Question

If A = $\left[\begin{array}{ccc}1&-2&3\\0&-1&4\\-2&2&1\end{array}\right]$, then find {A')⁻¹

Answer 1

Answer:

$[A^{'}]^{-1}=\left[\begin{array}{ccc}-9&-8&-2\\8&7&2\\-5&-4&-1\end{array}\right]$

Step-by-step explanation:

As we know that inverse of a matrix is calculated by Adjoint of matrix.

As

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

and adjoint of matrix A is calculated as

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

so here

$A=\left[\begin{array}{ccc}1&-2&3\\0&-1&4\\-2&2&1\end{array}\right]\\\\\\A^{'}=\left[\begin{array}{ccc}1&0&-2\\-2&-1&2\\3&4&1\end{array}\right]\\\\\\adj.A=\left[\begin{array}{ccc}-9&8&-5\\-8&7&-4\\-2&2&-1\end{array}\right] ^{'}\\\\\\adj.A=\left[\begin{array}{ccc}-9&-8&-2\\8&7&2\\-5&-4&-1\end{array}\right]$

Now

$|A^{'}|=\left |\begin{array}{ccc}1&0&2\\-2&-1&2\\3&4&1\end{array}\right |\\\\=1(-1)+(-2)(-2)(4)-3(-1)(-2)-4(2)(1)=1 \\\\\\(A^{'})^{-1} =\frac{1}{1}\left[\begin{array}{ccc}-9&-8&-2\\8&7&2\\-5&-4&-1\end{array}\right]$

$[A^{'}]^{-1}=\left[\begin{array}{ccc}-9&-8&-2\\8&7&2\\-5&-4&-1\end{array}\right]$