 QUESTION

# If A is a square matrix of order 3 such that $\left| {\text{A}} \right| = 2$ then ${\left| {adj{A^{ - 1}}} \right|^{ - 1}}$ is $\ldots$

Hint: To solve this type of question we should have knowledge of properties of matrices and also knowledge of taking determinants of any matrix. In this question we have to start from given and proceed further using properties.

We have
A is a square matrix of order 3 such that $\left| A \right| = 2$
We know that;
$\left| {{A^{ - 1}}} \right| = \dfrac{1}{{\left| A \right|}}$
$\therefore \left| {{A^{ - 1}}} \right| = \dfrac{1}{2}$
Now;
We can write
$\Rightarrow {\left| {adj{A^{ - 1}}} \right|^{ - 1}} = \dfrac{1}{{\left| {adj{A^{ - 1}}} \right|}}$
Now we will use this property.
$\left( {\because \left| {adjA} \right| = {{\left| A \right|}^{n - 1}}} \right)$
So we can write as
$\because {\left| {adj{A^{ - 1}}} \right|^{ - 1}} = \dfrac{1}{{{{\left| A^{ - 1} \right|}^{n - 1}}}}$
$\therefore {\left| {adj{A^{ - 1}}} \right|^{ - 1}} = \dfrac{1}{{{{\left| {{A^{ - 1}}} \right|}^{3 - 1}}}}$ ( order of matrix $n=3$)
$\Rightarrow {\left| {adj{A^{ - 1}}} \right|^{ - 1}} = \dfrac{1}{{{{\left| {{A^{ - 1}}} \right|}^2}}} = \dfrac{1}{{{{\left( {\dfrac{1}{2}} \right)}^2}}} = \dfrac{1}{{\dfrac{1}{4}}} = 4 \\$

Note: Whenever you get these types of questions the key concept of solving is you have to start from given and proceed from the formula of inverse of matrix and after that take determinant of that formula and apply properties of matrix to get an answer. Properties like $\left( {\because \left| {adjA} \right| = {{\left| A \right|}^{n - 1}}} \right)$ should be remembered.