Q52. - Algebra 2

Question

Find each product, if possible.

52.

$\left[\begin{array}{cc}\mathbf{2}& \mathbf{5}\\ & \end{array}\right]·\left[\begin{array}{cc}\mathbf{3}& \mathbf{1}\\ \mathbf{-}\mathbf{2}& \mathbf{6}\\ & \end{array}\right]$

Step-by-Step Solution

Answer

The product of the given matrices is $\left[\begin{array}{c}\mathbf{-}\mathbf{4}\\ \mathbf{32}\\ \end{array}\right]$.

Step 1 ­- Description of step.

The first matrix in the given product of matrices is:

$\left[\begin{array}{cc}2& 5\\ & \end{array}\right]$

In the matrix, there is one row and two columns.

Step 2 ­- Description of step.

The second matrix in the given product of matrices is:

$\left[\begin{array}{cc}3& 1\\ -2& 6\\ & \end{array}\right]$

In the matrix, there are two rows and two columns.

Step 3 ­-Description of step.

The product of the two matrices is possible when the number of columns in the first matrix is equal to the number of rows in second matrix.

Step 4 ­- Description of step.

As, the number of columns in the first matrix is 2 and the number of rows in the second matrix is 2, therefore the number of columns in the first matrix is equal to the number of rows in second matrix.

Therefore, the product of the given matrices is possible.

Step 5­- Description of step.

The elements $a_{ij}$ of $AB$ is the sum of the products of the corresponding elements in row $i$ of $A$ and column $j$ of $B$.

$\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\\ & \end{array}\right]·\left[\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\\ & \end{array}\right]=\left[\begin{array}{cc}{a}_1{x}_1+b_1{x}_2& a_1{y}_1+b_1{y}_2\\ a_2{x}_1+b_2{x}_2& a_2{y}_1+b_2{y}_2\\ & \end{array}\right]$

Step 6­- Description of step.

The product of the given matrices is:

$\begin{array}{c}\left[\begin{array}{cc}2& 5\end{array}\right]\cdot \left[\begin{array}{cc}3& 1\\ -2& 6\end{array}\right]=\left[\begin{array}{c}\left(2\right)\left(3\right)+\left(5\right)(-2)\\ \left(2\right)\left(1\right)+\left(5\right)\left(6\right)\end{array}\right]\\ =\left[\begin{array}{c}6-10\\ 2+30\end{array}\right]\\ =\left[\begin{array}{c}-4\\ 32\end{array}\right]\end{array}$

Therefore, the product of the given matrices is $\left[\begin{array}{c}-4\\ 32\\ \end{array}\right]$.