3],[2,3,4]], [1,0,2], [[1,0,0],[0,1,0],[2,-1,1]], [[1,2,3],[0,1,2],[0,0,0]]),
         ([[1,2,3],[2,4,0],[0,2,1]], [0,2,1], [[1,0,0],[0,1,0],[2,0,1]], [[1,2,3],[0,2,1],[0,0,-6]]),
         ([[0,1,2],[0,0,1],[1,3,5]], [2,0,1], np.eye(3), [[1,3,5],[0,1,2],[0,0,1]])]
for A, order, L, U in cases:
    A,L,U = map(np.array,(A,L,U)); P=np.eye(3)[order]
    check("PA", P@A, L@U)
    check("P^T LU", P.T@L@U, A)

============================================================
  Exercise 6-32 | Decomposition with Row Exchanges
============================================================

▶ Step 1: Check PA=LU and A=P^T LU separately
----------------------------------------

[PA]
  Computed value: [[1. 2. 3.]
 [0. 1. 2.]
 [2. 3. 4.]]
  Expected value: [[1 2 3]
 [0 1 2]
 [2 3 4]]
  check passed

[P^T LU]
  Computed value: [[0. 1. 2.]
 [1. 2. 3.]
 [2. 3. 4.]]
  Expected value: [[0 1 2]
 [1 2 3]
 [2 3 4]]
  check passed

[PA]
  Computed value: [[1. 2. 3.]
 [0. 2. 1.]
 [2. 4. 0.]]
  Expected value: [[1 2 3]
 [0 2 1]
 [2 4 0]]
  check passed

[P^T LU]
  Computed value: [[1. 2. 3.]
 [2. 4. 0.]
 [0. 2. 1.]]
  Expected value: [[1 2 3]
 [2 4 0]
 [0 2 1]]
  check passed

[PA]
  Computed value: [[1. 3. 5.]
 [0. 1. 2.]
 [0. 0. 1.]]
  Expected value: [[1. 3. 5.]
 [0. 1. 2.]
 [0. 0. 1.]]
  check passed

[P^T LU]
  Computed value: [[0. 1. 2.]
 [0. 0. 1.]
 [1. 3. 5.]]
  Expected value: [[0 1 2]
 [0 0 1]
 [1 3 5]]
  check passed

Exercise 33 (Row Exchanges for 2×2 Matrices)

Let $\mathbf{A}=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ .

If $a\ne0$ and the matrix is invertible, prove that the LU decomposition is unique.
Explain why the invertibility assumption is superfluous.
Prove that if there is no LU decomposition, then $a=0$ . Does the converse hold?
In the case where there is no LU decomposition, find a PLU decomposition.

Hint. The complete condition for an LU decomposition to exist is $a\ne0$ or $a=c=0$ . (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 34 (Block Decomposition)

Let $\mathbf{B}=(b_{ij})_{0\le i,j<3}$ with $b_{00}\ne0$ and $b_{00}b_{11}-b_{10}b_{01}\ne0$ . If $\mathbf{B}$ is invertible, use the $2\times2$ result and block multiplication to prove that its LU decomposition is unique.

Hint. Separate the first row, the first column, and the lower-right Schur complement. (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 35 (Solving via the Decomposition)

If a matrix $\mathbf{A}$ is known to satisfy $\mathbf{P}\mathbf{A}=\mathbf{L}\mathbf{U}$ , solving the system $\mathbf{A}\mathbf{x}=\mathbf{b}$ can often be sped up. Suppose

\mathbf{P}=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}, \mathbf{L}=\begin{pmatrix}1&0&0\\2&1&0\\-1&3&1\end{pmatrix}, \mathbf{U}=\begin{pmatrix}2&1&-1\\0&3&2\\0&0&4\end{pmatrix}.

(48)

Solve the nonhomogeneous system $\mathbf{A}\mathbf{x}=\mathbf{b}$ , where $\mathbf{b}=(1,2,5)^\top$ : in turn find $\mathbf{b}'=\mathbf{P}\mathbf{b}$ , solve $\mathbf{L}\mathbf{y}=\mathbf{b}'$ , and solve $\mathbf{U}\mathbf{x}=\mathbf{y}$ .
Explain why these steps solve $\mathbf{A}\mathbf{x}=\mathbf{b}$ .

Hint. First permute the constant vector, then perform forward substitution and back substitution.

# ============================================================
print_header("Exercise 6-35 | Solving via the Decomposition")
# ============================================================

# --- Step 1 ---
print_step(1, "Forward substitution, then back substitution")
P=np.eye(3)[[1,2,0]]; L=np.array([[1,0,0],[2,1,0],[-1,3,1]])
U=np.array([[2,1,-1],[0,3,2],[0,0,4]]); b=np.array([1,2,5])
y=np.linalg.solve(L,P@b); x=np.linalg.solve(U,y)
check("P b", P@b, [2,5,1]); check("y", y, [2,1,0]); check("x", x, [5/6,1/3,0])
# --- Step 2 ---
print_step(2, "Rebuild the coefficient matrix and substitute back")
A=P.T@L@U
check("A x", A@x, b)

============================================================
  Exercise 6-35 | Solving via the Decomposition
============================================================

▶ Step 1: Forward substitution, then back substitution
----------------------------------------

[P b]
  Computed value: [2. 5. 1.]
  Expected value: [2, 5, 1]
  check passed

[y]
  Computed value: [ 2.  1. -0.]
  Expected value: [2, 1, 0]
  check passed

[x]
  Computed value: [ 0.8333  0.3333 -0.    ]
  Expected value: [0.8333333333333334, 0.3333333333333333, 0]
  check passed

▶ Step 2: Rebuild the coefficient matrix and substitute back
----------------------------------------

[A x]
  Computed value: [1. 2. 5.]
  Expected value: [1 2 5]
  check passed

LDU and Symmetric Decompositions¶

Exercise 36 (LDU Decompositions of 2×2 Matrices)

LDU means that $\mathbf{L}$ is unit lower triangular, $\mathbf{D}$ is diagonal, and $\mathbf{U}$ is unit upper triangular. Classify each of the following matrices as having a unique decomposition, infinitely many decompositions, or no decomposition, and when a decomposition exists, find all the factors:

\begin{pmatrix}2&1\\5&3\end{pmatrix}, \begin{pmatrix}2&0\\0&7\end{pmatrix}, \begin{pmatrix}0&1\\1&0\end{pmatrix}, \begin{pmatrix}2&8\\1&4\end{pmatrix},

(49)

\begin{pmatrix}1&0\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\1&-1\end{pmatrix}, \begin{pmatrix}0&1\\0&-1\end{pmatrix}, \begin{pmatrix}0&0\\0&1\end{pmatrix}.

(50)

Hint. Expand the four entries of the product of the three factors.

Solution to Exercise 36

Step 1. Let $\mathbf{L}=\begin{pmatrix}1&0\\\ell&1\end{pmatrix}$ , $\mathbf{D}=\operatorname{diag}(p,q)$ , and $\mathbf{U}=\begin{pmatrix}1&u\\0&1\end{pmatrix}$ . The product is $\begin{pmatrix}p&pu\\\ell p&\ell pu+q\end{pmatrix}$ . If $a\ne0$ , the unique solution is $(p,q,\ell,u)=(a,d-cb/a,c/a,b/a)$ ; if $a=0$ , then necessarily $b=c=0$ , in which case $p=0,q=d$ , while $\ell,u$ are arbitrary.

Step 2. In the order of the original exercise:

Order	$(p,q,\ell,u)$ or reason	Classification
1	$(2,1/2,5/2,1/2)$	Unique
2	$(2,7,0,0)$	Unique
3	$a=0$ but $b,c\ne0$	No decomposition
4	$(2,0,1/2,4)$	Unique
5	$(1,0,0,0)$	Unique
6	$a=0$ but $c\ne0$	No decomposition
7	$a=0$ but $b\ne0$	No decomposition
8	$(0,1,\ell,u)$ , with $\ell,u\in\mathbb{R}$ arbitrary	Infinitely many

Hence $\underline{\text{a singular matrix may also have a unique LDU decomposition}}$ .

# ============================================================
print_header("Exercise 6-36 | LDU Decompositions of 2×2 Matrices")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the unique decompositions and the family with free parameters")
cases = [([[2,1],[5,3]],(2,.5,2.5,.5)), ([[2,0],[0,7]],(2,7,0,0)),
         ([[2,8],[1,4]],(2,0,.5,4)), ([[1,0],[0,0]],(1,0,0,0))]
for A,(p,q,ell,u) in cases:
    check("LDU", np.array([[1,0],[ell,1]])@np.diag([p,q])@np.array([[1,u],[0,1]]), A)
for ell,u in [(-2,3),(0,0),(4,-1)]:
    check("LDU with free parameters", np.array([[1,0],[ell,1]])@np.diag([0,1])@np.array([[1,u],[0,1]]), [[0,0],[0,1]])

============================================================
  Exercise 6-36 | LDU Decompositions of 2×2 Matrices
============================================================

▶ Step 1: Check the unique decompositions and the family with free parameters
----------------------------------------

[LDU]
  Computed value: [[2. 1.]
 [5. 3.]]
  Expected value: [[2, 1], [5, 3]]
  check passed

[LDU]
  Computed value: [[2 0]
 [0 7]]
  Expected value: [[2, 0], [0, 7]]
  check passed

[LDU]
  Computed value: [[2. 8.]
 [1. 4.]]
  Expected value: [[2, 8], [1, 4]]
  check passed

[LDU]
  Computed value: [[1 0]
 [0 0]]
  Expected value: [[1, 0], [0, 0]]
  check passed

[LDU with free parameters]
  Computed value: [[0 0]
 [0 1]]
  Expected value: [[0, 0], [0, 1]]
  check passed

[LDU with free parameters]
  Computed value: [[0 0]
 [0 1]]
  Expected value: [[0, 0], [0, 1]]
  check passed

[LDU with free parameters]
  Computed value: [[0 0]
 [0 1]]
  Expected value: [[0, 0], [0, 1]]
  check passed

Exercise 37 (Invertibility and the LDU Decomposition)

Prove that if an invertible $3\times3$ matrix has an LU decomposition, then it has a unique LDU decomposition. Does the statement still hold when the invertibility condition is removed?

Hint. Extract the diagonal matrix from the upper triangular factor, and check for zero pivots. (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 38 (LDU Decomposition with a Parameter)

We wish to express $\mathbf{A}=\begin{pmatrix}1&1&1\\1&\lambda^2&0\\1&0&0\end{pmatrix}$ as a unique LDU decomposition.

For which real numbers $\lambda$ must rows be exchanged?
When no row exchange is needed, find the LDU decomposition.

Hint. First set $t=\lambda^2-1$ , and note that although $\lambda=0$ gives a singular matrix, the decomposition can still be unique.

# ============================================================
print_header("Exercise 6-38 | LDU Decomposition with a Parameter")
# ============================================================

# --- Step 1 ---
print_step(1, "Check generic values and the singular value λ=0")
for lam in [0,2,-3]:
    t=lam**2-1
    A=np.array([[1,1,1],[1,lam**2,0],[1,0,0]])
    L=np.array([[1,0,0],[1,1,0],[1,-1/t,1]])
    D=np.diag([1,t,-lam**2/t]); U=np.array([[1,1,1],[0,1,-1/t],[0,0,1]])
    check(f"λ={lam}: LDU", L@D@U, A)
# --- Step 2 ---
print_step(2, "Check that elimination works for λ=±1 after a row exchange")
for lam in [-1,1]:
    A=np.array([[1,1,1],[1,lam**2,0],[1,0,0]])
    P=np.eye(3)[[0,2,1]]; L,U=lu_no_pivot(P@A)
    check("PA=LU", P@A, L@U)

============================================================
  Exercise 6-38 | LDU Decomposition with a Parameter
============================================================

▶ Step 1: Check generic values and the singular value λ=0
----------------------------------------

[λ=0: LDU]
  Computed value: [[1. 1. 1.]
 [1. 0. 0.]
 [1. 0. 0.]]
  Expected value: [[1 1 1]
 [1 0 0]
 [1 0 0]]
  check passed

[λ=2: LDU]
  Computed value: [[1. 1. 1.]
 [1. 4. 0.]
 [1. 0. 0.]]
  Expected value: [[1 1 1]
 [1 4 0]
 [1 0 0]]
  check passed

[λ=-3: LDU]
  Computed value: [[1. 1. 1.]
 [1. 9. 0.]
 [1. 0. 0.]]
  Expected value: [[1 1 1]
 [1 9 0]
 [1 0 0]]
  check passed

▶ Step 2: Check that elimination works for λ=±1 after a row exchange
----------------------------------------

[PA=LU]
  Computed value: [[1. 1. 1.]
 [1. 0. 0.]
 [1. 1. 0.]]
  Expected value: [[1. 1. 1.]
 [1. 0. 0.]
 [1. 1. 0.]]
  check passed

[PA=LU]
  Computed value: [[1. 1. 1.]
 [1. 0. 0.]
 [1. 1. 0.]]
  Expected value: [[1. 1. 1.]
 [1. 0. 0.]
 [1. 1. 0.]]
  check passed

Exercise 39 (Symmetric Decomposition)

What are the LU and LDU decompositions of the symmetric matrix $\mathbf{A}=\begin{pmatrix}1&2&1\\2&7&5\\1&5&10\end{pmatrix}$ ? Is the upper triangular factor in each equal to $\mathbf{L}^\top$ ?

Hint. Distinguish the upper triangular factor of the LU decomposition from the unit upper triangular factor of the LDU decomposition.

# ============================================================
print_header("Exercise 6-39 | Symmetric Decomposition")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the LU and LDL^T decompositions")
A=np.array([[1,2,1],[2,7,5],[1,5,10]])
L,V=lu_no_pivot(A); D=np.diag([1,3,6])
check("L", L, [[1,0,0],[2,1,0],[1,1,1]])
check("V", V, [[1,2,1],[0,3,3],[0,0,6]])
check("LDL^T", L@D@L.T, A)

============================================================
  Exercise 6-39 | Symmetric Decomposition
============================================================

▶ Step 1: Check the LU and LDL^T decompositions
----------------------------------------

[L]
  Computed value: [[1. 0. 0.]
 [2. 1. 0.]
 [1. 1. 1.]]
  Expected value: [[1, 0, 0], [2, 1, 0], [1, 1, 1]]
  check passed

[V]
  Computed value: [[1. 2. 1.]
 [0. 3. 3.]
 [0. 0. 6.]]
  Expected value: [[1, 2, 1], [0, 3, 3], [0, 0, 6]]
  check passed

[LDL^T]
  Computed value: [[ 1.  2.  1.]
 [ 2.  7.  5.]
 [ 1.  5. 10.]]
  Expected value: [[ 1  2  1]
 [ 2  7  5]
 [ 1  5 10]]
  check passed

Exercise 40 (A Degenerate Symmetric Form)

Let $\mathbf{A}=\begin{pmatrix}0&0\\0&3\end{pmatrix}$ . Explain how it can be written as $\mathbf{L}\mathbf{D}\mathbf{L}^\top$ , where $\mathbf{L}$ is unit lower triangular and $\mathbf{D}$ is diagonal. Is this representation unique?

Hint. Simply let the lower triangular entry be an arbitrary parameter.

# ============================================================
print_header("Exercise 6-40 | A Degenerate Symmetric Form")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the decomposition for different t")
for t in [-2,0,3]:
    L=np.array([[1,0],[t,1]])
    check(f"t={t}: LDL^T", L@np.diag([0,3])@L.T, [[0,0],[0,3]])

============================================================
  Exercise 6-40 | A Degenerate Symmetric Form
============================================================

▶ Step 1: Check the decomposition for different t
----------------------------------------

[t=-2: LDL^T]
  Computed value: [[0 0]
 [0 3]]
  Expected value: [[0, 0], [0, 3]]
  check passed

[t=0: LDL^T]
  Computed value: [[0 0]
 [0 3]]
  Expected value: [[0, 0], [0, 3]]
  check passed

[t=3: LDL^T]
  Computed value: [[0 0]
 [0 3]]
  Expected value: [[0, 0], [0, 3]]
  check passed

Exercise 41 (Classifying Symmetric Decompositions)

For a real symmetric matrix $\mathbf{A}=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ , classify its LDL $^\top$ decompositions with unit lower triangular $\mathbf{L}$ and diagonal $\mathbf{D}$ : when are they unique, infinitely many, or nonexistent? Explain the relation to the rank.

Hint. Compare $a=d_0$ , $b=\ell d_0$ , $c=\ell^2d_0+d_1$ . (This is a derivation/proof exercise; no Python verification is attached.)

Exercise 42 (Deriving the Symmetric Form)

If a real symmetric invertible matrix $\mathbf{A}$ has an LU decomposition, prove that it can be written as $\mathbf{A}=\mathbf{L}\mathbf{D}\mathbf{L}^\top$ , where $\mathbf{L}$ is unit lower triangular and $\mathbf{D}$ is diagonal. Does existence still hold when the invertibility condition is removed?

Hint. A matrix that is both symmetric and upper triangular must be diagonal. (This is a derivation/proof exercise; no Python verification is attached.)

Augmented Matrices and Inversion¶

Exercise 43 (Gauss–Jordan Inversion)

Let $\mathbf{A}=\begin{pmatrix}1&0&0&1\\0&1&-1&2\\2&-1&2&0\\1&0&2&2\end{pmatrix}$ .

Form $[\mathbf{A}\mid\mathbf{I}_4]$ .
Use Gauss–Jordan elimination to reduce the left half to $\mathbf{I}_4$ .
Write down the matrix $\mathbf{B}$ on the right.
Verify that $\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}=\mathbf{I}_4$ .

Hint. Apply the same row operations simultaneously to the left and right halves of the augmented matrix, eliminating downward first and then upward.

# ============================================================
print_header("Exercise 6-43 | Gauss–Jordan Inversion")
# ============================================================

# --- Step 1 ---
print_step(1, "Form the augmented matrix and eliminate downward")
A = np.array([[1, 0, 0, 1], [0, 1, -1, 2], [2, -1, 2, 0], [1, 0, 2, 2]])
G = np.column_stack([A, np.eye(4)])
print("[A | I_4] =\n", G)
G[2] -= 2 * G[0]
G[2] += G[1]
G[3] -= G[0]
G[3] -= 2 * G[2]
check("augmented matrix after downward elimination", G,
      [[1, 0, 0, 1, 1, 0, 0, 0], [0, 1, -1, 2, 0, 1, 0, 0],
       [0, 0, 1, 0, -2, 1, 1, 0], [0, 0, 0, 1, 3, -2, -2, 1]])
check("det(A)", np.linalg.det(A), 1)
check("rank(A)", np.linalg.matrix_rank(A), 4)

# --- Step 2 ---
print_step(2, "Eliminate upward and read off the inverse matrix")
G[0] -= G[3]
G[1] -= 2 * G[3]
G[1] += G[2]
B = G[:, 4:]
expected_inverse = np.array([[-2, 2, 2, -1], [-8, 6, 5, -2],
                             [-2, 1, 1, 0], [3, -2, -2, 1]])
check("left half", G[:, :4], np.eye(4))
check("A^{-1}", B, expected_inverse)

# --- Step 3 ---
print_step(3, "Check by multiplying on both sides")
check("AB", A @ B, np.eye(4))
check("BA", B @ A, np.eye(4))

============================================================
  Exercise 6-43 | Gauss–Jordan Inversion
============================================================

▶ Step 1: Form the augmented matrix and eliminate downward
----------------------------------------
[A | I_4] =
 [[ 1.  0.  0.  1.  1.  0.  0.  0.]
 [ 0.  1. -1.  2.  0.  1.  0.  0.]
 [ 2. -1.  2.  0.  0.  0.  1.  0.]
 [ 1.  0.  2.  2.  0.  0.  0.  1.]]

[augmented matrix after downward elimination]
  Computed value: [[ 1.  0.  0.  1.  1.  0.  0.  0.]
 [ 0.  1. -1.  2.  0.  1.  0.  0.]
 [ 0.  0.  1.  0. -2.  1.  1.  0.]
 [ 0.  0.  0.  1.  3. -2. -2.  1.]]
  Expected value: [[1, 0, 0, 1, 1, 0, 0, 0], [0, 1, -1, 2, 0, 1, 0, 0], [0, 0, 1, 0, -2, 1, 1, 0], [0, 0, 0, 1, 3, -2, -2, 1]]
  check passed

[det(A)]
  Computed value: 0.9999999999999998
  Expected value: 1
  check passed

[rank(A)]
  Computed value: 4
  Expected value: 4
  check passed

▶ Step 2: Eliminate upward and read off the inverse matrix
----------------------------------------

[left half]
  Computed value: [[1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]]
  Expected value: [[1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]]
  check passed

[A^{-1}]
  Computed value: [[-2.  2.  2. -1.]
 [-8.  6.  5. -2.]
 [-2.  1.  1.  0.]
 [ 3. -2. -2.  1.]]
  Expected value: [[-2  2  2 -1]
 [-8  6  5 -2]
 [-2  1  1  0]
 [ 3 -2 -2  1]]
  check passed

▶ Step 3: Check by multiplying on both sides
----------------------------------------

[AB]
  Computed value: [[1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]]
  Expected value: [[1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]]
  check passed

[BA]
  Computed value: [[1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]]
  Expected value: [[1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]]
  check passed

Exercise 44 (Inversion with a Parameter)

Let $\mathbf{B}_a=\begin{pmatrix}1&1&0\\1&2&1\\0&2&a\end{pmatrix}$ .

Perform elimination on $[\mathbf{B}_a\mid\mathbf{I}_3]$ . When does a pivot become zero in a way that no row exchange can remedy? In that case, use the rank to explain whether the inverse matrix exists.
When $a=3$ , find the inverse matrix by Gauss–Jordan elimination.

Hint. The first two pivots are fixed at 1, and the last one is $a-2$ .

# ============================================================
print_header("Exercise 6-44 | Inversion with a Parameter")
# ============================================================

# --- Step 1 ---
print_step(1, "Check the degenerate parameter")
check("rank for a=2", np.linalg.matrix_rank([[1,1,0],[1,2,1],[0,2,2]]), 2)
# --- Step 2 ---
print_step(2, "Check the inverse matrix on both sides")
B=np.array([[1,1,0],[1,2,1],[0,2,3]])
C=np.array([[4,-3,1],[-3,3,-1],[2,-2,1]])
check("B C", B@C, np.eye(3)); check("C B", C@B, np.eye(3))

============================================================
  Exercise 6-44 | Inversion with a Parameter
============================================================

▶ Step 1: Check the degenerate parameter
----------------------------------------

[rank for a=2]
  Computed value: 2
  Expected value: 2
  check passed

▶ Step 2: Check the inverse matrix on both sides
----------------------------------------

[B C]
  Computed value: [[1 0 0]
 [0 1 0]
 [0 0 1]]
  Expected value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
  check passed

[C B]
  Computed value: [[1 0 0]
 [0 1 0]
 [0 0 1]]
  Expected value: [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
  check passed

✓ All exercises completed.