Q.If $A = \begin{pmatrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{pmatrix}$ and $B^T = \begin{pmatrix} 1 & -2 & 5 \\ -2 & 4 & 1 \\ 3 & -6 & -1 \end{pmatrix}$, then find the rank of $AB$.

Transpose $B^T$ to get $B$, compute $AB$, find $\det(AB)=0$ but a $2\times2$ minor $e 0$; hence the rank is $2$.

Step 1 — recover $B$. Since $B^T = \begin{pmatrix} 1 & -2 & 5 \\ -2 & 4 & 1 \\ 3 & -6 & -1 \end{pmatrix}$,

$B = (B^T)^T = \begin{pmatrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{pmatrix}.$

Step 2 — compute $AB$ with $A = \begin{pmatrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{pmatrix}$:

$AB = \begin{pmatrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{pmatrix}.$

(For example, row 1: $1\cdot1+1\cdot(-2)+(-1)\cdot5 = -6$; $1\cdot(-2)+1\cdot4+(-1)\cdot1 = 1$; $1\cdot3+1\cdot(-6)+(-1)\cdot(-1) = -2$.)

Step 3 — determinant. …