Question

If A is a $3\times 3$ matrix such that $\left|A\right|=4than\left|{\left(adjA\right)}^{-1}\right|=$

• A. $16$
• B. $64$
• C. $\frac{1}{16}$
• D. None

Answer 1

Answer: (A)

$16$

We know, $A^{-1}=\frac{adjA}{|A|}$

Multiplying above equation with A both sides,

$A{A}^{-1}=\frac{A\times adjA}{|A|}\Rightarrow |A|=A\times adjA$

Multiplying with $(adjA{)}^{-1}$ both sides ,

$|A|\times (adjA{)}^{-1}=A\times adjA\times (adjA{)}^{-1}\Rightarrow |A|\times (adjA{)}^{-1}=A$

Taking determinant both sides,$||A|\times (adjA{)}^{-1}|=|A|\Rightarrow ||A||\times |(adjA{)}^{-1}|=|A|\Rightarrow |A{|}^n\times |(adjA{)}^{-1}|=|A|$

Where n is the order of matrix A, i.e. $n=3$ and $|A|=4$

Thus, $|(adjA{)}^{-1}|=\frac{4}{4^3}=\frac{1}{16}$