Question

# If ${\text{A}}\left( {{\text{adjA}}} \right) = 5{\text{I}}$ where ${\text{I}}$ is the identity matrix of order 3, then $|adjA|$ is equal toA. 125B. 25C. 5D. 10

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Hint: Use property of inverse of A and determinant of adjoint of A. Also two matrices are equal to each other then, the order of both the matrices will be equal.

Given, ${\text{A}}\left( {{\text{adjA}}} \right) = 5{\text{I}}$ where order of identity matrix is 3.
Clearly, the order of matrix A and that of identity matrix are equal.
So, the order of matrix A is also 3.
As we know that inverse of any matrix A is given by ${{\text{A}}^{ - 1}} = \dfrac{1}{{|A|}}\left( {{\text{adjA}}} \right)$ where |A| is the determinant of matrix A and adjA is the adjoint matrix of matrix A.
$\therefore {\text{ A}}\left[ {{{\text{A}}^{ - 1}}} \right] = {\text{A}}\left[ {\dfrac{1}{{|A|}}\left( {{\text{adjA}}} \right)} \right] = \dfrac{{{\text{A}}\left( {{\text{adjA}}} \right)}}{{|A|}} = \dfrac{{5{\text{I}}}}{{|A|}}$
Also, we know that ${\text{ A}}\left[ {{{\text{A}}^{ - 1}}} \right] = {\text{I}}$ where ${\text{I}}$ is the identity matrix order 3
Therefore, $\Rightarrow {\text{I}} = \dfrac{{5{\text{I}}}}{{|A|}} \\ \Rightarrow |{\text{A}}|I = 5I \\$
On comparing the above equation, we get
Determinant of the matrix A, $|{\text{A}}| = 5$
Using the identity, $|{\text{adjA}}| = {\left[ {|A|} \right]^{n - 1}}$ where n is the order of the matrix of A
Put $|{\text{A}}| = 5$ and ${\text{n}} = 3$ in the above identity, we have
$\Rightarrow |{\text{adjA}}| = {\left[ 5 \right]^{3 - 1}} = {5^2} = 25$
Therefore, the determinant of matrix adjA is 25.
Option B is correct.

Note- Here, the inverse matrix only exists for non-singular matrices (i.e., determinant of that matrix whose inverse is required should always be non-zero). Also if in an equation two matrices are equal to each other then, order of both the matrices will be equal.