Question

1. \\(\

$$\begin{bmatrix}-3&5\\\\-2&1\\end{bmatrix}$$

\cdot\

$$\begin{bmatrix}6&-2\\\\1&-5\\end{bmatrix}$$

\\)

Explanation:

Step1: Recall matrix multiplication rule

For two matrices \( A =

$$\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$

\) and \( B =

$$\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$$

\), the product \( AB \) is given by:
\[
AB=

$$\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{bmatrix}$$

\]
Here, \( A=

$$\begin{bmatrix}-3&5\\-2&1\end{bmatrix}$$

\) and \( B =

$$\begin{bmatrix}6&-2\\1&-5\end{bmatrix}$$

\)

Step2: Calculate the first element of the first row

\( a_{11}b_{11}+a_{12}b_{21}=(-3)\times6 + 5\times1=-18 + 5=-13 \)

Step3: Calculate the second element of the first row

\( a_{11}b_{12}+a_{12}b_{22}=(-3)\times(-2)+5\times(-5) = 6-25=-19 \)

Step4: Calculate the first element of the second row

\( a_{21}b_{11}+a_{22}b_{21}=(-2)\times6+1\times1=-12 + 1=-11 \)

Step5: Calculate the second element of the second row

\( a_{21}b_{12}+a_{22}b_{22}=(-2)\times(-2)+1\times(-5)=4 - 5=-1 \)

Answer:

\(

$$\begin{bmatrix}-13&-19\\-11&-1\end{bmatrix}$$

\)