Chapter 4, Problem 15MCQ

To determine

Tofind:the product of the two matrices given below.

Answer to Problem 15MCQ

The side $\text{a}$ is equals to $\text{1}\text{.71}$ , $\text{c=}3.45$ and $\angle \text{B}={60}^0$ .

Explanation of Solution

Given:

  $\left[\begin{array}{ccc}4& 0& -2\end{array}\right]\text{ and }\left[\begin{array}{cc}2& -1\\ -3& 0\\ 0& 4\end{array}\right]$ .

Concept used:

Steps to solve the inverse of matrix by

Step1: calculating the matrix of minors.

Step2:turn that into the matrix of cofactors.

Step3: then ad jugate by transposing the cofactor.

Step4: last multiply by the $\frac{1}{\text{determinant}}$ .

Formula for inverse of matrix A is:

  ${\text{A}}^{\text{-1}}\text{=}\frac{\text{adjA}}{\left|\text{A}\right|}$

Calculation:

Here $\left[\begin{array}{ccc}4& 0& -2\end{array}\right]\text{ and }\left[\begin{array}{cc}2& -1\\ -3& 0\\ 0& 4\end{array}\right]$ .

The product will have dimension $1\times 2$ since the first matrix has dimension $1\times 3$ and second matrix has dimension of $3\times 2$ .

  $\left[\begin{array}{ccc}4& 0& -2\end{array}\right]\text{. }\left[\begin{array}{cc}2& -1\\ -3& 0\\ 0& 4\end{array}\right]=\left[\begin{array}{cc}8+0+0& -4+0-8\end{array}\right]=\left[\begin{array}{cc}8& -12\end{array}\right]$ .

Hence the product of the two matrices is $\left[\begin{array}{cc}8& -12\end{array}\right]$ .