Evaluate ${\begin{bmatrix} -2 & 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & 4 \\ 3 & -1 \\ \end{bmatrix}}$

In the given matrix problem, two square matrices of the order $2$ are involved in multiplication.

1. In one matrix, the entries in first row are $-2$ and $3$, and the elements in second row are $-1$ and $4$.
2. In the second matrix, the elements in the first column are $6$ and $3$, and the entries in the second column are $4$ and $-1$.

The multiplication of the $2 \times 2$ matrices is expressed in mathematics as follows.

${\begin{bmatrix} -2 & 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & 4 \\ 3 & -1 \\ \end{bmatrix}}$

Now, let’s start the process for multiplying the matrices of the order $2 \times 2$.

Multiply the entries in columns by the elements of first row

Multiply the first column’s entries of the second matrix $6$ and $3$ by the first row’s elements of the first matrix $-2$ and $3$ respectively. After that, add the products of them to evaluate the first row first column for the multiplication of the given two matrices.

$\implies$ ${\begin{bmatrix} \color{red} -2 & \color{red} 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} \color{blue} 6 & 4 \\ \color{blue} 3 & -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} (-2) \times 6+3 \times 3 & \, \\ \, & \, \\ \end{bmatrix}}$

$\implies$ ${\begin{bmatrix} \color{red} -2 & \color{red} 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} \color{blue} 6 & 4 \\ \color{blue} 3 & -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -12+9 & \, \\ \, & \, \\ \end{bmatrix}}$

$\implies$ ${\begin{bmatrix} \color{red} -2 & \color{red} 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} \color{blue} 6 & 4 \\ \color{blue} 3 & -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & \, \\ \, & \, \\ \end{bmatrix}}$

Similarly, multiply the elements in the second column of the second matrix $4$ and $-1$ by the entries in the first row of the first matrix $-2$ and $3$ respectively. Now, add the products of them to find the element in the first row first column for the multiplication of the given two by two square matrices.

$\implies$ ${\begin{bmatrix} \color{red} -2 & \color{red} 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & \color{blue} 4 \\ 3 & \color{blue} -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & (-2) \times 4+3 \times (-1) \\ \, & \, \\ \end{bmatrix}}$

$\implies$ ${\begin{bmatrix} \color{red} -2 & \color{red} 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & \color{blue} 4 \\ 3 & \color{blue} -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & -8-3 \\ \, & \, \\ \end{bmatrix}}$

$\implies$ ${\begin{bmatrix} \color{red} -2 & \color{red} 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & \color{blue} 4 \\ 3 & \color{blue} -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & -11 \\ \, & \, \\ \end{bmatrix}}$

Multiply the entries in columns by the elements of second row

Multiply the elements in the first column of the second matrix $6$ and $3$ by the entries in the second row of the first matrix $-1$ and $4$ respectively. Later, find the sum of the products of them to calculate the entry in the second row first column for the multiplication of the given second order square matrices.

$\implies$ ${\begin{bmatrix} -2 & 3 \\ \color{red} -1 & \color{red} 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} \color{blue} 6 & 4 \\ \color{blue} 3 & -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & -11 \\ (-1) \times 6+4 \times 3 & \, \\ \end{bmatrix}}$

$\implies$ ${\begin{bmatrix} -2 & 3 \\ \color{red} -1 & \color{red} 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} \color{blue} 6 & 4 \\ \color{blue} 3 & -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & -11 \\ -6+12 & \, \\ \end{bmatrix}}$

$\implies$ ${\begin{bmatrix} -2 & 3 \\ \color{red} -1 & \color{red} 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} \color{blue} 6 & 4 \\ \color{blue} 3 & -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & -11 \\ 6 & \, \\ \end{bmatrix}}$

In the same way, multiply the entries in the second column of the second matrix $4$ and $-1$ by the elements in the second row of the first matrix $-1$ and $4$ respectively. Finally, calculate the products of them and then add them for evaluating the element in the second row second column for the multiplication of the matrices of the order $2 \times 2$.

$\implies$ ${\begin{bmatrix} -2 & 3 \\ \color{red} -1 & \color{red} 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & \color{blue} 4 \\ 3 & \color{blue} -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & -11 \\ 6 & (-1) \times 4+4 \times (-1) \\ \end{bmatrix}}$

$\implies$ ${\begin{bmatrix} -2 & 3 \\ \color{red} -1 & \color{red} 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & \color{blue} 4 \\ 3 & \color{blue} -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & -11 \\ 6 & -4-4 \\ \end{bmatrix}}$

$\implies$ ${\begin{bmatrix} -2 & 3 \\ \color{red} -1 & \color{red} 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & \color{blue} 4 \\ 3 & \color{blue} -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & -11 \\ 6 & -8 \\ \end{bmatrix}}$

Thus, the matrix ${\begin{bmatrix} 6 & 4 \\ 3 & -1 \\ \end{bmatrix}}$ can be multiplied by the matrix ${\begin{bmatrix} -2 & 3 \\ -1 & 4 \\ \end{bmatrix}}$ in the mathematics.

$\,\,\,\therefore\,\,\,\,\,\,$ ${\begin{bmatrix} -2 & 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & 4 \\ 3 & -1 \\ \end{bmatrix}}$ $\,=\,$ ${\begin{bmatrix} -3 & -11 \\ 6 & -8 \\ \end{bmatrix}}$