Q. 36 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $31-40$, each matrix is nonsingular. Find the inverse of each matrix.

$\left[\begin{array}{ll}b& 3\\ b& 2\end{array}\right] b\ne 0$

Step-by-Step Solution

Answer

The inverse of the matrix $\left[\begin{array}{ll}b& 3\\ b& 2\end{array}\right] b\ne 0$ is, $\left[\begin{array}{cc}\frac{-2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]$.

Step 1 . Given information

$\left[\begin{array}{ll}b& 3\\ b& 2\end{array}\right] b\ne 0$

We have to find the inverse of the given matrix.

Step 2 . First find the augmented matrix A | I 3 .

$\left[A|I_3\right]\to \left[\begin{array}{llll}b& 3& 1& 0\\ b& 2& 0& 1\end{array}\right]$

Now transform it into $\left[I_3|A^{-1}\right]$ using the row transformations.

$$\begin{gathered} \left[\begin{array}{llll}b& 3& 1& 0\\ b& 2& 0& 1\end{array}\right]\to \left\{R_2\to R_2-R_1\right\}\left[\begin{array}{cccc}b& 3& 1& 0\\ 0& -1& -1& 1\end{array}\right] \\ \to \left\{R_1\to \frac{R_1}{b}\right\} \left[\begin{array}{cccc}1& \frac{3}{b}& \frac{1}{b}& 0\\ 0& -1& -1& 1\end{array}\right] \\ \to \left\{R_2\to -R_2\right\}\left[\begin{array}{cccc}1& \frac{3}{b}& \frac{1}{b}& 0\\ 0& 1& 1& -1\end{array}\right] \\ \to \left\{R_1\to R_1-\frac{3}{b}{R}_2\right\}\left[\begin{array}{cccc}1& 0& -\frac{2}{b}& \frac{3}{b}\\ 0& 1& 1& -1\end{array}\right] \end{gathered}$$

Step 3 . The augmented matrix A | I n is now in the reduced row echelon form and the identity matrix src="data:image/svg+xml;base64,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" role="math" localid="1647448841999" I 2 is on the left side of the vertical bar and A - 1 is on the right side of the vertical bar.

$A^{-1}=\left[\begin{array}{cc}-\frac{2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]$.