Question

If $A$ is a square matrix of order $3 \times 3$, such that $\left| A \right| = 5$, then the value of $\left| {{\text{adj}}A} \right|$ is?

Answer 1

Hint: We will use the identity $A.{\text{adj}}A = \left| A \right|{I_n}$, where ${I_n}$ is an identity matrix of order $n$, which is equal to the order of matrix $A$. We will take determinant on both sides and will substitute the value of order and determinant of $A$ to find the value of $\left| {{\text{adj}}A} \right|$.

Complete step-by-step answer:
We are given that $A$ is a square matrix of order $3 \times 3$ and $\left| A \right| = 5$, where $\left| A \right|$ represents determinant of $A$.
We have to calculate the determinant of adjoint of $A$
Now, we know that $A.{\text{adj}}A = \left| A \right|{I_n}$, where ${I_n}$ is an identity matrix of order $n$, which is equal to the order of matrices $A$.
On substituting the value of $\left| A \right| = 5$, we will get,
$A.{\text{adj}}A = 5{I_3}$
We will take determinant on both sides,
$\left| {A.{\text{adj}}A} \right| = \left| {5{I_3}} \right|$
Now, $\left| {MN} \right| = \left| M \right|\left| N \right|$ and $\left| {kM} \right| = {k^n}\left| M \right|$, where $n$ is the order of the matrix.
$
  \left| {A.{\text{adj}}A} \right| = \left| {5{I_3}} \right| \\
   \Rightarrow \left| A \right|\left| {{\text{adj}}A} \right| = {5^3}\left| {{I_3}} \right| \\
$
Determinant of the identity matrix is always 1.
$
  5\left| {{\text{adj}}A} \right| = {5^3} \\
   \Rightarrow \left| {{\text{adj}}A} \right| = \dfrac{{{5^3}}}{5} \\
   \Rightarrow \left| {{\text{adj}}A} \right| = 25 \\
$
Hence, the value of $\left| {{\text{adj}}A} \right|$ is equal to 25.

Note: We can also calculate the $\left| {{\text{adj}}A} \right|$ using the direct formula, $\left| {{\text{adj}}A} \right| = {\left| A \right|^{n - 1}}$, where $n$ is the order of matrix $A$. Determinant is always calculated as a square matrix. Adjoint of a matrix is the transpose of the cofactors of the matrix.