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MA207 - Quantitative Methods — Exercise 212 problems · 1 page

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Problems0/12 done

About this problem set

This set covers the fundamental operations of linear algebra, focusing on Row Echelon Form (REF), matrix rank, and solving systems of linear equations ( $A\mathbf{x} = \mathbf{b}$ and $A\mathbf{x} = \mathbf{0}$ ). You will also explore determinants as a tool for checking linear independence and matrix invertibility. Use these notes to master the connection between a matrix's structure and the nature of its solution space. Pay close attention to how a single row reduction reveals the rank, basis, and solvability.

Keep scrolling — every problem is worked below.

Problem 1(a)

Recognition — You are given a $4 \times 4$ matrix and asked for its rank and null space (the solution space to $A\mathbf{x} = \mathbf{0}$ ). The cue for the rank is the number of non-zero rows after reduction, while the basis for the solution space comes from identifying the “free” variables.

Plan — 1. Reduce matrix $A$ to Row Echelon Form (REF).

2. Count the pivots to find the Rank.

3. Express the variables in $A\mathbf{x} = \mathbf{0}$ in terms of free variables to find the basis.

4. Use the definition of matrix-vector multiplication to find $\mathbf{p}$ and the general solution.

Worked path — To find the rank, we perform row operations. Swapping Row 1 and Row 2 makes the pivot 1, which simplifies the arithmetic:

R_1 \leftrightarrow R_2 \Rightarrow \begin{pmatrix} 1 & 3 & 4 & 2 \\ 5 & 3 & 2 & 1 \\ -1 & 1 & 2 & 1 \\ 2 & 2 & 2 & 1 \end{pmatrix}

Now, eliminate the values below the first pivot using $R_2 \to R_2 - 5R_1$ , $R_3 \to R_3 + R_1$ , and $R_4 \to R_4 - 2R_1$ :

\begin{pmatrix} 1 & 3 & 4 & 2 \\ 0 & -12 & -18 & -9 \\ 0 & 4 & 6 & 3 \\ 0 & -4 & -6 & -3 \end{pmatrix}

Notice that $R_2$ , $R_3$ , and $R_4$ are multiples of each other ( $R_2 = -3R_3$ and $R_4 = -R_3$ ). This means only two rows are linearly independent.

\text{Rank}(A) = 2

For the solution space $A\mathbf{x} = \mathbf{0}$ , we have two free variables ( $x_3$ and $x_4$ ) because there are 4 columns and the rank is 2 ( $4 - 2 = 2$ ). Using $R_3$ , we get the equation $4x_2 + 6x_3 + 3x_4 = 0$ . Solving for $x_2$ :

x_2 = -1.5x_3 - 0.75x_4

Substitute this into $R_1$ ( $x_1 + 3x_2 + 4x_3 + 2x_4 = 0$ ) to solve for $x_1$ :

x_1 + 3(-1.5x_3 - 0.75x_4) + 4x_3 + 2x_4 = 0

x_1 = 0.5x_3 + 0.25x_4

To find the basis vectors, set $(x_3, x_4)$ to $(1, 0)$ and $(0, 1)$ respectively. Scaling to clear fractions gives the basis vectors.

The vector $\mathbf{b}$ is the sum of the first two columns, meaning $\mathbf{b} = 1 \cdot (\text{col}_1) + 1 \cdot (\text{col}_2)$ . By the definition of matrix multiplication, this means $\mathbf{p} = (1, 1, 0, 0)^T$ is a specific solution.

Checkpoint — Rank is 2. Basis for solution space is $\{ (1, -3, 2, 0)^T, (1, -3, 0, 4)^T \}$ . Vector $\mathbf{p} = (1, 1, 0, 0)^T$ . General solution is $\mathbf{x} = \mathbf{p} + c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2$ .

Problem 1(b)

Recognition — Same logic as (a), but for matrix $B$ , starting with a determinant check. A non-zero determinant implies full rank.

Plan — 1. Calculate $\det(B)$ .

2. If $\det(B) \neq 0$ , the rank is 4 and the solution space to $B\mathbf{x} = \mathbf{0}$ is just the zero vector.

3. Apply the same logic for $\mathbf{b}$ and $\mathbf{p}$ .

Worked path — Performing row operations on $B$ to reach REF: $R_2 \to R_2 - 2R_1$ , $R_3 \to R_3 - R_1$ , $R_4 \to R_4 - 2R_1$ results in:

\begin{pmatrix} 1 & -4 & 3 & 2 \\ 0 & 1 & -1 & -3 \\ 0 & 6 & 3 & -2 \\ 0 & -2 & 8 & 0 \end{pmatrix}

Continuing the reduction, $R_3 \to R_3 - 6R_2$ and $R_4 \to R_4 + 2R_2$ :

\begin{pmatrix} 1 & -4 & 3 & 2 \\ 0 & 1 & -1 & -3 \\ 0 & 0 & 9 & 16 \\ 0 & 0 & 6 & -6 \end{pmatrix}

The last two rows are clearly independent. Thus, all 4 rows are independent. The determinant is the product of the pivots (after accounting for row swaps, though we didn't use any here):

\det(B) = 1 \times 1 \times (9 \times -6 - 16 \times 6) = -150

Since $\det(B) \neq 0$ , the rank is 4. The only solution to $B\mathbf{x} = \mathbf{0}$ is the trivial solution $\mathbf{x} = \mathbf{0}$ . Therefore, the basis for the solution space is the empty set (or just the origin). For $B\mathbf{p} = \mathbf{b}$ , the logic remains that $\mathbf{p} = (1, 1, 0, 0)^T$ because $\mathbf{b}$ is the sum of the first two columns.

Checkpoint — $\det(B) = -150$ , Rank is 4, Basis for solution space is $\{ \}$ , General solution is $\mathbf{x} = (1, 1, 0, 0)^T$ .

Problem 2

Recognition — This is a routine Rank and Reduced Row-Echelon Form (RREF) calculation for a $3 \times 4$ matrix.

Plan — 1. Use Gaussian elimination to find REF.

2. Back-substitute to reach RREF (pivots are 1, zeros above and below pivots).

Worked path — Starting with matrix $C$ :

\begin{pmatrix} 3 & 2 & 1 & 3 \\ 2 & 1 & 1 & 0 \\ 6 & 2 & 4 & 6 \end{pmatrix}

$R_1 \to R_1 - R_2$ gives a pivot of 1:

\begin{pmatrix} 1 & 1 & 0 & 3 \\ 2 & 1 & 1 & 0 \\ 6 & 2 & 4 & 6 \end{pmatrix}

Eliminate below: $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 6R_1$ :

\begin{pmatrix} 1 & 1 & 0 & 3 \\ 0 & -1 & 1 & -6 \\ 0 & -4 & 4 & -12 \end{pmatrix}

$R_3 \to R_3 - 4R_2$ :

\begin{pmatrix} 1 & 1 & 0 & 3 \\ 0 & -1 & 1 & -6 \\ 0 & 0 & 0 & 12 \end{pmatrix}

The rank is 3 (three pivots). To reach RREF, divide $R_3$ by 12, $R_2$ by -1, then eliminate above pivots.

Checkpoint — $\text{Rank} = 3$ .

\text{RREF}(C) = \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}

Problem 2(a)

Recognition — You are asked to interpret the RREF of $C$ if it represents an augmented matrix $[A|\mathbf{b}]$ .

Plan — Observe the last row of the RREF. If it represents an equation like $0 = 1$ , the system is inconsistent.

Worked path — The last row of our RREF is $(0, 0, 0, 1)$ . In an augmented matrix context, this represents the equation:

0x_1 + 0x_2 + 0x_3 = 1

Since $0$ can never equal $1$ , there are no solutions.

Checkpoint — No solutions.

Problem 2(b)

Recognition — You are asked to interpret the solutions if $C$ is the coefficient matrix for a homogeneous system $C\mathbf{x} = \mathbf{0}$ .

Plan — Use the Rank-Nullity Theorem: $\text{number of columns} - \text{rank} = \text{dimension of solution space}$ .

Worked path — The matrix $C$ has 4 columns (variables) and we found the rank is 3.

4 - 3 = 1 \text{ free variable}

The solutions form a 1-dimensional subspace (a line through the origin) in $\mathbb{R}^4$ .

Checkpoint — Infinitely many solutions forming a 1D subspace in $\mathbb{R}^4$ .

Problem 3

Recognition — A standard non-homogeneous system $A\mathbf{x} = \mathbf{b}$ followed by conceptual questions about rank and independence.

Plan — 1. Solve $A\mathbf{x} = \mathbf{b}$ using an augmented matrix.

2. Use the RREF to identify the rank and null space basis.

3. Check if $\text{rank} = \text{number of columns}$ for linear independence.

Worked path — Augmented matrix $[A|\mathbf{b}]$ :

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

$R_3 \to R_3 - R_1$ :

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

$R_3 \to R_3 + R_2$ yields a row of zeros:

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

This confirms the rank is 2. For the general solution, let $x_3$ be the free variable $t$ . From $R_2$ : $x_2 + 2t = 3 \Rightarrow x_2 = 3 - 2t$ . From $R_1$ : $x_1 + 2(3-2t) + t = 5 \Rightarrow x_1 = -1 + 3t$ .

Checkpoint — General solution $\mathbf{x} = (-1, 3, 0)^T + t(3, -2, 1)^T$ .

Problem 3(i)

Recognition — Result from the row reduction in the main part of Problem 3.

Checkpoint — Rank = 2.

Problem 3(ii)

Recognition — The solution to $A\mathbf{x} = \mathbf{0}$ is the “homogeneous part” of the general solution found above.

Checkpoint — Basis is $\{ (3, -2, 1)^T \}$ .

Problem 3(iii)

Recognition — Linear independence of columns.

Plan — If rank < number of columns, the vectors are linearly dependent.

Worked path — Rank is 2, but there are 3 column vectors. Therefore, they are linearly dependent.

Checkpoint — No, they are linearly dependent.

Problem 3 (Linear Combination)

Recognition — Expressing $\mathbf{b}$ as combinations of columns.

Plan — Use the general solution $\mathbf{x}$ found earlier. Each specific value of $t$ gives a valid linear combination.

Worked path — The general solution is $\mathbf{x} = (-1, 3, 0)^T + t(3, -2, 1)^T$ . For $t = 0$ : $\mathbf{b} = -1\mathbf{a}_1 + 3\mathbf{a}_2 + 0\mathbf{a}_3$ . For $t = 1$ : $\mathbf{x} = (2, 1, 1)^T$ , so $\mathbf{b} = 2\mathbf{a}_1 + 1\mathbf{a}_2 + 1\mathbf{a}_3$ .

Checkpoint — Two ways: $\mathbf{b} = -\mathbf{a}_1 + 3\mathbf{a}_2$ and $\mathbf{b} = 2\mathbf{a}_1 + \mathbf{a}_2 + \mathbf{a}_3$ .

Ask the chatbot — I have the general solution $\mathbf{x} = \mathbf{p} + t\mathbf{v}$ . Explain why every different value of $t$ provides a different way to write $\mathbf{b}$ as a linear combination of columns, and give me a practice problem where the rank is 1 so I have two free variables to play with.

Problem 4

Recognition — A $3 \times 3$ matrix with a variable $a$ . The determinant determines the existence of non-trivial solutions and linear independence.

Plan — 1. Calculate $\det(B)$ using cofactor expansion.

2. $B\mathbf{x} = \mathbf{0}$ has non-trivial solutions if $\det(B) = 0$ .

3. Columns are linearly independent if $\det(B) \neq 0$ .

Worked path — Expand $\det(B)$ along the first row:

\det(B) = a(1(2) - (-1)(2)) - 1(2(2) - (-1)(0)) + 4(2(2) - 1(0))

\det(B) = a(4) - 4 + 16 = 4a + 12

For non-trivial solutions to $B\mathbf{x} = \mathbf{0}$ , we need $\det(B) = 0$ :

4a + 12 = 0 \Rightarrow a = -3

The columns $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are independent if the determinant is non-zero.

Watch out — Students often confuse “independent” with “det = 0”. Remember: $\det \neq 0$ means the matrix is full rank, and full rank means independence.

Checkpoint — $\det(B) = 4a + 12$ . Non-trivial solutions when $a = -3$ . Linearly independent when $a \neq -3$ .

Problem 5

Recognition — A matrix is “not invertible” if its determinant is zero. This is a characteristic equation problem.

Plan — Set $\det(A) = 0$ and solve for $\lambda$ .

Worked path — Calculate the determinant of the $2 \times 2$ matrix:

(2 - \lambda)(1 - \lambda) - (3)(2) = 0

Expand the quadratic:

2 - 2\lambda - \lambda + \lambda^2 - 6 = 0

\lambda^2 - 3\lambda - 4 = 0

Factor the quadratic: $(\lambda - 4)(\lambda + 1) = 0$ .

Checkpoint — $\lambda = 4$ and $\lambda = -1$ .

What This Set Is Training

See a row of zeros in an augmented matrix $[A|\mathbf{b}]$ where the last entry is non-zero → the system is inconsistent (no solution).
See more columns than the rank → the columns are linearly dependent and $A\mathbf{x} = \mathbf{0}$ has non-trivial solutions.
Calculate the basis for a solution space → identify free variables in RREF and set them to 1 and 0 sequentially (Problem 1(a), p. 1).
Relate a specific solution $\mathbf{p}$ and the null space → the general solution to $A\mathbf{x} = \mathbf{b}$ is always $\mathbf{x} = \mathbf{p} + \text{null space}$ (Problem 3, p. 1).
Use a determinant to find parameter values for invertibility or independence → set $\det(A) = 0$ to find where the matrix “collapses” (Problem 4 and 5, p. 1).

Redo Problem 1, p. 1: It covers every major concept—rank, basis, specific solutions, and the general solution—in one go.

Redo Problem 3, p. 1: This is the best test of whether you understand how the solution to $A\mathbf{x} = \mathbf{b}$ and $A\mathbf{x} = \mathbf{0}$ are geometrically related.