Question

If $A$ is an $n \times n$ non-singular matrix, then $\left| {adj\;A} \right|$ is
A. ${\left| A \right|^n}$
B. ${\left| A \right|^{n + 1}}$
C. ${\left| A \right|^{n - 1}}$
D. ${\left| A \right|^{n - 2}}$

Answer 1

Hint: In this problem, given that $A$ is an $n \times n$ non-singular (invertible) matrix. To find the determinant of the matrix $adj\;A$, first we will use the result $A\left( {adj\;A} \right) = \left| A \right|I$ where $I$ is $n \times n$ identity matrix. Then, we will use the following properties of determinants
.$
  \left( 1 \right)\quad \left| {AB} \right| = \left| A \right|\left| B \right| \\
  \left( 2 \right)\quad \left| {kA} \right| = {k^n}\left| A \right| \\
 $

Complete step-by-step solution:
In this problem, we have $n \times n$ non-singular (invertible) matrix $A$. The matrix $A$ is nonsingular if and only if $\left| A \right| \ne 0$.
We know the result $A\left( {adj\;A} \right) = \left| A \right|I \cdots \cdots \left( 1 \right)$. Let us assume $\left| A \right| = k$ where $k$ is non-zero constant.
Therefore, from $\left( 1 \right)$ we can write $A\left( {adj\;A} \right) = kI \cdots \cdots \left( 2 \right)$.
Let us take determinant on both sides of $\left( 2 \right)$. Therefore, we get $\left| {A\left( {adj\;A} \right)} \right| = \left| {kI} \right| \cdots \cdots \left( 3 \right)$.
Now first we will solve the LHS part of equation $\left( 3 \right)$. For this, we will use the property of determinant which is given by $\left| {AB} \right| = \left| A \right|\left| B \right|$. Therefore, LHS part of equation $\left( 3 \right)$ is $
  \left| {A\left( {adj\;A} \right)} \right| = \left| A \right|\left| {adj\;A} \right| \\
   \Rightarrow \left| {A\left( {adj\;A} \right)} \right| = k\left| {adj\;A} \right|\quad \left[ {\because \left| A \right| = k} \right] \cdots \cdots \left( 4 \right) \\
 $
Now we will solve the RHS part of equation $\left( 3 \right)$. For this, we will use another property of determinant which is given by $\left| {kA} \right| = {k^n}\left| A \right|$. Therefore, RHS part of equation $\left( 3 \right)$ is $\left| {kI} \right| = {k^n}\left| I \right|$. We know that the determinant of the identity matrix is $1$. Therefore, now RHS part of equation $\left( 3 \right)$ is $\left| {kI} \right| = {k^n}\left( 1 \right) = {k^n} \cdots \cdots \left( 5 \right)$.
Now from $\left( 3 \right),\left( 4 \right)$ and $\left( 5 \right)$, we can write $k\left| {adj\;A} \right| = {k^n}$. Simplify this equation, we will get $
  \left| {adj\;A} \right| = \dfrac{{{k^n}}}{k} \\
   \Rightarrow \left| {adj\;A} \right| = {k^{n - 1}} \\
   \Rightarrow \left| {adj\;A} \right| = {\left| A \right|^{n - 1}}\quad \left[ {\because k = \left| A \right|} \right] \\
 $
Therefore, if $A$ is an $n \times n$ non-singular matrix, then $\left| {adj\;A} \right| = {\left| A \right|^{n - 1}}$.
Therefore, option C is true.

Note: Determinant of the matrix $A$ is denoted by $\det \left( A \right)$ or $\left| A \right|$. If $\left| A \right| = 0$ then matrix $A$ is singular (not invertible). That is, the inverse matrix of $A$ does not exist. Adjoint of matrix is matrix and determinant of matrix is a number (constant). If we know the determinant and order of matrix then to find determinant of adjoint of that matrix we can use the property which is given by $\left| {adj\;A} \right| = {\left| A \right|^{n - 1}}$. If we use this property then we can find the determinant of the adjoint of the matrix without finding the adjoint of the matrix.