Question

1. \\(\

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

\

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

\\)

Explanation:

To multiply two matrices, we use the rule for matrix multiplication: if we have a matrix \( A \) of size \( m \times n \) and a matrix \( B \) of size \( n \times p \), the resulting matrix \( C = A \times B \) will be of size \( m \times p \), and each element \( c_{ij} \) is calculated as the dot product of the \( i \)-th row of \( A \) and the \( j \)-th column of \( B \), i.e., \( c_{ij}=\sum_{k = 1}^{n}a_{ik}b_{kj} \)

Let \( A=

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

\) (a \( 3\times2 \) matrix) and \( B=

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

\) (a \( 2\times2 \) matrix). The product \( A\times B \) will be a \( 3\times2 \) matrix.

Step 1: Calculate the first row of the product matrix

The first row of \( A \) is \( [0,5] \)

• For the first element of the first row (\( i = 1,j=1 \)):

\( c_{11}=0\times(-4)+5\times(-2)=0 - 10=- 10 \)

• For the second element of the first row (\( i = 1,j = 2 \)):

\( c_{12}=0\times4+5\times(-4)=0-20 = - 20 \)

Step 2: Calculate the second row of the product matrix

The second row of \( A \) is \( [-3,1] \)

• For the first element of the second row (\( i=2,j = 1 \)):

\( c_{21}=(-3)\times(-4)+1\times(-2)=12 - 2=10 \)

• For the second element of the second row (\( i = 2,j=2 \)):

\( c_{22}=(-3)\times4+1\times(-4)=-12-4=-16 \)

Step 3: Calculate the third row of the product matrix

The third row of \( A \) is \( [-5,1] \)

• For the first element of the third row (\( i = 3,j=1 \)):

\( c_{31}=(-5)\times(-4)+1\times(-2)=20 - 2 = 18 \)

• For the second element of the third row (\( i=3,j = 2 \)):

\( c_{32}=(-5)\times4+1\times(-4)=-20-4=-24 \)

Answer:

The product of the two matrices is \(

$$\begin{bmatrix}-10&-20\\10&-16\\18&-24\end{bmatrix}$$

\)