College Algebra, Chapter 7, 7.3, Section 7.3, Problem 8

Determine the inverse of the matrix and verify that $B^{-1} B = BB^{-1} = I_3$

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

We first add the identity matrix to the right of our matrix

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

Using Gauss-Jordan Elimination

$\displaystyle R_3 + 2R_1 \to R_3$

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


$\displaystyle \frac{1}{2} R_2$

$\left[ \begin{array}{ccc|ccc}
1 & 3 & 2 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & \displaystyle \frac{1}{2} & 0 \\
0 & 5 & 4 & 2 & 0 & 1
\end{array} \right]$


$\displaystyle R_3 - 5 R_2 \to R_3$

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

$\displaystyle - R_3$

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

$\displaystyle R_2 - R_3 \to R_2$

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

$\displaystyle R_1 - 2 R_3 \to R_1$

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

$\displaystyle R_1 - 3 R_2 \to R_1$

$\left[ \begin{array}{ccc|ccc}
1 & 0 & 0 & -1 & 1 & -1 \\
0 & 1 & 0 & 2 & -2 & 1 \\
0 & 0 & 1 & -2 & \displaystyle \frac{5}{2} & -1
\end{array} \right]$

The right half is now $B^{-1}$

$\displaystyle B^{-1} = \left[ \begin{array}{ccc}
-1 & 1 & -1 \\
2 & -2 & 1 \\
-2 & \displaystyle \frac{5}{2} & -1
\end{array} \right]$

We calculate $BB^{-1}$ and $B^{-1} B$ and verify that both products give the identity matrix $I_3$


$
\begin{equation}
\begin{aligned}

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

\left[ \begin{array}{ccc}
-1 & 1 & -1 \\
2 & -2 & 1 \\
-2 & \displaystyle \frac{5}{2} & -1
\end{array} \right]

=&

\left[ \begin{array}{ccc}
1 \cdot (-1) + 3 \cdot 2 + 2 \cdot (-2) & \displaystyle 1 \cdot 1 + 3 \cdot (-2) \cdot \frac{5}{2} & 1 \cdot (-1) + 3 \cdot 1 + 2 \cdot (-1) \\
0 \cdot (-1) + 2 \cdot 2 + 2 \cdot (-2) & \displaystyle 0 \cdot 1 + 2 \cdot (-2) + 2 \cdot \frac{5}{2} & 0 \cdot (-1) + 2 \cdot 1 + 2 \cdot (-1) \\
-2 \cdot (-1) + (-1) \cdot 2 + 0 \cdot (-2) & \displaystyle -2 \cdot 1 + (-1) \cdot (-2) + 0 \cdot \frac{5}{2} & -2 \cdot (-1) + (-1) \cdot 1 + 0 \cdot (-1)
\end{array} \right]

\\
\\
\\

=& \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right]

\\
\\
\\

B^{-1} B =\left[ \begin{array}{ccc}
-1 & 1 & -1 \\
2 & -2 & 1 \\
-2 & \displaystyle \frac{5}{2} & -1
\end{array} \right]

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

=& \left[ \begin{array}{ccc}
-1 \cdot 1 + 1 \cdot 0 + (-1) \cdot (-2) & -1 \cdot 3 + 1 \cdot 2 + (-1) \cdot (-1) & -1 \cdot 2 + 1 \cdot 2 + (-1) \cdot 0 \\
2 \cdot 1 + (-2) \cdot 0 + 1 \cdot (-2) & 2 \cdot 3 + (-2) \cdot 2 + 1 \cdot (-1) & 2 \cdot 2 + (-2) \cdot 2 + 1 \cdot 0 \\
\displaystyle -2 \cdot 1 + \frac{5}{2} \cdot 0 + (-1) \cdot (-2) & \displaystyle -2 \cdot 3 + \frac{5}{2} \cdot 2 + (-1) \cdot (-1) & \displaystyle -2 \cdot 2 + \frac{5}{2} \cdot 2 + (-1) \cdot 0
\end{array} \right]

\\
\\
\\

=& \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right]

\end{aligned}
\end{equation}
$

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