Question

If A is a square matrix of order n, then $adj\left( {adjA} \right)$ is equal to
$
  {\text{A}}{\text{. }}{\left| A \right|^{n - 1}}A \\
  {\text{B}}{\text{. }}{\left| A \right|^n}A \\
  {\text{C}}{\text{. }}{\left| A \right|^{n - 2}}A \\
  {\text{D}}{\text{. none of these}} \\
$

Answer 1

Hint: -To solve this question we have to have remembered all the properties of matrices and determinant like $A\left( {adjA} \right) = \left| A \right|{I_n}$ and $\left| {adjA} \right| = {\left| A \right|^{n - 1}}$ apply these properties to get an answer easily.

Complete Step-by-Step solution:
We have
A square matrix A of order n and we have to find $adj\left( {adjA} \right)$
We know from the properties of matrices and determinant
$x\left( {adjx} \right) = \left| x \right|{I_n}$
Now we will put $x = adjA$ then we get,
$\left( {adjA} \right)\left[ {adj\left( {adjA} \right)} \right] = \left| {adjA} \right|{I_n}$
Now using the property $\left( {\because \left| {adjA} \right| = {{\left| A \right|}^{n - 1}}} \right)$ we get,
$\left( {adjA} \right)\left[ {adj\left( {adjA} \right)} \right] = {\left| A \right|^{n - 1}}{I_n}$
Now multiplying by A on both side we get,
$\left( {AadjA} \right)\left[ {adj\left( {adjA} \right)} \right] = {\left| A \right|^{n - 1}}A{\text{ }}\left( {\because A{I_n} = A} \right)$
We know $A\left( {adjA} \right) = \left| A \right|{I_n}$ using this property we get
$
  \left| A \right|{I_n}\left[ {adj\left( {adjA} \right)} \right] = {\left| A \right|^{n - 1}}A \\
  \therefore adj\left( {adjA} \right) = \dfrac{{{{\left| A \right|}^{n - 1}}A}}{{\left| A \right|}} = {\left| A \right|^{n - 2}}A \\
 $
Hence option C is the correct option.

Note: -Whenever we get this type of question the key concept of solving is we have to care about where to start solving and think about which property would be perfect for proceeding. Here ${I_n}$ is the unit matrix of order n . In this type of question we have to use our brain to identify which property should be used.