Given the matrices:

• $A = \begin{pmatrix} -3 & 0 \\ 1 & 5 \end{pmatrix}$
• $B = \begin{pmatrix} 4 & -1 \\ 3 & 1 \end{pmatrix}$

The matrix $C = A \times B$ is calculated by performing matrix multiplication of $A$ and $B$.

Step 1: Matrix Multiplication

To compute $C = AB$, we use the rule for matrix multiplication, where each element of the resulting matrix is obtained by summing the products of corresponding entries from rows of $A$ and columns of $B$.

$C_{11} = (-3 \times 4) + (0 \times 3) = -12$ $C_{12} = (-3 \times -1) + (0 \times 1) = 3$ $C_{21} = (1 \times 4) + (5 \times 3) = 4 + 15 = 19$ $C_{22} = (1 \times -1) + (5 \times 1) = -1 + 5 = 4$

Thus, matrix $C$ is:

$C = \begin{pmatrix} -12 & 3 \\ 19 & 4 \end{pmatrix}$

Step 2: Inverse of Matrix $C$

To find the inverse of $C$, we need to compute the determinant of $C$ and then apply the formula for the inverse of a $2 \times 2$ matrix.

The formula for the inverse of a $2 \times 2$ matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is:

$M^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

For matrix $C$:

$\text{det}(C) = (-12 \times 4) - (3 \times 19) = -48 - 57 = -105$

Since the determinant is non-zero, the inverse of $C$ exists. The inverse is:

$C^{-1} = \frac{1}{-105} \begin{pmatrix} 4 & -3 \\ -19 & -12 \end{pmatrix}$

Thus, the inverse of matrix $C$ is:

= \begin{pmatrix} -\frac{4}{105} & \frac{1}{35} \\ \frac{19}{105} & \frac{4}{35} \end{pmatrix}$$ Would you like more details, or have any questions? Here are 5 related questions: 1. How do you find the determinant of a $$3 \times 3$$ matrix? 2. What conditions must hold for a matrix to have an inverse? 3. How is matrix multiplication defined for $$3 \times 3$$ matrices? 4. What is the geometric interpretation of matrix inverses? 5. How are matrix operations used in solving systems of equations? **Tip**: The inverse of a matrix exists only if its determinant is non-zero.