Chapter 11, Problem 12CT

To determine

The inverse of a nonsingular matrix $B=\left[\begin{array}{ccc}1& -1& 1\\ 2& 5& -1\\ 2& 3& 0\end{array}\right]$.

Answer to Problem 12CT

Solution:

The inverse of the matrix $B=\left[\begin{array}{ccc}1& -1& 1\\ 2& 5& -1\\ 2& 3& 0\end{array}\right]$ is $\left[\begin{array}{ccc}3& 3& -4\\ -2& -2& 3\\ -4& -5& 7\end{array}\right]$.

Explanation of Solution

Given information:

The matrix, $B=\left[\begin{array}{ccc}1& -1& 1\\ 2& 5& -1\\ 2& 3& 0\end{array}\right]$.

Explanation:

Consider a matrix, $B=\left[\begin{array}{ccc}1& -1& 1\\ 2& 5& -1\\ 2& 3& 0\end{array}\right]$.

To find inverse of matrix $A$:

Step 1: Form the matrix $\left[A|I_n\right]$.

Step 2: Transform the matrix $\left[A|I_n\right]$ into reduced row echelon form.

Step 3: The reduced row echelon form of $\left[A|I_n\right]$ contains the identity matrix $I_n$ on the left of the vertical bar, the $n$ by $n$ matrix on the right vertical bar is an inverse of $A$.

Consider the matrix $\left[A|I_2\right]=\left[\begin{array}{ccc}1& -1& 1\\ 2& 5& -1\\ 2& 3& 0\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Use $R_2\to -2r_1+r_2,R_3\to -2r_1+r_3$,

$\left[A|I_2\right]=\left[\begin{array}{ccc}1& -1& 1\\ 0& 7& -3\\ 0& 5& -2\end{array}|\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ -2& 0& 1\end{array}\right]$

Use row operation $R_2\to \frac{1}{7}{r}_2$,

$\left[A|I_2\right]=\left[\begin{array}{ccc}1& -1& 1\\ 0& 1& -\frac{3}{7}\\ 0& 5& -2\end{array}|\begin{array}{ccc}1& 0& 0\\ -\frac{2}{7}& \frac{1}{7}& 0\\ -2& 0& 1\end{array}\right]$

Use $R_1\to r_2+r_1,R_3\to -5r_2+r_3$

$\left[A|I_2\right]=\left[\begin{array}{ccc}1& 0& \frac{4}{7}\\ 0& 1& -\frac{3}{7}\\ 0& 0& \frac{1}{7}\end{array}|\begin{array}{ccc}\frac{5}{7}& \frac{1}{7}& 0\\ -\frac{2}{7}& \frac{1}{7}& 0\\ -\frac{4}{7}& -\frac{5}{7}& 1\end{array}\right]$

Use $R_3\to 7r_2$

$\left[A|I_2\right]=\left[\begin{array}{ccc}1& 0& \frac{4}{7}\\ 0& 1& -\frac{3}{7}\\ 0& 0& 1\end{array}|\begin{array}{ccc}\frac{5}{7}& \frac{1}{7}& 0\\ -\frac{2}{7}& \frac{1}{7}& 0\\ -4& -5& 7\end{array}\right]$

Use $R_1\to -\frac{4}{7}{r}_3+r_1,r_2\to \frac{3}{7}{r}_3+r_2$,

$\left[A|I_2\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}3& 3& -4\\ -2& -2& 3\\ -4& -5& 7\end{array}\right]$

Thus, the inverse of the matrix $B$ is $\left[\begin{array}{ccc}3& 3& -4\\ -2& -2& 3\\ -4& -5& 7\end{array}\right]$.