# Find the inverse of the matrix [[6,-2,-3],[-1,8,-7],[4,-4,6]]?

##### 2 Answers
Sep 2, 2017

The inverse is $= \left(\begin{matrix}\frac{5}{62} & \frac{3}{31} & \frac{19}{124} \\ - \frac{11}{124} & \frac{6}{31} & \frac{45}{248} \\ - \frac{7}{62} & \frac{2}{31} & \frac{23}{124}\end{matrix}\right)$

#### Explanation:

The matrix is $A = \left(\begin{matrix}6 & - 2 & - 3 \\ - 1 & 8 & - 7 \\ 4 & - 4 & 6\end{matrix}\right)$

We calculate the inverse of the matrix as follows

$\left(\begin{matrix}6 & - 2 & - 3 & : & 1 & 0 & 0 \\ - 1 & 8 & - 7 & : & 0 & 1 & 0 \\ 4 & - 4 & 6 & : & 0 & 0 & 1\end{matrix}\right)$

$R 3 \leftarrow R 3 + 4 R 2$, $\implies$, $\left(\begin{matrix}6 & - 2 & - 3 & : & 1 & 0 & 0 \\ - 1 & 8 & - 7 & : & 0 & 1 & 0 \\ 0 & 28 & - 22 & : & 0 & 4 & 1\end{matrix}\right)$

$R 2 \leftarrow R 1 + 6 R 2$, $\implies$, $\left(\begin{matrix}6 & - 2 & - 3 & : & 1 & 0 & 0 \\ 0 & 46 & - 45 & : & 1 & 6 & 0 \\ 0 & 28 & - 22 & : & 0 & 4 & 1\end{matrix}\right)$

$R 1 \leftarrow \frac{R 1}{6}$, $\implies$, $\left(\begin{matrix}1 & - \frac{1}{3} & - \frac{1}{2} & : & \frac{1}{6} & 0 & 0 \\ 0 & 46 & - 45 & : & 1 & 6 & 0 \\ 0 & 28 & - 22 & : & 0 & 4 & 1\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{46}$, $\implies$, $\left(\begin{matrix}1 & - \frac{1}{3} & - \frac{1}{2} & : & \frac{1}{6} & 0 & 0 \\ 0 & 1 & - \frac{45}{46} & : & \frac{1}{46} & \frac{3}{23} & 0 \\ 0 & 28 & - 22 & : & 0 & 4 & 1\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{28}$, $\implies$, $\left(\begin{matrix}1 & - \frac{1}{3} & - \frac{1}{2} & : & \frac{1}{6} & 0 & 0 \\ 0 & 1 & - \frac{45}{46} & : & \frac{1}{46} & \frac{3}{23} & 0 \\ 0 & 1 & - \frac{11}{14} & : & 0 & \frac{1}{7} & \frac{1}{28}\end{matrix}\right)$

$R 3 \leftarrow \left(R 3 - R 2\right)$, $\implies$, $\left(\begin{matrix}1 & - \frac{1}{3} & - \frac{1}{2} & : & \frac{1}{6} & 0 & 0 \\ 0 & 1 & - \frac{45}{46} & : & \frac{1}{46} & \frac{3}{23} & 0 \\ 0 & 0 & \frac{31}{161} & : & - \frac{1}{46} & \frac{2}{161} & \frac{1}{28}\end{matrix}\right)$

$R 3 \leftarrow \left(R 3\right) \cdot \left(\frac{161}{31}\right)$, $\implies$, $\left(\begin{matrix}1 & - \frac{1}{3} & - \frac{1}{2} & : & \frac{1}{6} & 0 & 0 \\ 0 & 1 & - \frac{45}{46} & : & \frac{1}{46} & \frac{3}{23} & 0 \\ 0 & 0 & 1 & : & - \frac{7}{62} & \frac{2}{31} & \frac{23}{124}\end{matrix}\right)$

$R 2 \leftarrow \left(R 2\right) + \left(\frac{45}{46}\right) R 3$, $\implies$, $\left(\begin{matrix}1 & - \frac{1}{3} & - \frac{1}{2} & : & \frac{1}{6} & 0 & 0 \\ 0 & 1 & 0 & : & - \frac{11}{124} & \frac{6}{31} & \frac{45}{248} \\ 0 & 0 & 1 & : & - \frac{7}{62} & \frac{2}{31} & \frac{23}{124}\end{matrix}\right)$

$R 1 \leftarrow \left(R 1\right) + \left(\frac{1}{3}\right) R 2$, $\implies$, $\left(\begin{matrix}1 & 0 & - \frac{1}{2} & : & \frac{17}{124} & \frac{2}{31} & \frac{15}{248} \\ 0 & 1 & 0 & : & - \frac{11}{124} & \frac{6}{31} & \frac{45}{248} \\ 0 & 0 & 1 & : & - \frac{7}{62} & \frac{2}{31} & \frac{23}{124}\end{matrix}\right)$

$R 1 \leftarrow \left(R 1\right) + \left(\frac{1}{2}\right) R 3$, $\implies$, $\left(\begin{matrix}1 & 0 & 0 & : & \frac{5}{62} & \frac{3}{31} & \frac{19}{124} \\ 0 & 1 & 0 & : & - \frac{11}{124} & \frac{6}{31} & \frac{45}{248} \\ 0 & 0 & 1 & : & - \frac{7}{62} & \frac{2}{31} & \frac{23}{124}\end{matrix}\right)$

Sep 2, 2017

${\left(\begin{matrix}6 & - 2 & - 3 \\ - 1 & 8 & - 7 \\ 4 & - 4 & 6\end{matrix}\right)}^{- 1} = \frac{1}{248} \left(\begin{matrix}20 & 24 & 38 \\ - 22 & 48 & 45 \\ - 28 & 16 & 46\end{matrix}\right)$

#### Explanation:

Given:

$\left(\begin{matrix}6 & - 2 & - 3 \\ - 1 & 8 & - 7 \\ 4 & - 4 & 6\end{matrix}\right)$

The nice thing about the augmented matrix approach is that it works for matrices of any size, but in the case of $3 \times 3$ matrices it is practical to calculate the adjunct matrix...

Consider the matrix formed by replacing each entry by the determinant of the $2 \times 2$ matrix formed by the entries of the other rows and columns, reckoned cyclically:

$\left(\begin{matrix}\left\mid \begin{matrix}8 & - 7 \\ - 4 & 6\end{matrix} \right\mid & \left\mid \begin{matrix}- 7 & - 1 \\ 6 & 4\end{matrix} \right\mid & \left\mid \begin{matrix}- 1 & 8 \\ 4 & - 4\end{matrix} \right\mid \\ \left\mid \begin{matrix}- 4 & 6 \\ - 2 & - 3\end{matrix} \right\mid & \left\mid \begin{matrix}6 & 4 \\ - 3 & 6\end{matrix} \right\mid & \left\mid \begin{matrix}4 & - 4 \\ 6 & - 2\end{matrix} \right\mid \\ \left\mid \begin{matrix}- 2 & - 3 \\ 8 & - 7\end{matrix} \right\mid & \left\mid \begin{matrix}- 3 & 6 \\ - 7 & - 1\end{matrix} \right\mid & \left\mid \begin{matrix}6 & - 2 \\ - 1 & 8\end{matrix} \right\mid\end{matrix}\right)$

$= \left(\begin{matrix}48 - 28 & - 28 + 6 & 4 - 32 \\ 12 + 12 & 36 + 12 & - 8 + 24 \\ 14 + 24 & 3 + 42 & 48 - 2\end{matrix}\right)$

$= \left(\begin{matrix}20 & - 22 & - 28 \\ 24 & 48 & 16 \\ 38 & 45 & 46\end{matrix}\right)$

Transpose this matrix to get:

$\left(\begin{matrix}20 & 24 & 38 \\ - 22 & 48 & 45 \\ - 28 & 16 & 46\end{matrix}\right)$

Multiplying by the original we find:

$\left(\begin{matrix}6 & - 2 & - 3 \\ - 1 & 8 & - 7 \\ 4 & - 4 & 6\end{matrix}\right) \left(\begin{matrix}20 & 24 & 38 \\ - 22 & 48 & 45 \\ - 28 & 16 & 46\end{matrix}\right) = \left(\begin{matrix}248 & 0 & 0 \\ 0 & 248 & 0 \\ 0 & 0 & 248\end{matrix}\right)$

So:

${\left(\begin{matrix}6 & - 2 & - 3 \\ - 1 & 8 & - 7 \\ 4 & - 4 & 6\end{matrix}\right)}^{- 1} = \frac{1}{248} \left(\begin{matrix}20 & 24 & 38 \\ - 22 & 48 & 45 \\ - 28 & 16 & 46\end{matrix}\right)$