First Principles AI

Solution

A=\begin{bmatrix} 1&2&-3\\ 0&-5&0\\ -2&-2&7 \end{bmatrix}

Cofactors are defined by:

A_{ij}=(-1)^{i+j}M_{ij}

where $M_{ij}$ is the minor.

Compute the minors of the 3rd row:

\begin{aligned} M_{31}&=\left|\begin{matrix}2&-3\\-5&0\end{matrix}\right|=2\cdot 0-(-3)(-5)=-15\\ A_{31}&=(-1)^{3+1}M_{31}=(-1)^4(-15)=-15 \end{aligned}

\begin{aligned} M_{32}&=\left|\begin{matrix}1&-3\\0&0\end{matrix}\right|=0\\ A_{32}&=(-1)^{3+2}M_{32}=(-1)^5\cdot 0=0 \end{aligned}

\begin{aligned} M_{33}&=\left|\begin{matrix}1&2\\0&-5\end{matrix}\right|=1(-5)-2\cdot 0=-5\\ A_{33}&=(-1)^{3+3}M_{33}=(-1)^6(-5)=-5 \end{aligned}

Now expand $|A|$ along the 3rd row:

\begin{aligned} |A|&=a_{31}A_{31}+a_{32}A_{32}+a_{33}A_{33}\\ &=(-2)(-15)+(-2)(0)+7(-5)\\ &=30-35=-5 \end{aligned}

\boxed{A_{31}=-15,\;A_{32}=0,\;A_{33}=-5,\;|A|=-5}

B=\begin{bmatrix} -5&-2&5\\ -3&-1&4\\ -2&-1&2 \end{bmatrix}

\begin{aligned} M_{31}&=\left|\begin{matrix}-2&5\\-1&4\end{matrix}\right|=(-2)(4)-5(-1)=-8+5=-3\\ B_{31}&=(-1)^{3+1}M_{31}=(-1)^4(-3)=-3 \end{aligned}

\begin{aligned} M_{32}&=\left|\begin{matrix}-5&5\\-3&4\end{matrix}\right|=(-5)(4)-5(-3)=-20+15=-5\\ B_{32}&=(-1)^{3+2}M_{32}=(-1)^5(-5)=5 \end{aligned}

\begin{aligned} M_{33}&=\left|\begin{matrix}-5&-2\\-3&-1\end{matrix}\right|=(-5)(-1)-(-2)(-3)=5-6=-1\\ B_{33}&=(-1)^{3+3}M_{33}=(-1)^6(-1)=-1 \end{aligned}

Expand along the 3rd row:

\begin{aligned} |B|&=b_{31}B_{31}+b_{32}B_{32}+b_{33}B_{33}\\ &=(-2)(-3)+(-1)(5)+2(-1)\\ &=6-5-2=-1 \end{aligned}

\boxed{B_{31}=-3,\;B_{32}=5,\;B_{33}=-1,\;|B|=-1}