Question

If we are given the following data about a matrix A, that is, \[{{A}_{3\times 3}}\] and \[\det A=5\], then \[\det \left( adjA \right)=\]
A. 5
B. 25
C. 125
D. \[\dfrac{1}{5}\]

Answer 1

Hint: We know that if we have a matrix A of order n and its determinant value is equal to ‘k’ then the determinant of adjoint of A is given as follows:
\[\det \left( adjA \right)={{\left| A \right|}^{n-1}}={{k}^{n-1}}\]

Complete step-by-step answer:

In this question, we have been given a matrix A of the order \[\left( 3\times 3 \right)\] and det A = 5.
Now, we know that \[{{A}^{-1}}=\dfrac{adjA}{\left| A \right|}\].
Also, we know that \[A.{{A}^{-1}}=I\].
By substituting the value of \[{{A}^{-1}}\] in the above expression, we will get as follows:
\[\begin{align}
  & A\left( \dfrac{adjA}{\left| A \right|} \right)=I \\
 & \Rightarrow \left| A.adj\left( A \right) \right|=\left| A \right|.I \\
 & \Rightarrow \left| A \right|.\left| adj\left( A \right) \right|={{\left| A \right|}^{n}} \\
 & \Rightarrow \left| adj\left( A \right) \right|=\dfrac{{{\left| A \right|}^{n}}}{\left| A \right|} \\
 & \Rightarrow \left| adj\left( A \right) \right|={{\left| A \right|}^{n-1}} \\
\end{align}\]
Hence we can see that if we have a matrix A of order n and its determinant is equal to \[\left| A \right|\] then the determinant of adjoint of A i.e. adjA is equal to \[{{\left| A \right|}^{n-1}}\].
Here, we have n = 3 and \[\left| A \right|=5\]
\[\Rightarrow \det \left[ adj\left( A \right) \right]={{\left| A \right|}^{n-1}}={{\left( 5 \right)}^{3-1}}={{5}^{2}}=25\]
Therefore the correct answer of the question is option B.

Note: Remember that the adjoint of any matrix A is the transpose of the cofactor matrix of A and it is denoted by “adj(A)”. It is also known as an adjugate matrix. Also, remember the formula of determinant of adjoint of matrix A since it will take your time to find the expression so it is better to memorize the formula which is \[\det \left( adjA \right)={{\left| A \right|}^{n-1}}={{k}^{n-1}}\]