Question

Find the adjoint and the inverse of the matrix $\left[\begin{array}{ccc}1&0&2\\2&1&0\\3&2&1\end{array}\right]$

Answer 1

Answer:

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

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

Step-by-step explanation:

As we know that Adjoint of matrix is calculated as Minor × Co-factor of each element and taking transpose of it.

or

$adj.A=[A_{ji}]_{n\times n}$

$A=\left[\begin{array}{ccc}1&0&2\\2&1&0\\3&2&1\end{array}\right] \\\\adj.A=\left[\begin{array}{ccc}1&-2&1\\4&-5&-2\\-2&4&1\end{array}\right] ^{'}\\\\adj.A=\left[\begin{array}{ccc}1&4&-2\\-2&-5&4\\1&-2&1\end{array}\right]$

Now

$A^{-1} =\frac{adj.A}{|A|} \\\\|A|= \left|\begin{array}{ccc}1&0&2\\2&1&0\\3&2&1\end{array}\right|\\\\|A|=1(1-0)-0(2)+2(4-3)=3 \\\\\\A^{-1} =\frac{1}{3} \left[\begin{array}{ccc}1&4&-2\\-2&-5&4\\1&-2&1\end{array}\right]$

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