Given the two matrices, A=begin{bmatrix}1 & 2&3 1 & 1&20&1&2 end{bmatrix} text{ and } B=begin{bmatrix}1 & 1&1 2 & 1&23&1&2 end{bmatrix} (a) Find det A

Step 1
We have given the matrices
$A=\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& 2\\ 0& 1& 2\end{array}\right]\text{ and }B=\left[\begin{array}{ccc}1& 1& 1\\ 2& 1& 2\\ 3& 1& 2\end{array}\right]$
Step 2
Part(a)
Find det A:
$detA=det\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& 2\\ 0& 1& 2\end{array}\right]=1\cdot det\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right]-2\cdot det\left[\begin{array}{cc}1& 2\\ 0& 2\end{array}\right]+3\cdot det\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
$=1\cdot 0-2\cdot 2+3\cdot 1$
=-1 Find det B:
$detB=det\left[\begin{array}{ccc}1& 1& 1\\ 2& 1& 2\\ 3& 1& 2\end{array}\right]$ $=1\cdot det\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right]-1\cdot det\left[\begin{array}{cc}2& 2\\ 3& 2\end{array}\right]+1\cdot det\left[\begin{array}{cc}2& 1\\ 3& 1\end{array}\right]$
$=1\cdot 0-1\cdot (-2)+1\cdot (-1)$
=1
Step 3
Find det(AB) and det(BA)
According to determinant properties,
$det(AB)=detA\times detB$
$=-1\times 1$
=-1
$det(BA)=detB\times detA$
$=1\times -1$
=-1
Step 4
Find det(5A)
$det(5A)=5^3\times detA$
$=125\times -1$
=-125
Find $det{A}^T:$
$det{A}^T=detA$
=-1
Find $det(B^6):$
$det(B^6)=(detB{)}^6$
$=1^6$
=1
Step 5
Part (b)
Find adj A and adj B
$A=\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& 2\\ 0& 1& 2\end{array}\right]$
The cofactors matrix is
$C=\left[\begin{array}{ccc}+det\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right]& -det\left[\begin{array}{cc}1& 2\\ 0& 2\end{array}\right]& +det\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\\ -det\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]& +det\left[\begin{array}{cc}1& 3\\ 0& 2\end{array}\right]& -det\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\\ +det\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]& -det\left[\begin{array}{cc}1& 3\\ 1& 2\end{array}\right]& +det\left[\begin{array}{cc}1& 2\\ 1& 1\end{array}\right]\end{array}\right]$