Question

1. \\(\

$$\begin{bmatrix} 5 & 3 & 5 \\\\ 1 & 5 & 0 \\end{bmatrix}$$

\cdot \

$$\begin{bmatrix} -4 & 2 \\\\ -3 & 4 \\\\ 3 & -5 \\end{bmatrix}$$

\\)

Explanation:

Step1: Recall matrix multiplication rule

For two matrices \( A = [a_{ij}] \) (of size \( m \times n \)) and \( B = [b_{ij}] \) (of size \( n \times p \)), the product \( AB = [c_{ij}] \) where \( c_{ij}=\sum_{k = 1}^{n}a_{ik}b_{kj} \).

The first matrix \( A=

$$\begin{bmatrix}5&3&5\\1&5&0\end{bmatrix}$$

\) (size \( 2\times3 \)) and the second matrix \( B=

$$\begin{bmatrix}-4&2\\-3&4\\3&-5\end{bmatrix}$$

\) (size \( 3\times2 \)). So the product will be a \( 2\times2 \) matrix.

Step2: Calculate the first element of the first row (\( c_{11} \))

\( c_{11}=5\times(-4)+3\times(-3)+5\times3 \)
\( = - 20-9 + 15=-14 \)

Step3: Calculate the second element of the first row (\( c_{12} \))

\( c_{12}=5\times2+3\times4+5\times(-5) \)
\( = 10 + 12-25=-3 \)

Step4: Calculate the first element of the second row (\( c_{21} \))

\( c_{21}=1\times(-4)+5\times(-3)+0\times3 \)
\( = - 4-15+0=-19 \)

Step5: Calculate the second element of the second row (\( c_{22} \))

\( c_{22}=1\times2+5\times4+0\times(-5) \)
\( = 2 + 20+0=22 \)

Answer:

\(

$$\begin{bmatrix}-14&-3\\-19&22\end{bmatrix}$$

\)