Matrices — practice questions

First, check the orders. $\mathbf{A}$ is $3 \times 2$ and $\mathbf{B}$ is $2 \times 2$. The inner dimensions match (2 and 2), so the product is defined. The resulting matrix will be of order $3 \times 2$.

$\mathbf{AB} = \begin{pmatrix} 1 & 3 \\ -2 & 0 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 5 & 2 \\ -1 & 6 \end{pmatrix}$

To find the element in row 1, column 1 of the product: $(1)(5) + (3)(-1) = 5 - 3 = 2$

Row 1, column 2: $(1)(2) + (3)(6) = 2 + 18 = 20$

Row 2, column 1: $(-2)(5) + (0)(-1) = -10 + 0 = -10$

Row 2, column 2: $(-2)(2) + (0)(6) = -4 + 0 = -4$

Row 3, column 1: $(4)(5) + (1)(-1) = 20 - 1 = 19$

Row 3, column 2: $(4)(2) + (1)(6) = 8 + 6 = 14$

So, $\mathbf{AB} = \begin{pmatrix} 2 & 20 \\ -10 & -4 \\ 19 & 14 \end{pmatrix}$.

Step 1: Find the determinant of $\mathbf{M}$. $\det(\mathbf{M}) = 2 \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} + (-1) \begin{vmatrix} 1 & 3 \\ -1 & 2 \end{vmatrix}$ $= 2(3-4) - 1(1 - (-2)) - 1(2 - (-3))$ $= 2(-1) - 1(3) - 1(5) = -2 - 3 - 5 = -10$. Since $\det(\mathbf{M}) \neq 0$, the inverse exists.

Step 2: Find the matrix of cofactors. First, the matrix of minors: $\begin{pmatrix} \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} & \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} & \begin{vmatrix} 1 & 3 \\ -1 & 2 \end{vmatrix} \\ \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} & \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} & \begin{vmatrix} 2 & 1 \\ -1 & 2 \end{vmatrix} \\ \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} & \begin{vmatrix} 2 & -1 \\ 1 & 2 \end{vmatrix} & \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} \end{pmatrix} = \begin{pmatrix} -1 & 3 & 5 \\ 3 & 1 & 5 \\ 5 & 5 & 5 \end{pmatrix}$ Now apply the sign 'checkerboard' $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$ to get the matrix of cofactors: $\mathbf{C} = \begin{pmatrix} -1 & -3 & 5 \\ -3 & 1 & -5 \\ 5 & -5 & 5 \end{pmatrix}$

Step 3: Find the adjugate matrix by transposing $\mathbf{C}$. $\text{adj}(\mathbf{M}) = \mathbf{C}^T = \begin{pmatrix} -1 & -3 & 5 \\ -3 & 1 & -5 \\ 5 & -5 & 5 \end{pmatrix}$

Step 4: Calculate the inverse. $\mathbf{M}^{-1} = \frac{1}{\det(\mathbf{M})} \text{adj}(\mathbf{M}) = -\frac{1}{10} \begin{pmatrix} -1 & -3 & 5 \\ -3 & 1 & -5 \\ 5 & -5 & 5 \end{pmatrix} = \begin{pmatrix} 1/10 & 3/10 & -5/10 \\ 3/10 & -1/10 & 5/10 \\ -5/10 & 5/10 & -5/10 \end{pmatrix}$.