Question

Assertion:
If A = \left[ {\begin{array}{*{20}{c}} 2&0&{ - 1} \\\ 5&1&0 \\\ 0&1&3 \end{array}} \right] , then $adj\left( {adj{\text{ }}A} \right) = A$
Reason:
$\left| {adj\left( {adjA} \right)} \right| = {\left| A \right|^{{{\left( {n - 1} \right)}^2}}}$ , A is a nonsingular matrix of order n.
A.Both Assertion and Reason are individually true and Reason is correct explanation of Assertion
B.Both Assertion and Reason are individually true and Reason is not correct(proper) explanation of Assertion
C.Assertion is true but Reason is false
D.Assertion is false but Reason is true

Answer 1

Firstly, find the adjoint of A i.e. adj A. Then, find adjoint of adjoint of A i.e. adj (adj A) and compare whether $adj\left( {adj{\text{ }}A} \right) = A$ or not.
Then, find the value of determinant of adj (adj A) i.e. |adj (adj A| and also find the value of determinant of A i.e. |A| and find ${\left| A \right|^{{{\left( {n - 1} \right)}^2}}}$ . Thus, check whether $\left| {adj\left( {adjA} \right)} \right| = {\left| A \right|^{{{\left( {n - 1} \right)}^2}}}$ or not.
Also, verify if the Assertion and Reason are true or false. If both are true, then check whether Reason is the correct explanation of Assertion or not.
Thus, choose the correct option.

Complete step-by-step answer:
It is given in assertion that, If A = \left[ {\begin{array}{*{20}{c}} 2&0&{ - 1} \\\ 5&1&0 \\\ 0&1&3 \end{array}} \right] , then $adj\left( {adj{\text{ }}A} \right) = A$ .
Now, to verify that $adj\left( {adj{\text{ }}A} \right) = A$ , we have to first find the adjoint of A.
Adjoint of A can be found using the formula adjA = \left[ {\begin{array}{*{20}{c}} { + \left| {\begin{array}{*{20}{c}} {{a_{22}}}&{{a_{23}}} \\\ {{a_{32}}}&{{a_{33}}} \end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}} {{a_{12}}}&{{a_{13}}} \\\ {{a_{32}}}&{{a_{33}}} \end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}} {{a_{12}}}&{{a_{13}}} \\\ {{a_{22}}}&{{a_{23}}} \end{array}} \right|} \\\ { - \left| {\begin{array}{*{20}{c}} {{a_{21}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{33}}} \end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{13}}} \\\ {{a_{31}}}&{{a_{33}}} \end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{23}}} \end{array}} \right|} \\\ { + \left| {\begin{array}{*{20}{c}} {{a_{21}}}&{{a_{22}}} \\\ {{a_{31}}}&{{a_{32}}} \end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}} \\\ {{a_{31}}}&{{a_{32}}} \end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}} \\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right|} \end{array}} \right]
So, adjA = \left[ {\begin{array}{*{20}{c}} { + \left| {\begin{array}{*{20}{c}} 1&0 \\\ 1&3 \end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}} 0&{ - 1} \\\ 1&3 \end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}} 0&{ - 1} \\\ 1&0 \end{array}} \right|} \\\ { - \left| {\begin{array}{*{20}{c}} 5&0 \\\ 0&3 \end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}} 2&{ - 1} \\\ 0&3 \end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}} 2&{ - 1} \\\ 5&0 \end{array}} \right|} \\\ { + \left| {\begin{array}{*{20}{c}} 5&1 \\\ 0&1 \end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}} 2&0 \\\ 0&1 \end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}} 2&0 \\\ 5&1 \end{array}} \right|} \end{array}} \right]
\Rightarrow adjA = \left[ {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\\ { - 15}&6&{ - 5} \\\ 5&{ - 2}&2 \end{array}} \right]
Similarly, we will find adj (adj A).

\Rightarrow adj\left( {adjA} \right) = \left[ {\begin{array}{*{20}{c}} { + \left| {\begin{array}{*{20}{c}} 6&{ - 5} \\\ { - 2}&2 \end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}} { - 1}&1 \\\ { - 2}&2 \end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}} { - 1}&1 \\\ 6&{ - 5} \end{array}} \right|} \\\ { - \left| {\begin{array}{*{20}{c}} { - 15}&{ - 5} \\\ 5&2 \end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}} 3&1 \\\ 5&2 \end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}} 3&1 \\\ { - 15}&{ - 5} \end{array}} \right|} \\\ { + \left| {\begin{array}{*{20}{c}} { - 15}&6 \\\ 5&{ - 2} \end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}} 3&{ - 1} \\\ 5&{ - 2} \end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}} 3&{ - 1} \\\ { - 15}&6 \end{array}} \right|} \end{array}} \right] \\\ \Rightarrow adj\left( {adjA} \right) = \left[ {\begin{array}{*{20}{c}} 2&0&{ - 1} \\\ 5&1&0 \\\ 0&1&3 \end{array}} \right] \\\

Thus, we get $adj\left( {adjA} \right) = A$ .
So, Assertion given here is true.
Also, in reason, it is given that, $\left| {adj\left( {adjA} \right)} \right| = {\left| A \right|^{{{\left( {n - 1} \right)}^2}}}$ , A is a non-singular matrix of order n.
Now, to prove the above relation, we will take the help of the given matrix in assertion. So, we will find the value of $\left| {adj\left( {adjA} \right)} \right|$ .
$\Rightarrow \left| {adj\left( {adjA} \right)} \right| = 2\left( {3 - 0} \right) - 0\left( {15 - 0} \right) - 1\left( {5 - 0} \right) \\\ \Rightarrow \left| {adj\left( {adjA} \right)} \right| = 2\left( 3 \right) - 1\left( 5 \right) \\\ \Rightarrow \left| {adj\left( {adjA} \right)} \right| = 6 - 5 \\\ \Rightarrow \left| {adj\left( {adjA} \right)} \right| = 1 \\\$
Then, we have to find the value of $\left| A \right|$ .

$$\Rightarrow \left| A \right| = 2\left( {3 - 0} \right) - 0\left( {15 - 0} \right) - 1\left( {5 - 0} \right) \\\ \Rightarrow \left| A \right| = 2\left( 3 \right) - 1\left( 5 \right) \\\ \Rightarrow \left| A \right| = 6 - 5 \\\ \Rightarrow \left| A \right| = 1 \\\$$

Here, in assertion, A is a matrix of order 3.
So, ${\left( {n - 1} \right)^2} = {\left( {3 - 1} \right)^2} = {2^2} = 4 \Rightarrow {\left| A \right|^{{{\left( {n - 1} \right)}^2}}} = {\left| A \right|^4} = {1^4} = 1$
Thus, we get $\left| {adj\left( {adjA} \right)} \right| = {\left| A \right|^{{{\left( {n - 1} \right)}^2}}}$ .
Hence, the Reason given here is true and it is also a correct explanation of Assertion.
So, option (A) is correct.

Note: On observing the above calculations, we get the property as follows:
For any non-singular matrix B of order m, if $adj\left( {adj{\text{ }}B} \right) = B$ , then $\left| {adj\left( {adjB} \right)} \right| = {\left| B \right|^{{{\left( {m - 1} \right)}^2}}}$ .
Thus, as per the question, if $adj\left( {adjA} \right) = A$ , then $\left| {adj\left( {adjA} \right)} \right| = {\left| A \right|^{{{\left( {n - 1} \right)}^2}}}$ .