Question

If the matrices $A=\left[\begin{array}{ccc}1& 1& 2\\ 1& 3& 4\\ 1& -1& 3\end{array}\right]$, $B=adjA$,$C=3A$ then,$\frac{\left|adjB\right|}{\left|C\right|}$ is equal to

• A. $16$
• B. $2$
• C. $8$
• D. $72$

Answer 1

The correct option is C

$8$

Explanation for the correct answer:

$A=\left[\begin{array}{ccc}1& 1& 2\\ 1& 3& 4\\ 1& -1& 3\end{array}\right]$

Hence, the determinant of $A$ is calculated as

$\left|A\right|$$=\left|\begin{array}{ccc}1& 1& 2\\ 1& 3& 4\\ 1& -1& 3\end{array}\right|$$=\left(3\times 3-4\times (-1)\right)-\left(3-4\right)+2\left(-1-3\right)$

$\Rightarrow$ $\left|A\right|$$=13+1-8$

$\Rightarrow$ $\left|A\right|$$=6$

By the property of determinants we know that

$\left|adjA\right|={\left|A\right|}^{n-1}$ where $n$ is the order of the matrix

$B=adjA$

$\Rightarrow$$\left|adjB\right|=\left|adj\left(adjA\right)\right|={\left|A\right|}^{{\left(n-1\right)}^2}$

Substituting the values we get

$\Rightarrow$$\left|adjB\right|={\left(6\right)}^{{\left(3-1\right)}^2}$

$\Rightarrow$$\left|adjB\right|=6^4$

By the property of determinants we know that

$\left|kA\right|=k^n\left|A\right|$ where $n$ is the order of a matrix and $k$ is a constant.

$C=3A$

$\Rightarrow$$\left|C\right|=\left|3A\right|$

$\Rightarrow$$\left|C\right|=3^3\times \left|A\right|$

$\Rightarrow$$\left|C\right|=3^3\times 6$

Substituting all the required values in the required expression we get

$\frac{\left|adjB\right|}{\left|C\right|}=\frac{6^4}{3^3\times 6}$

$\Rightarrow$$\frac{\left|adjB\right|}{\left|C\right|}={\left(\frac{6}{3}\right)}^3$

$\Rightarrow$$\frac{\left|adjB\right|}{\left|C\right|}=8$

For the given matrices , the value of $\frac{\left|adjB\right|}{\left|C\right|}$ is $8$.

Hence, option (C) is the correct answer.