Chapter 12, Problem 21MCQ

Solution

To find: The inverse of the matrix, if it exists

  $\left[\begin{array}{ccc}\frac{6}{25}& -\frac{1}{5}& \frac{2}{25}\\ -\frac{1}{25}& \frac{1}{5}& \frac{8}{25}\\ \frac{3}{25}& \frac{2}{5}& \frac{1}{25}\end{array}\right]$

Given information:

Given,matrix is:

  $\left[\begin{array}{ccc}3& -1& 2\\ -1& 0& 2\\ 1& 3& -1\end{array}\right]$

Calculation:

The determinant of the matrix is given by:

  $\begin{array}{l}D=\left(3\right)\left[\left(0\right)\left(-1\right)-\left(2\right)\left(3\right)\right]-\left(-1\right)\left[\left(-1\right)\left(-1\right)-\left(2\right)\left(1\right)\right]+\left(2\right)\left[\left(-1\right)\left(3\right)-\left(0\right)\left(1\right)\right]\\ D=\left(3\right)\left(0-6\right)-\left(-1\right)\left(1-2\right)+\left(2\right)\left(-3-0\right)\\ D=\left(3\right)\left(-6\right)-\left(-1\right)\left(-1\right)+\left(2\right)\left(-3\right)\\ D=-18-1-6\\ D=-25\end{array}$

Since the determinant is not 0, the inverse exists.

The inverse of a 3X3 matrix $A$ is given by $A^{-1}=\frac{\text{adj }A}{\text{det }A}$ .

The cofactor matrix is given by:

  $\begin{array}{l}\text{cofactor }A=\left[\begin{array}{ccc}\left(0\right)\left(-1\right)-\left(2\right)\left(3\right)& -\left[\left(-1\right)\left(-1\right)-\left(2\right)\left(1\right)\right]& \left(-1\right)\left(3\right)-\left(0\right)\left(1\right)\\ -\left[\left(-1\right)\left(-1\right)-\left(2\right)\left(3\right)\right]& \left(3\right)\left(-1\right)-\left(2\right)\left(1\right)& -\left[\left(3\right)\left(3\right)-\left(-1\right)\left(1\right)\right]\\ \left(-1\right)\left(2\right)-\left(2\right)\left(0\right)& -\left[\left(3\right)\left(2\right)-\left(2\right)\left(-1\right)\right]& \left(3\right)\left(0\right)-\left(-1\right)\left(-1\right)\end{array}\right]\\ \text{cofactor }A=\left[\begin{array}{ccc}0-6& -\left(1-2\right)& -3-0\\ -\left(1-6\right)& -3-2& -\left(9+1\right)\\ -2-0& -\left(6+2\right)& 0-1\end{array}\right]\\ \text{cofactor }A=\left[\begin{array}{ccc}-6& 1& -3\\ 5& -5& -10\\ -2& -8& -1\end{array}\right]\end{array}$

The adjoint matrix is the transpose of cofactor matrix.

Hence the adjoint matrix is given by:

  $\text{adj }A=\left[\begin{array}{ccc}-6& 5& -2\\ 1& -5& -8\\ -3& -10& -1\end{array}\right]$

Hence the inverse matrix of the given matrix is given by:

  $\begin{array}{l}{A}^{-1}=\frac{\text{adj }A}{\text{det }A}\\ A^{-1}=\frac{1}{-25}\left[\begin{array}{ccc}-6& 5& -2\\ 1& -5& -8\\ -3& -10& -1\end{array}\right]\\ A^{-1}=\left[\begin{array}{ccc}\frac{-6}{-25}& \frac{5}{-25}& \frac{-2}{-25}\\ \frac{1}{-25}& \frac{-5}{-25}& \frac{-8}{-25}\\ \frac{-3}{-25}& \frac{-10}{-25}& \frac{-1}{-25}\end{array}\right]\\ A^{-1}=\left[\begin{array}{ccc}\frac{6}{25}& -\frac{1}{5}& \frac{2}{25}\\ -\frac{1}{25}& \frac{1}{5}& \frac{8}{25}\\ \frac{3}{25}& \frac{2}{5}& \frac{1}{25}\end{array}\right]\end{array}$

Conclusion:

The inverse of the matrix is $\left[\begin{array}{ccc}\frac{6}{25}& -\frac{1}{5}& \frac{2}{25}\\ -\frac{1}{25}& \frac{1}{5}& \frac{8}{25}\\ \frac{3}{25}& \frac{2}{5}& \frac{1}{25}\end{array}\right]$