Math Doubts

Evaluate $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & 7\\ 6 & 5 & 4\\ 3 & 2 & 1\\ \end{bmatrix}$

Two square matrices of the order $3$ are participated in the multiplication in the given matrix problem.

  1. In the first matrix, $1$, $2$ and $3$ are entries in the first row, $4$, $5$ and $6$ are elements in the second row, and $7$, $8$ and $9$ are the entries in the third row.
  2. In the second matrix, $9$, $6$ and $3$ are elements in the first column, $8$, $5$ and $2$ are entries in the second column, and $7$, $4$ and $1$ are the elements in the third column.

The multiplication of the two $3 \times 3$ matrices is written in the following mathematical form.

$\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & 7\\ 6 & 5 & 4\\ 3 & 2 & 1\\ \end{bmatrix}$

Now, let’s learn how to multiply the matrices of the order $3 \times 3$ in mathematics.

Multiply the entries in columns by the elements of first row

The multiplication process of the $3 \times 3$ matrices can be started by multiplying the entries in the first column of the second matrix $9$, $6$ and $3$ with the elements in the first row of the first matrix $1$, $2$ and $3$ respectively. Now, add the products of them for evaluating the entry in the first row first column for the multiplication of the two square matrices of the order $3$.

$\implies$ $\begin{bmatrix} \color{red} 1 & \color{red} 2 & \color{red} 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} \color{magenta} 9 & 8 & 7\\ \color{magenta} 6 & 5 & 4\\ \color{magenta} 3 & 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 1 \times 9 + 2 \times 6 + 3 \times 3 & \, & \,\\ \, & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} \color{red} 1 & \color{red} 2 & \color{red} 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} \color{magenta} 9 & 8 & 7\\ \color{magenta} 6 & 5 & 4\\ \color{magenta} 3 & 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 9+12+9 & \, & \,\\ \, & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} \color{red} 1 & \color{red} 2 & \color{red} 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} \color{magenta} 9 & 8 & 7\\ \color{magenta} 6 & 5 & 4\\ \color{magenta} 3 & 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & \, & \,\\ \, & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

Now, multiply the elements in the second column of the second matrix $8$, $5$ and $2$ by the elements in the first row of the first matrix $1$, $2$ and $3$ respectively. The element in the first row second column can be evaluated for the multiplication of them by adding their products.

$\implies$ $\begin{bmatrix} \color{red} 1 & \color{red} 2 & \color{red} 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & \color{magenta} 8 & 7\\ 6 & \color{magenta} 5 & 4\\ 3 & \color{magenta} 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 1 \times 8+2 \times 5+3 \times 2 & \,\\ \, & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} \color{red} 1 & \color{red} 2 & \color{red} 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & \color{magenta} 8 & 7\\ 6 & \color{magenta} 5 & 4\\ 3 & \color{magenta} 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 8+10+6 & \,\\ \, & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} \color{red} 1 & \color{red} 2 & \color{red} 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & \color{magenta} 8 & 7\\ 6 & \color{magenta} 5 & 4\\ 3 & \color{magenta} 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & \,\\ \, & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

Finally, multiply the entries in the third column of the second matrix $7$, $4$ and $1$ by the elements in the first row of the first matrix $1$, $2$ and $3$ respectively. After that, add the products of them and it is the entry in the first row third column for the multiplication of the square matrices of the order $3 \times 3$.

$\implies$ $\begin{bmatrix} \color{red} 1 & \color{red} 2 & \color{red} 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & \color{magenta} 7\\ 6 & 5 & \color{magenta} 4 \\ 3 & 2 & \color{magenta} 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 1 \times 7+2 \times 4+3 \times 1 \\ \, & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} \color{red} 1 & \color{red} 2 & \color{red} 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & \color{magenta} 7\\ 6 & 5 & \color{magenta} 4 \\ 3 & 2 & \color{magenta} 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 7+8+3 \\ \, & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} \color{red} 1 & \color{red} 2 & \color{red} 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & \color{magenta} 7\\ 6 & 5 & \color{magenta} 4 \\ 3 & 2 & \color{magenta} 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ \, & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

Multiply the elements in columns by the entries of second row

In this step, multiply the elements in the second row of the first matrix $4$, $5$ and $6$ with the entries in the first column of the second matrix $9$, $6$ and $3$ respectively. The products of them can be added to calculate the element in the second row first column for the matrix that respects the multiplication of the two $3 \times 3$ matrices.

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ \color{red} 4 & \color{red} 5 & \color{red} 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} \color{magenta} 9 & 8 & 7\\ \color{magenta} 6 & 5 & 4 \\ \color{magenta} 3 & 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 4 \times 9+ 5 \times 6+ 6 \times 3 & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ \color{red} 4 & \color{red} 5 & \color{red} 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} \color{magenta} 9 & 8 & 7\\ \color{magenta} 6 & 5 & 4 \\ \color{magenta} 3 & 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 36+30+18 & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ \color{red} 4 & \color{red} 5 & \color{red} 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} \color{magenta} 9 & 8 & 7\\ \color{magenta} 6 & 5 & 4 \\ \color{magenta} 3 & 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & \, & \,\\ \, & \, & \, \\ \end{bmatrix}$

Now, multiply the second row’s entries $4$, $5$ and $6$ of the first matrix with the second column’s elements $8$, $5$ and $2$ of the second matrix respectively, and then find the sum of the products of them for finding the element in the second row second column for the matrix, which is the multiplication of the two matrices of the order $3$.

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ \color{red} 4 & \color{red} 5 & \color{red} 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & \color{magenta} 8 & 7\\ 6 & \color{magenta} 5 & 4 \\ 3 & \color{magenta} 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 4 \times 8+5 \times 5+6 \times 2 & \,\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ \color{red} 4 & \color{red} 5 & \color{red} 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & \color{magenta} 8 & 7\\ 6 & \color{magenta} 5 & 4 \\ 3 & \color{magenta} 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 32+25+12 & \,\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ \color{red} 4 & \color{red} 5 & \color{red} 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & \color{magenta} 8 & 7\\ 6 & \color{magenta} 5 & 4 \\ 3 & \color{magenta} 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & \,\\ \, & \, & \, \\ \end{bmatrix}$

Finally, multiply the second row’s elements $4$, $5$ and $6$ of the first matrix with the third column’s entries $7$, $4$ and $1$ of the second matrix respectively. The products of the elements is added to find the element in the second row third column of the matrix, which expresses the multiplication of the given two square matrices.

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ \color{red} 4 & \color{red} 5 & \color{red} 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & \color{magenta} 7\\ 6 & 5 & \color{magenta} 4 \\ 3 & 2 & \color{magenta} 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 4 \times 7+5 \times 4+6 \times 1\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ \color{red} 4 & \color{red} 5 & \color{red} 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & \color{magenta} 7\\ 6 & 5 & \color{magenta} 4 \\ 3 & 2 & \color{magenta} 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 28+20+6\\ \, & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ \color{red} 4 & \color{red} 5 & \color{red} 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & \color{magenta} 7\\ 6 & 5 & \color{magenta} 4 \\ 3 & 2 & \color{magenta} 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ \, & \, & \, \\ \end{bmatrix}$

Multiply the entries in columns by the elements of third row

In this step, the elements in the third row of the first matrix are multiplied with the entries in each column of the second matrix.

Multiply the entries $7$, $8$ and $9$ in the third row of the first matrix with the elements $9$, $6$ and $3$ in the first column of the second matrix for calculating the element in the third row first column respectively for the matrix that represents multiplication of the given $3 \times 3$ matrices.

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ \color{red} 7 & \color{red} 8 & \color{red}9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} \color{magenta} 9 & 8 & 7\\ \color{magenta} 6 & 5 & 4 \\ \color{magenta} 3 & 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ 7 \times 9+8 \times 6+9 \times 3 & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ \color{red} 7 & \color{red} 8 & \color{red}9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} \color{magenta} 9 & 8 & 7\\ \color{magenta} 6 & 5 & 4 \\ \color{magenta} 3 & 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ 63+48+27 & \, & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ \color{red} 7 & \color{red} 8 & \color{red}9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} \color{magenta} 9 & 8 & 7\\ \color{magenta} 6 & 5 & 4 \\ \color{magenta} 3 & 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ 138 & \, & \, \\ \end{bmatrix}$

Now, multiply the elements $8$, $5$ and $2$ in the second column of the second matrix by the elements $7$, $8$ and $9$ in the third row of the first matrix respectively for evaluating the entry in third row second column of the matrix that expresses the multiplication of the matrices.

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ \color{red} 7 & \color{red} 8 & \color{red}9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & \color{magenta} 8 & 7\\ 6 & \color{magenta} 5 & 4 \\ 3 & \color{magenta} 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ 138 & 7 \times 8+8 \times 5+9 \times 2 & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ \color{red} 7 & \color{red} 8 & \color{red}9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & \color{magenta} 8 & 7\\ 6 & \color{magenta} 5 & 4 \\ 3 & \color{magenta} 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ 138 & 56+40+18 & \, \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ \color{red} 7 & \color{red} 8 & \color{red}9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & \color{magenta} 8 & 7\\ 6 & \color{magenta} 5 & 4 \\ 3 & \color{magenta} 2 & 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ 138 & 114 & \, \\ \end{bmatrix}$

Lastly, multiply the entries $7$, $4$ and $1$ in the third column of the second matrix by the entries $7$, $8$ and $9$ in the third row of the first matrix respectively for finding the element in the third row third column for the matrix that expresses the multiplication of the given two matrices.

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ \color{red} 7 & \color{red} 8 & \color{red}9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & \color{magenta} 7\\ 6 & 5 & \color{magenta} 4 \\ 3 & 2 & \color{magenta} 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ 138 & 114 & 7 \times 7+8 \times 4+9 \times 1 \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ \color{red} 7 & \color{red} 8 & \color{red}9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & \color{magenta} 7\\ 6 & 5 & \color{magenta} 4 \\ 3 & 2 & \color{magenta} 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ 138 & 114 & 49+32+9 \\ \end{bmatrix}$

$\implies$ $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ \color{red} 7 & \color{red} 8 & \color{red}9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & \color{magenta} 7\\ 6 & 5 & \color{magenta} 4 \\ 3 & 2 & \color{magenta} 1\\ \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 30 & 24 & 18 \\ 84 & 69 & 54 \\ 138 & 114 & 90 \\ \end{bmatrix}$

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