Question

How do you multiply a 3x3 matrix?
Consider the example of a 3x3 matrix:
Let A = \left( {\begin{array}{*{20}{c}} 1&2&3 \\\ 4&6&7 \\\ 5&8&0 \end{array}} \right) and B = \left( {\begin{array}{*{20}{c}} 1&5&4 \\\ 2&6&0 \\\ 3&8&7 \end{array}} \right)

Answer 1

In order to multiply the two matrices, first check that the matrix can be multiplied or not, by checking the number of columns and rows of the two matrices. Then if possible to multiply, then move forward to multiply the digits inside the matrix.

Complete step by step answer:
We are given two matrices A and B, and we need to know about the steps to be followed for multiplying them.
The first basic and important rule of multiplying the two matrices is to check out the rows and columns. If the number of columns of the first matrix is equal to the number of rows to the second matrix, then only they can be multiplied, otherwise not.
Let’s check it out for the given matrices A and B. Since, the number of columns of the matrix A and the number of rows of the Matrix B is equal, that is $3$, so we can multiply them, and since they are both $3$. Therefore, after multiplication the resultant matrix will also be in 3x3 matrix.

The basic representation of a 3x3 matrix is {\left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right)_{3 \times 3}}, considering it to be AB that is equal to AB = {\left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right)_{3 \times 3}} or written as AB = {\left( {\begin{array}{*{20}{c}} 1&2&3 \\\ 4&6&7 \\\ 5&8&0 \end{array}} \right)_{3 \times 3}}{\left( {\begin{array}{*{20}{c}} 1&5&4 \\\ 2&6&0 \\\ 3&8&7 \end{array}} \right)_{3 \times 3}} = {\left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right)_{3 \times 3}}

The first row, 1st digit of the resultant matrix will be the sum of the products of the row of the first matrix and 1st column of the second matrix, numerically written as:
${a_{11}} = 1 \times 1 + 2 \times 2 + 3 \times 3 = 1 + 4 + 9 = 14$
Similarly, for the second value ${a_{12}}$, the first row will be multiplied with the values of the second column of the second matrix and then add them as:
${a_{12}} = 1 \times 5 + 2 \times 6 + 3 \times 8 = 5 + 12 + 24 = 41$
And, for ${a_{13}}$, the first row will be multiplied with the values of the third column of the second matrix and then add them as:
${a_{13}} = 1 \times 4 + 2 \times 0 + 3 \times 7 = 4 + 0 + 21 = 25$
Same steps to be followed for the other rows and we get:
${a_{21}} = 4 \times 1 + 6 \times 2 + 7 \times 3 = 4 + 12 + 21 = 37$
${a_{22}} = 4 \times 5 + 6 \times 6 + 7 \times 8 = 20 + 36 + 56 = 112$
${a_{23}} = 4 \times 4 + 6 \times 0 + 7 \times 7 = 16 + 0 + 49 = 65$
${a_{31}} = 5 \times 1 + 8 \times 2 + 0 \times 3 = 5 + 16 + 0 = 21$
${a_{32}} = 5 \times 5 + 8 \times 6 + 0 \times 8 = 25 + 48 + 0 = 73$
${a_{33}} = 5 \times 4 + 8 \times 0 + 0 \times 7 = 20 + 0 + 0 = 20$
Substituting the values in the resultant matrix, we get:

{{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right)_{3 \times 3}} = {\left( {\begin{array}{*{20}{c}} {14}&{41}&{25} \\\ {37}&{112}&{65} \\\ {21}&{73}&{20} \end{array}} \right)_{3 \times 3}}$$ Therefore, the product or multiplication of matrix A and B is: $AB = {\left( {\begin{array}{*{20}{c}} 1&2&3 \\\ 4&6&7 \\\ 5&8&0 \end{array}} \right)_{3 \times 3}}{\left( {\begin{array}{*{20}{c}} 1&5&4 \\\ 2&6&0 \\\ 3&8&7 \end{array}} \right)_{3 \times 3}} = {\left( {\begin{array}{*{20}{c}} {14}&{41}&{25} \\\ {37}&{112}&{65} \\\ {21}&{73}&{20} \end{array}} \right)_{3 \times 3}}$ **Note:** 1\. Always remember to check the first method that is checking the number of columns of the first matrix and number of rows of the second matrix, they need to be equal otherwise it would result in wrong values and a proper matrix would not be formed. 2\. To remember the step 1 order, we can remember it as the CR method, that is Column for first and Row for second.