Matrix Operations Exercise

Matrix Operations Exercise

The following matrices are given:

$$A = \begin{bmatrix} 3 & -1 \\ -2 & 4 \\ 0 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 3 \\ 3 & 2 \end{bmatrix}, \quad C = \begin{bmatrix} -3 & -2 \\ 1 & 2 \end{bmatrix}, \quad D = \begin{bmatrix} 2 & -3 & 1 \\ 1 & 4 & 0 \\ -1 & 5 & -4 \end{bmatrix}$$

Questions

(a) Perform the following operations where possible:

(i) $A – 2B$

(ii) $4B – 3C$

(iii) $AC^T$

(iv) $BA$

(b) Calculate the determinant of matrix $D$ and explain whether the matrix is invertible.

Solutions

(a) Matrix operations

(i) $A – 2B$

First, compute $2B$:

$$2B = 2 \times \begin{bmatrix} -1 & 3 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 6 \\ 6 & 4 \end{bmatrix}$$

Matrix $A$ is $3 \times 2$, $2B$ is $2 \times 2$.
🚫 Operation not possible (cannot subtract matrices of different dimensions).

(ii) $4B – 3C$

First, compute $4B$:

$$4B = 4 \times \begin{bmatrix} -1 & 3 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} -4 & 12 \\ 12 & 8 \end{bmatrix}$$

Compute $3C$:

$$3C = 3 \times \begin{bmatrix} -3 & -2 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} -9 & -6 \\ 3 & 6 \end{bmatrix}$$

Now subtract:

$$4B – 3C = \begin{bmatrix} -4 & 12 \\ 12 & 8 \end{bmatrix} – \begin{bmatrix} -9 & -6 \\ 3 & 6 \end{bmatrix} = \begin{bmatrix} 5 & 18 \\ 9 & 2 \end{bmatrix}$$

✅ Answer:

$$4B – 3C = \begin{bmatrix} 5 & 18 \\ 9 & 2 \end{bmatrix}$$

(iii) $AC^T$

First, compute $C^T$:

$$C^T = \begin{bmatrix} -3 & 1 \\ -2 & 2 \end{bmatrix}$$

Now multiply $A$ (3×2) by $C^T$ (2×2):

$$AC^T = \begin{bmatrix} 3 & -1 \\ -2 & 4 \\ 0 & 5 \end{bmatrix} \times \begin{bmatrix} -3 & 1 \\ -2 & 2 \end{bmatrix}$$

Multiply:

First row:

$$(3)(-3) + (-1)(-2) = -9 + 2 = -7$$ $$(3)(1) + (-1)(2) = 3 – 2 = 1$$

Second row:

$$(-2)(-3) + (4)(-2) = 6 – 8 = -2$$ $$(-2)(1) + (4)(2) = -2 + 8 = 6$$

Third row:

$$(0)(-3) + (5)(-2) = 0 – 10 = -10$$ $$(0)(1) + (5)(2) = 0 + 10 = 10$$

So:

$$AC^T = \begin{bmatrix} -7 & 1 \\ -2 & 6 \\ -10 & 10 \end{bmatrix}$$

✅ Answer:

$$AC^T = \begin{bmatrix} -7 & 1 \\ -2 & 6 \\ -10 & 10 \end{bmatrix}$$

(iv) $BA$

Matrix $B$ is $2 \times 2$, matrix $A$ is $3 \times 2$.

🚫 Operation not possible (inner dimensions do not match: $2 \times \underline{2}$ and $\underline{3} \times 2$).

(b) Determinant of matrix $D$

Matrix $D$:

$$D = \begin{bmatrix} 2 & -3 & 1 \\ 1 & 4 & 0 \\ -1 & 5 & -4 \end{bmatrix}$$

Use cofactor expansion (along the first row):

$$\det(D) = 2 \cdot \det \begin{bmatrix} 4 & 0 \\ 5 & -4 \end{bmatrix} – (-3) \cdot \det \begin{bmatrix} 1 & 0 \\ -1 & -4 \end{bmatrix} + 1 \cdot \det \begin{bmatrix} 1 & 4 \\ -1 & 5 \end{bmatrix}$$

Calculate minors:

$$\det \begin{bmatrix} 4 & 0 \\ 5 & -4 \end{bmatrix} = (4)(-4) – (0)(5) = -16$$ $$\det \begin{bmatrix} 1 & 0 \\ -1 & -4 \end{bmatrix} = (1)(-4) – (0)(-1) = -4$$ $$\det \begin{bmatrix} 1 & 4 \\ -1 & 5 \end{bmatrix} = (1)(5) – (4)(-1) = 5 + 4 = 9$$

Now compute:

$$\det(D) = 2(-16) + 3(-4) + 1(9) = -32 -12 + 9 = -35$$

✅ Answer:

$$\det(D) = -35$$

Since $\det(D) \neq 0$, matrix $D$ is invertible.

Final Answers

• (a):
  • (i) $A – 2B$: Not possible
  • (ii) $4B – 3C = \begin{bmatrix} 5 & 18 \\ 9 & 2 \end{bmatrix}$
  • (iii) $AC^T = \begin{bmatrix} -7 & 1 \\ -2 & 6 \\ -10 & 10 \end{bmatrix}$
  • (iv) $BA$: Not possible
• (b):
  • $\det(D) = -35$, so matrix $D$ is invertible.