Chapter 12.4, Problem 51PPSE

Solution

To find: To solve the matrix equation.

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

Given information:

Given,matrix equation is:

  $-2\left[\begin{array}{cc}-2& 0\\ 0& -1\end{array}\right]+\left[\begin{array}{cc}0& -3\\ 5& -4\end{array}\right]X+\left[\begin{array}{cc}0& -3\\ 5& -4\end{array}\right]=\left[\begin{array}{cc}19& -27\\ 10& -24\end{array}\right]$

Calculation:

Simplifying the equation:

  $\begin{array}{l}-2\left[\begin{array}{cc}-2& 0\\ 0& -1\end{array}\right]+\left[\begin{array}{cc}0& -3\\ 5& -4\end{array}\right]X=\left[\begin{array}{cc}19& -27\\ 10& -24\end{array}\right]-\left[\begin{array}{cc}0& -3\\ 5& -4\end{array}\right]\\ \left[\begin{array}{cc}4& 0\\ 0& 2\end{array}\right]+\left[\begin{array}{cc}0& -3\\ 5& -4\end{array}\right]X=\left[\begin{array}{cc}19& -24\\ 5& -20\end{array}\right]\\ \left[\begin{array}{cc}0& -3\\ 5& -4\end{array}\right]X=\left[\begin{array}{cc}19& -24\\ 5& -20\end{array}\right]-\left[\begin{array}{cc}4& 0\\ 0& 2\end{array}\right]\\ \left[\begin{array}{cc}0& -3\\ 5& -4\end{array}\right]X=\left[\begin{array}{cc}15& -24\\ 5& -22\end{array}\right]\end{array}$

Multiplying both sides by the inverse of the matrix on the left gives the required solution.

The determinant of the coefficent matrix is given by:

  $\begin{array}{l}D=\left(0\right)\left(-4\right)-\left(-3\right)\left(5\right)\\ D=0+15\\ D=15\end{array}$

Since the determinant is not 0, the inverse exists.

The inverse of a 2X2 matrix $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is $\frac{1}{D}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ .

Plugging the values, the inverse matrix is given by:

  $\frac{1}{15}\left[\begin{array}{cc}-4& 3\\ -5& 0\end{array}\right]=\left[\begin{array}{cc}-\frac{4}{15}& \frac{3}{15}\\ -\frac{5}{15}& 0\end{array}\right]$

Multiplying both sides of the equation by the inverse:

  $\left[\begin{array}{cc}-\frac{4}{15}& \frac{3}{15}\\ -\frac{5}{15}& 0\end{array}\right]\left[\begin{array}{cc}0& -3\\ 5& -4\end{array}\right]X=\left[\begin{array}{cc}-\frac{4}{15}& \frac{3}{15}\\ -\frac{5}{15}& 0\end{array}\right]\left[\begin{array}{cc}15& -24\\ 5& -22\end{array}\right]$

  $\left[\begin{array}{cc}\left(-\frac{4}{15}\right)\left(0\right)+\left(\frac{3}{15}\right)\left(5\right)& \left(-\frac{4}{15}\right)\left(-3\right)+\left(\frac{3}{15}\right)\left(-4\right)\\ \left(-\frac{5}{15}\right)\left(0\right)+\left(0\right)\left(5\right)& \left(-\frac{5}{15}\right)\left(-3\right)+\left(0\right)\left(-4\right)\end{array}\right]X=\left[\begin{array}{cc}\left(-\frac{4}{15}\right)\left(15\right)+\left(\frac{3}{15}\right)\left(5\right)& \left(-\frac{4}{15}\right)\left(-24\right)+\left(\frac{3}{15}\right)\left(-22\right)\\ \left(-\frac{5}{15}\right)\left(15\right)+\left(0\right)\left(5\right)& \left(-\frac{5}{15}\right)\left(-24\right)+\left(0\right)\left(-22\right)\end{array}\right]$

  $\begin{array}{l}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]X=\left[\begin{array}{cc}-3& 2\\ -5& 8\end{array}\right]\\ X=\left[\begin{array}{cc}-3& 2\\ -5& 8\end{array}\right]\end{array}$

Conclusion:

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