Question

Let $A$ be a matrix of order $3\times 3$ such that $\left|\overrightarrow{A}\right|=1$. Let $B=2A^{-1}$ and $C=\frac{adj.A}{2}$. Then the value of $\left|{AB}^2{C}^3\right|$, is ( where $\left|A\right|$ represent det. $A$)

• A. $1$
• B. $\frac{1}{2}$
• C. $8$
• D. $64$

Answer 1

Answer: (B)

$\frac{1}{2}$

$adj\left(A\right)=|\overrightarrow{A}{|}^{n-1}$

$n=3,\therefore adj\left(A\right)=|\overrightarrow{A}{|}^{3-1}=|\overrightarrow{A}{|}^2$

$|\overrightarrow{A}|=1,\therefore adj\left(A\right)=1$

$adj\left(A\right)=|\overrightarrow{A}{|}^{n-1}$

$n=3,\therefore adj\left(A\right)=|\overrightarrow{A}{|}^{3-1}=|\overrightarrow{A}{|}^2$

$|\overrightarrow{A}|=1,\therefore adj\left(A\right)=1$

$|\overrightarrow{A}|=1$

$B=2|\overrightarrow{A}|=1\Rightarrow B=2\times 1=2$

$C=\frac{adjA}{2}=\frac{1}{2}$

$|A{B}^2{C}^3|=|1\times 2^2\times {\left(\frac{1}{2}\right)}^3|$

$=\frac{1}{2}$