Question

Find the the rank of matrix 0 1 -3 -1
1 0 1 1
3 1 0 2
1 1 -2 0​

Answer 1

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

Given :

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

To Find :

$\text{Inverse of matrix}$

Formula Applied :

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

Solution :

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

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

|A|=-3(0-2)-1(0-0)+2(2-0)∣A∣=−3(0−2)−1(0−0)+2(2−0)

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

|A| = 10∣A∣=10

$\begin{gathered}A= \left[\begin{array}{ccc}-3&1&2\\1&0&1\\0&2&0\end{array}\right]\end{gathered}</p><p> </p><p></p><p>[tex]\begin{gathered}adjA = \left[\begin{array}{cccc}0&1&1&0\\2&0&0&2\\1&2&-3&1\\0&1&1&0 \end{array}\right]^T\end{gathered}$

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

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

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

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

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

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

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

Answer 2

Step-by-step explanation:

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

Given :

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

To Find :

$\text{Inverse of matrix}$

Formula Applied :

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

Solution :

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

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

|A|=-3(0-2)-1(0-0)+2(2-0)∣A∣=−3(0−2)−1(0−0)+2(2−0)

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

|A| = 10∣A∣=10

$\begin{gathered}A= \left[\begin{array}{ccc}-3&1&2\\1&0&1\\0&2&0\end{array}\right]\end{gathered}</p><p></p><p>[tex]\begin{gathered}adjA = \left[\begin{array}{cccc}0&1&1&0\\2&0&0&2\\1&2&-3&1\\0&1&1&0 \end{array}\right]^T\end{gathered}$

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

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

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

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

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

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

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