Chapter 12, Problem 10MCQ

Solution

To find: To solve the given matrix equation.

  $X=\left[\begin{array}{cc}2& -2\\ -5& 4\end{array}\right]$

Given information:

Given,matrix equation is:

  $3X+\left[\begin{array}{cc}-2& 1\\ 7& -3\end{array}\right]=\left[\begin{array}{cc}4& -5\\ -8& 9\end{array}\right]$

Calculation:

Simplifying the given matrix equation and solving for the matrix X :

  $\begin{array}{l}3X+\left[\begin{array}{cc}-2& 1\\ 7& -3\end{array}\right]=\left[\begin{array}{cc}4& -5\\ -8& 9\end{array}\right]\\ 3X=\left[\begin{array}{cc}4& -5\\ -8& 9\end{array}\right]-\left[\begin{array}{cc}-2& 1\\ 7& -3\end{array}\right]\\ 3X=\left[\begin{array}{cc}4-\left(-2\right)& -5-1\\ -8-7& 9-\left(-3\right)\end{array}\right]\\ 3X=\left[\begin{array}{cc}4+2& -5-1\\ -8-7& 9+3\end{array}\right]\\ 3X=\left[\begin{array}{cc}6& -6\\ -15& 12\end{array}\right]\\ X=\frac{1}{3}\left[\begin{array}{cc}6& -6\\ -15& 12\end{array}\right]\\ X=\left[\begin{array}{cc}\frac{6}{3}& -\frac{6}{3}\\ -\frac{15}{3}& \frac{12}{3}\end{array}\right]\\ X=\left[\begin{array}{cc}2& -2\\ -5& 4\end{array}\right]\end{array}$

Conclusion:

  $X=\left[\begin{array}{cc}2& -2\\ -5& 4\end{array}\right]$