Question

Verify A(adj A)=(adj A)A=\(\mid A \mid I\).

\(\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}\)

Answer 1

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

IAI=1(0-0)+1(9+2)+2(0-0)=11

IAII=\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}\)

Now,A₁₁=0,A₁₂=-(9+2)=-11, A₁₃=0
A₂₁=-(-3-0)=3, A₂₂=3-2=1,A₂₃=-(0+1)=-1
A₃₁=2-0=2, A₃₂=-(-2-6)=8, A₃₃=0+3=3
therefore adj A=[032 -1118 0-13]
Now,A(adjA)= \(\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}\begin{bmatrix}0&3&2\\-11&1&18\\0&-1&3\end{bmatrix}\)

=\(\begin{bmatrix}0+11+0&3-1-2&2-8+6\\0+0+0&9+0+2&6+0-6\\0+0+0&3+0-3&2+0+9\end{bmatrix}\)

=\(\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}\)

Also ,(adjA).A=\(\begin{bmatrix}0&3&2\\-11&1&18\\0&-1&3\end{bmatrix}\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}\)

=\(\begin{bmatrix}0+9+2&0+0+0&0-6+6\\-11+3+8&11+0+0&-22-2+24\\0-3+3&0+0+0&0+2+9\end{bmatrix}\)

=\(\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}\)

Hence A(adjA)=(adj A)A=IAII.