Question

Find the inverse of a matrix
(-3 1 2
1 0 1
0 2 0)​

Answer 1

Answer:

(-3 1 0

1 0 2

2 1 0) is this question answer

Answer 2

Inverse of Given matrix is $\left[\begin{array}{ccc} \frac{-1}{5}&\frac{2}{5}&\frac{1}{10}\\0&0&\frac{1}{2}\\\frac{1}{5}&\frac{3}{5}&\frac{-1}{10}\end{array}\right]$

Given :

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

To Find :

$\text{Inverse of matrix}$

Formula Applied :

$A^{-1}=\frac{1}{|A|}adjA$

Solution :

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

$|A| = -3\left[\begin{array}{cc}0&1\\2&0\end{array}\right] -1\left[\begin{array}{cc}1&1\\0&0\end{array}\right] +2 \left[\begin{array}{cc}1&0\\0&2\end{array}\right]$

$|A|=-3(0-2)-1(0-0)+2(2-0)$

$|A|=-3(-2)-1(0) +2(2)\implies 6-0+4$

$|A| = 10$

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

$adjA = \left[\begin{array}{cccc}0&1&1&0\\2&0&0&2\\1&2&-3&1\\0&1&1&0 \end{array}\right]^T$

$adjA=\left[\begin{array}{ccc}(0-2)&(0-0)&(2-0)\\(4-0)&(0-0)&(0+6)\\(1-0)&(2+3)&(0-1)\end{array}\right]^T$

$adjA = \left[\begin{array}{ccc}-2&0&2\\4&0&6\\1&5&1\end{array}\right] ^T$

$adj A = \left[\begin{array}{ccc}-2&4&1\\0&0&5\\2&6& -1\end{array}\right]$

$A^{-1}=\frac{1}{|A|}adjA$

$A^{-1}= \frac{1}{10}\left[\begin{array}{ccc}-2&4&1\\0&0&5\\2&6& -1\end{array}\right]$

$A^{-1} = \left[\begin{array}{ccc}\frac{-2}{10}&\frac{4}{10}&\frac{1}{10}\\\frac{0}{10}&\frac{0}{10}&\frac{5}{10}\\\frac{2}{10}&\frac{6}{10}&\frac{-1}{10}\end{array}\right]$

$A^{-1}= \left[\begin{array}{ccc} \frac{-1}{5}&\frac{2}{5}&\frac{1}{10}\\0&0&\frac{1}{2}\\\frac{1}{5}&\frac{3}{5}&\frac{-1}{10}\end{array}\right]$