Question

If A is an $n \times n$ non-singular matrix, then $\left| {AdjA} \right|$ 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

It is given in the question that if A is an $n \times n$ non-singular matrix.
Then, what is the value of $\left| {adjA} \right|$ . The matrix A is nonsingular if and only if $\left| A \right| \ne 0$ .
First, we know that $A\left( {adjA} \right) = \left| A \right|I$ . Then, we will assume $\left| A \right| = k$ .
Finally, after using the properties of determinants we will get the answer.
Properties:
$\left| {AB} \right| = \left| A \right|\left| B \right|$
$\left| {kA} \right| = {k^n}A$

Complete step-by-step answer:
It is given in the question that if A is an $n \times n$ non-singular matrix.
Then, what is the value of $\left| {adjA} \right|$.
The matrix A is non-singular if and only if $\left| A \right| \ne 0$ .
Since, we know that $A\left( {adjA} \right) = \left| A \right|I$ (I)
Now, let us assume that $\left| A \right| = k$ where k is non zero constant.
Therefore, from equation (I), we can write $A\left( {adjA} \right) = kI$ (II)
Now, let us take determinant on both the side of equation (II), we get,
$\left| {A\left( {adjA} \right)} \right| = \left| {kI} \right|$ (III)
Now, first we solve the L.H.S part of the equation (III).
For this, we use the property of determinant $\left| {AB} \right| = \left| A \right|\left| B \right|$ .
Therefore, L.H.S of equation (III) is
$\left| {A\left( {adjA} \right)} \right| = \left| A \right|\left| {adjA} \right|$
$\left| {A\left( {adjA} \right)} \right| = k\left| {adjA} \right|$ $\left( {\because \left| A \right| = k} \right)$ (IV)
Now, we will solve the R.H.S part of equation (III).
For this, we will use another property of determinant i.e. $\left| {kA} \right| = {k^n}A$ .
Therefore, R.H.S part of equation (III) is $\left| {kI} \right| = {k^n}I$ .
Since, we know that the determinant of the identity matrix is 1.
Therefore, now R.H.S part of equation (III) is $\left| {kI} \right| = {k^n}\left( 1 \right) = {k^n}$ . (V)
Now, from equation (III), (Iv), and (v), we can write $k\left| {adjA} \right| = {k^n}$ .
Now, when we simplify the above equation, we get,
$\left| {adjA} \right| = \dfrac{{{k^n}}}{k}$
$\left| {adjA} \right| = {k^{n - 1}}$
$\left| {adjA} \right| = {\left| A \right|^{n - 1}}$ $\left( {\because \left| A \right| = k} \right)$
Hence option C is correct .
Note: Some properties of determinant:
1. $\left| {{A^T}} \right| = \left| A \right|$
2. $\left| {AB} \right| = \left| A \right|\left| B \right|$
3. $\left| {{A^{ - 1}}} \right| = \dfrac{1}{A}$
4. $\left| {kA} \right| = {k^n}\left| A \right|$ , where n is an order of matrices.
5. $\left| { - A} \right| = \left| { - 1 \times A} \right| = \left( { - 1} \right)n \times \left| A \right|$ .