Question:

Find adjoint of each of the matrices \(\begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}\)

Updated On: Aug 29, 2023
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

Let A=\(\begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}\)
we have 

A11=\(\begin{vmatrix}3&5\\0&1\end{vmatrix}\)=3-0=3

A12=-\(\begin{vmatrix}2&5\\-2&1\end{vmatrix}\)=-12

A13=\(\begin{vmatrix}2&3\\-2&0\end{vmatrix}\)=0+6=6

A21=-\(\begin{vmatrix}-1&2\\0&1\end{vmatrix}\)=-(-1-0)=1

A22=\(\begin{vmatrix}1&2\\-2&1\end{vmatrix}\)=1+4=5

A23=-\(\begin{vmatrix}1&-1\\-2&0\end{vmatrix}\)=-(0-2)=2

A31=\(\begin{vmatrix}-1&2\\3&5\end{vmatrix}\)=-5-6=-11

A32=-\(\begin{vmatrix}1&2\\2&5\end{vmatrix}\)=-(5-4)=-1

A33=\(\begin{vmatrix}1&-1\\2&3\end{vmatrix}\)=3+2=5

Hence adj A=\(\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{23}\\A_{13}&A_{23}&A_{33}\end{bmatrix}\)=\(\begin{bmatrix}3&1&-11\\ -12&5&-1\\ 6&2&5\end{bmatrix}\)

Was this answer helpful?
0
0