Question

ASSERTION
If A = \left( {\begin{array}{*{20}{c}} 2&0&{ - 1} \\\ 5&1&0 \\\ 0&1&3 \end{array}} \right) then $adj(adj\,A) = A$
REASON
$\left| {adj \cdot (adj \cdot A)} \right| = {\left| A \right|^{{{(n - 1)}^2}}}$, A is a nonsingular matrix of order N.
(A)Both (A) & (R) are individually true & (R) is correct explanation of (A),
(B)Both (A) & (R) are individually true but (R) is not the correct (proper explanation of (A).
(C)(A) is true but (R) is false,
(D) (A) is false but (R) is true.

Answer 1

Hint : Use properties and prove the assertion and reason relations.
First, we are going to use the property of adjoint which is $adj(adjA) = \mid A{\mid ^{(n - 2)}}A$, with that we are going to find $\left| A \right|$, such that it proves the assertion statement and with the help of the same property we are going to prove the reason statement and then find the relation between them and choose the option.

Complete step-by-step answer :
First, we are going to the given in the assertion statement which is
A = \left( {\begin{array}{*{20}{c}} 2&0&{ - 1} \\\ 5&1&0 \\\ 0&1&3 \end{array}} \right)
To find the adjoint(A), we are going to find the cofactor matrix of A and Then apply transpose the cofactor matrix, which will give us the adjoint matrix.
First, let is find the cofactor of each element of the given matrix and find the cofactor matrix

1&0 \\\ 1&3 \end{array}} \right| = 3$$

{C_{12}} = {\left( { - 1} \right)^{1 + 2}}\left| {\begin{array}{{20}{c}}
5&0 \\
0&3
\end{array}} \right| = - 15 \\
{C_{13}} = {\left( { - 1} \right)^{1 + 3}}\left| {\begin{array}{{20}{c}}
5&1 \\
0&1
\end{array}} \right| = 5 \\
{C_{21}} = {\left( { - 1} \right)^{2 + 1}}\left| {\begin{array}{{20}{c}}
0&{ - 1} \\
1&3
\end{array}} \right| = - 1 \\
{C_{22}} = {\left( { - 1} \right)^{2 + 2}}\left| {\begin{array}{{20}{c}}
2&{ - 1} \\
0&3
\end{array}} \right| = 6 \\
{C_{23}} = {\left( { - 1} \right)^{2 + 3}}\left| {\begin{array}{{20}{c}}
2&0 \\
0&1
\end{array}} \right| = - 2 \\
{C_{31}} = {\left( { - 1} \right)^{3 + 1}}\left| {\begin{array}{{20}{c}}
0&{ - 1} \\
1&0
\end{array}} \right| = 1 \\
{C_{32}} = {\left( { - 1} \right)^{3 + 2}}\left| {\begin{array}{{20}{c}}
2&{ - 1} \\
5&0
\end{array}} \right| = - 5 \\
{C_{11}} = {\left( { - 1} \right)^{3 + 3}}\left| {\begin{array}{{20}{c}}
2&0 \\
5&1
\end{array}} \right| = 2 \\

Now from this write the cofactor matrix $$\left[ {\begin{array}{*{20}{c}} 3&{ - 15}&5 \\\ { - 1}&6&{ - 2} \\\ 1&{ - 5}&2 \end{array}} \right]$$ Now apply transpose to the matrix, we get $$\left[ {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\\ { - 15}&6&{ - 5} \\\ 5&{ - 2}&2 \end{array}} \right]$$ This is adjoint A, We have to perform to adjoint to the obtained matrix So first find the cofactors and then the cofactor matrix

{C_{11}} = {\left( { - 1} \right)^{1 + 1}}\left| {\begin{array}{{20}{c}}
6&{ - 5} \\
{ - 2}&2
\end{array}} \right| = 2 \\
{C_{12}} = {\left( { - 1} \right)^{1 + 2}}\left| {\begin{array}{{20}{c}}
{ - 15}&{ - 5} \\
5&2
\end{array}} \right| = 5 \\
{C_{13}} = {\left( { - 1} \right)^{1 + 3}}\left| {\begin{array}{{20}{c}}
{ - 15}&6 \\
5&{ - 2}
\end{array}} \right| = 0 \\
{C_{21}} = {\left( { - 1} \right)^{2 + 1}}\left| {\begin{array}{{20}{c}}
{ - 1}&1 \\
{ - 2}&2
\end{array}} \right| = 0 \\
{C_{22}} = {\left( { - 1} \right)^{2 + 2}}\left| {\begin{array}{{20}{c}}
3&1 \\
5&2
\end{array}} \right| = 1 \\
{C_{23}} = {\left( { - 1} \right)^{2 + 3}}\left| {\begin{array}{{20}{c}}
3&{ - 1} \\
5&{ - 2}
\end{array}} \right| = 1 \\
{C_{31}} = {\left( { - 1} \right)^{3 + 1}}\left| {\begin{array}{{20}{c}}
{ - 1}&1 \\
6&{ - 5}
\end{array}} \right| = - 1 \\
{C_{32}} = {\left( { - 1} \right)^{3 + 2}}\left| {\begin{array}{{20}{c}}
3&1 \\
{ - 15}&{ - 5}
\end{array}} \right| = 0 \\
{C_{33}} = {\left( { - 1} \right)^{3 + 3}}\left| {\begin{array}{*{20}{c}}
3&{ - 1} \\
{ - 15}&6
\end{array}} \right| = 3 \\

We get the cofactor matrix which is $$\left[ {\begin{array}{*{20}{c}} 2&5&0 \\\ 0&1&1 \\\ { - 1}&0&3 \end{array}} \right]$$ We have to appy transpose now, then we get the adjoint $adj(adj\,a) = \left[ {\begin{array}{*{20}{c}} 2&0&{ - 1} \\\ 5&1&0 \\\ 0&1&3 \end{array}} \right]$ Hence proved Therefore, the assertion(A) statement is true. Also given that $adj(adj\,A) = A$, we are going to prove this with the help of a property of the adjoint which is $$adj(adjA) = \mid A{\mid ^{(n - 2)}}A$$, so for that we will find $\left| A \right|$. Solving the given matrix, we will get

\left| A \right| = \left( {\begin{array}{*{20}{c}}
2&0&{ - 1} \\
5&1&0 \\
0&1&3
\end{array}} \right) = 2 \times (3 - 1) + 0 \times (15 - 0) - 1 \times (5 - 0) \\
= 6 - 5 = 1 ;

Next, we have to prove the reason (R) is true. Using the same property which is $$adj(adjA) = \mid A{\mid ^{(n - 2)}}A$$ On applying mod on RHS side, we get $$\left| {adj(adjA)} \right| = \mid A{\mid ^{n(n - 2)}}\left| A \right|$$ Which implies that $\left| {adj(adjA)} \right| = {\left| A \right|^{{{(n - 1)}^2}}}$ Now, even though the reason (R) statement is also true it does not give the correct explanation of Assertion A. Therefore, option B is the correct answer **Note** : We have to be familiar with the properties of matrices and adjoints such that we can easily prove these statements easily and we have to be careful to apply mod on both sides and know what are changes that will happen on applying them.