Question

If $A =\begin{bmatrix} 3& 5 \\\[0.3em] 2 & 0 \\\[0.3em] \end{bmatrix}$ and $\begin{bmatrix} 1& 17 \\\[0.3em] 0 & -10 \\\[0.3em]\end{bmatrix},$ then $|AB|$ is equal t

• A. 80
• B. 100
• C. -110
• D. 92

Answer 1

Answer: (B)

100

Answer 2

$A = \begin{bmatrix} 3& 5 \\\ 2 & 0 \end{bmatrix}$ and $B=\begin{bmatrix} 1& 17 \\\ 0 & -10 \end{bmatrix},$
$\therefore AB= \begin{bmatrix} 3& 5 \\\ 2 & 0 \\\ \end{bmatrix} \, \begin{bmatrix} 1& 17 \\\ 0 & -10 \\\ \end{bmatrix}, \begin{bmatrix} 3+0& 51-50 \\\ 2+0 & 34-0 \end{bmatrix} = \begin{bmatrix} 3& 1 \\\ 2 & 34 \end{bmatrix},$
$\Rightarrow |AB|= \begin{vmatrix} 1& 1 \\\ 2 & 34 \end{vmatrix},$
$=102-2$
$=100$