# Evaluate ${\begin{bmatrix} 2 & -5 \\ 0 & -3 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 1 & -1 \\ 3 & 2 \\ \end{bmatrix}}$

$2$, $-5$, $0$ and $-3$ are the entries in the first square matrix of order $2$. Similarly, $1$, $-1$, $3$ and $2$ are the elements in the second $2 \times 2$ matrix. The two matrices of order $2$ should be multiplied to find the product of them in this matrix multiplication question. Now, let us learn how to multiply the $2 \times 2$ matrix with elements $2$, $-5$, $0$ and $-3$ by another matrix of order $2$ with entries $1$, $-1$, $3$ and $2$.

### Find sum after multiplying first row with first column

$\left[\begin{array}{rr} \hline 2 & -5 \\ \hline 0 & -3 \\ \end{array}\right]$ $\times$ $\left[\begin{array}{|r|r} 1 & -1 \\ 3 & 2 \end{array}\right]$

$2$ and $–5$ are the entries in the first row of the first matrix. $1$ and $3$ are elements in the first column of the second matrix.

1. Multiply the entries $2$ and $–5$ in the first row of first matrix by the elements $1$ and $3$ in the first column of the second matrix respectively.
2. Find the sum of the products of $2$ and $1$, and $-5$ and $3$ by adding their products.

In this step, $2$ and $–5$ belong to first row and $1$ and $3$ belong to first column. Hence, the sum of products of them will be an entry in the first row and first column of the matrix.

$\implies$ $\left[\begin{array}{rr} \hline 2 & -5 \\ \hline 0 & -3 \\ \end{array}\right]$ $\times$ $\left[\begin{array}{|r|r} 1 & -1 \\ 3 & 2 \end{array}\right]$ $\,=\,$ ${\begin{bmatrix} 2 \times 1+(-5) \times 3 & \,\,\, \\ \,\,\, & \,\,\, \\ \end{bmatrix}}$

Now, simplify the arithmetic expression in the first row and first column of the $2 \times 2$ matrix.

$=\,\,$ ${\begin{bmatrix} 2-15 & \,\,\, \\ \,\,\, & \,\,\, \\ \end{bmatrix}}$

$=\,\,$ ${\begin{bmatrix} -13 & \,\,\, \\ \,\,\, & \,\,\, \\ \end{bmatrix}}$

### Find sum after multiplying first row with second column

$\left[\begin{array}{rr} \hline 2 & -5 \\ \hline 0 & -3 \\ \end{array}\right]$ $\times$ $\left[\begin{array}{r|r|} 1 & -1 \\ 3 & 2 \end{array}\right]$

$2$ and $–5$ are the elements in the first row of the first matrix. $-1$ and $2$ are the entries in the second column of the second matrix.

1. Multiply the entries $2$ and $–5$ in the first row of first matrix by the elements $-1$ and $2$ in the second column of the second matrix respectively.
2. Calculate the sum of the products of $2$ and $-1$, and $-5$ and $2$ by adding their products.

In this step, $2$ and $–5$ belong to first row and $-1$ and $2$ belong to second column. Therefore, the sum of products of them will be an entry in the first row and second column of the matrix.

$\implies$ $\left[\begin{array}{rr} \hline 2 & -5 \\ \hline 0 & -3 \\ \end{array}\right]$ $\times$ $\left[\begin{array}{r|r|} 1 & -1 \\ 3 & 2 \end{array}\right]$ $\,=\,$ ${\begin{bmatrix} -13 & 2 \times (-1)+(-5) \times 2 \\ \,\,\, & \,\,\, \\ \end{bmatrix}}$

Now, simplify the arithmetic expression in the first row and second column of the matrix of order $2$.

$=\,\,$ ${\begin{bmatrix} -13 & -2-10 \\ \,\,\, & \,\,\, \\ \end{bmatrix}}$

$=\,\,$ ${\begin{bmatrix} -13 & -12 \\ \,\,\, & \,\,\, \\ \end{bmatrix}}$

### Find sum after multiplying second row with first column

$\left[\begin{array}{rr} 2 & -5 \\ \hline 0 & -3 \\ \hline \end{array}\right]$ $\times$ $\left[\begin{array}{|r|r} 1 & -1 \\ 3 & 2 \end{array}\right]$

$0$ and $–3$ are the elements in the second row of the first matrix. $1$ and $3$ are the entries in the first column of the second matrix.

1. Multiply the elements $0$ and $–3$ in the second row of first matrix by the entries $1$ and $3$ in the first column of the second matrix respectively.
2. Evaluate the sum of the products of $0$ and $1$, and $-3$ and $3$ by adding their products.

In this step, $0$ and $–3$ belong to second row and $1$ and $3$ belong to first column. So, the sum of products of them will be an entry in the second row and first column of the matrix.

$\implies$ $\left[\begin{array}{rr} 2 & -5 \\ \hline 0 & -3 \\ \hline \end{array}\right]$ $\times$ $\left[\begin{array}{|r|r} 1 & -1 \\ 3 & 2 \end{array}\right]$ $\,=\,$ ${\begin{bmatrix} -13 & -12 \\ 0 \times 1+(-3) \times 3 & \,\,\, \\ \end{bmatrix}}$

Now, simplify the arithmetic expression in the second row and first column of the square matrix of order $2$.

$=\,\,$ ${\begin{bmatrix} -13 & -12 \\ 0-9 & \,\,\, \\ \end{bmatrix}}$

$=\,\,$ ${\begin{bmatrix} -13 & -12 \\ -9 & \,\,\, \\ \end{bmatrix}}$

### Find sum after multiplying second row with second column

$\left[\begin{array}{rr} 2 & -5 \\ \hline 0 & -3 \\ \hline \end{array}\right]$ $\times$ $\left[\begin{array}{r|r|} 1 & -1 \\ 3 & 2 \end{array}\right]$

$0$ and $–3$ are the entries in the second row of the first matrix. $-1$ and $2$ are the elements in the second column of the second matrix.

1. Multiply the elements $0$ and $–3$ in the second row of first matrix by the elements $-1$ and $2$ in the second column of the second matrix respectively.
2. Find the sum of the products of $0$ and $-1$, and $-3$ and $2$ by adding their products.

In this step, $0$ and $–3$ belong to second row and $-1$ and $2$ belong to second column. Hence, the sum of products of them will be an entry in the second row and second column of the matrix.

$\implies$ $\left[\begin{array}{rr} 2 & -5 \\ \hline 0 & -3 \\ \hline \end{array}\right]$ $\times$ $\left[\begin{array}{r|r|} 1 & -1 \\ 3 & 2 \end{array}\right]$ $\,=\,$ ${\begin{bmatrix} -13 & -12 \\ -9 & 0 \times (-1)+(-3) \times 2 \\ \end{bmatrix}}$

Now, simplify the arithmetic expression in the second row and second column of the $2$ by $2$ square matrix.

$=\,\,$ ${\begin{bmatrix} -13 & -12 \\ -9 & 0-6 \\ \end{bmatrix}}$

$=\,\,$ ${\begin{bmatrix} -13 & -12 \\ -9 & -6 \\ \end{bmatrix}}$

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