
Assertion: $\left| adj\left( adj\left( adjA \right) \right) \right|={{\left| A \right|}^{{{\left( n-1 \right)}^{3}}}}$, where ‘n’ is the order of matrix A.
Reason: $\left| adjA \right|={{\left| A \right|}^{n}}$
check if the assertion and reason are correct or incorrect.
Answer
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Hint: We will check if the assertion and reason are correct or incorrect. This can be done using the formula of inverse matrix, which is given by: ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$. If both of them are correct, then we will check if reason is the correct explanation of assertion or not. In all other cases, our answer will be complete at the first step only. We shall proceed in this manner in our solution.
Complete step by step answer:
The first step in the problem is to find if assertion and reason are correct or not. Let us start by deriving the value for left-hand side of our reason. This can be done as follows:
$\Rightarrow {{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$
Pre-multiply the above equation by ‘A’ on both the sides, we get:
$\begin{align}
& \Rightarrow A\times {{A}^{-1}}=A\times \dfrac{adj\left( A \right)}{\left| A \right|} \\
& \Rightarrow I=A\times \dfrac{adj\left( A \right)}{\left| A \right|} \\
& \Rightarrow I\times \left| A \right|=Aadj\left( A \right) \\
\end{align}$
Taking determinant on both sides of our equation, we get:
$\begin{align}
& \Rightarrow \left| I\times \left| A \right| \right|=\left| Aadj\left( A \right) \right| \\
& \Rightarrow \left| I \right|\times \left| \left| A \right| \right|=\left| A \right|\left| adj\left( A \right) \right| \\
\end{align}$
Now, using the property of determinant, that is:
$\Rightarrow \left\| A \right\|={{\left| A \right|}^{n}}$
Our equation becomes:
$\begin{align}
& \Rightarrow 1\times {{\left| A \right|}^{n}}=\left| A \right|\left| adj\left( A \right) \right| \\
& \Rightarrow \left| adj\left( A \right) \right|={{\left| A \right|}^{n-1}} \\
\end{align}$
Therefore, $\left| adj\left( A \right) \right|\ne {{\left| A \right|}^{n}}$.
Thus, our reason is incorrect.
Now, we will check for our assertion. This can be done as follows:
We know from our above expression for determinant of adjoint of a matrix A that:
$\Rightarrow \left| adj\left( A \right) \right|={{\left| A \right|}^{n-1}}$
Here, replacing $A\to adjA$, we get:
$\begin{align}
& \Rightarrow \left| adj\left( adjA \right) \right|={{\left| adjA \right|}^{n-1}} \\
& \Rightarrow \left| adj\left( adjA \right) \right|={{\left[ {{\left| A \right|}^{\left( n-1 \right)}} \right]}^{\left( n-1 \right)}} \\
& \Rightarrow \left| adj\left( adjA \right) \right|={{\left| A \right|}^{\left( n-1 \right)\times \left( n-1 \right)}} \\
& \Rightarrow \left| adj\left( adjA \right) \right|={{\left| A \right|}^{{{\left( n-1 \right)}^{2}}}} \\
\end{align}$
Again replacing $A\to adjA$, we get:
$\begin{align}
& \Rightarrow \left| adj\left( adj\left( adjA \right) \right) \right|={{\left| adjA \right|}^{{{\left( n-1 \right)}^{2}}}} \\
& \Rightarrow \left| adj\left( adj\left( adjA \right) \right) \right|={{\left[ {{\left| A \right|}^{\left( n-1 \right)}} \right]}^{{{\left( n-1 \right)}^{2}}}} \\
& \Rightarrow \left| adj\left( adj\left( adjA \right) \right) \right|={{\left| A \right|}^{\left( n-1 \right)\times {{\left( n-1 \right)}^{2}}}} \\
& \Rightarrow \left| adj\left( adj\left( adjA \right) \right) \right|={{\left| A \right|}^{{{\left( n-1 \right)}^{3}}}} \\
\end{align}$
Therefore, we can see that the assertion is true.
Hence, we can say that the assertion is correct but the reason is incorrect.
Note: Whenever solving a problem based on assertion and reason, we have to be careful when both the assertion and reason are correct. In such a case, the reason should be a proper explanation of the assertion. If in our problem, the reason would have been correct, then it would have been a correct explanation of our assertion.
Complete step by step answer:
The first step in the problem is to find if assertion and reason are correct or not. Let us start by deriving the value for left-hand side of our reason. This can be done as follows:
$\Rightarrow {{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$
Pre-multiply the above equation by ‘A’ on both the sides, we get:
$\begin{align}
& \Rightarrow A\times {{A}^{-1}}=A\times \dfrac{adj\left( A \right)}{\left| A \right|} \\
& \Rightarrow I=A\times \dfrac{adj\left( A \right)}{\left| A \right|} \\
& \Rightarrow I\times \left| A \right|=Aadj\left( A \right) \\
\end{align}$
Taking determinant on both sides of our equation, we get:
$\begin{align}
& \Rightarrow \left| I\times \left| A \right| \right|=\left| Aadj\left( A \right) \right| \\
& \Rightarrow \left| I \right|\times \left| \left| A \right| \right|=\left| A \right|\left| adj\left( A \right) \right| \\
\end{align}$
Now, using the property of determinant, that is:
$\Rightarrow \left\| A \right\|={{\left| A \right|}^{n}}$
Our equation becomes:
$\begin{align}
& \Rightarrow 1\times {{\left| A \right|}^{n}}=\left| A \right|\left| adj\left( A \right) \right| \\
& \Rightarrow \left| adj\left( A \right) \right|={{\left| A \right|}^{n-1}} \\
\end{align}$
Therefore, $\left| adj\left( A \right) \right|\ne {{\left| A \right|}^{n}}$.
Thus, our reason is incorrect.
Now, we will check for our assertion. This can be done as follows:
We know from our above expression for determinant of adjoint of a matrix A that:
$\Rightarrow \left| adj\left( A \right) \right|={{\left| A \right|}^{n-1}}$
Here, replacing $A\to adjA$, we get:
$\begin{align}
& \Rightarrow \left| adj\left( adjA \right) \right|={{\left| adjA \right|}^{n-1}} \\
& \Rightarrow \left| adj\left( adjA \right) \right|={{\left[ {{\left| A \right|}^{\left( n-1 \right)}} \right]}^{\left( n-1 \right)}} \\
& \Rightarrow \left| adj\left( adjA \right) \right|={{\left| A \right|}^{\left( n-1 \right)\times \left( n-1 \right)}} \\
& \Rightarrow \left| adj\left( adjA \right) \right|={{\left| A \right|}^{{{\left( n-1 \right)}^{2}}}} \\
\end{align}$
Again replacing $A\to adjA$, we get:
$\begin{align}
& \Rightarrow \left| adj\left( adj\left( adjA \right) \right) \right|={{\left| adjA \right|}^{{{\left( n-1 \right)}^{2}}}} \\
& \Rightarrow \left| adj\left( adj\left( adjA \right) \right) \right|={{\left[ {{\left| A \right|}^{\left( n-1 \right)}} \right]}^{{{\left( n-1 \right)}^{2}}}} \\
& \Rightarrow \left| adj\left( adj\left( adjA \right) \right) \right|={{\left| A \right|}^{\left( n-1 \right)\times {{\left( n-1 \right)}^{2}}}} \\
& \Rightarrow \left| adj\left( adj\left( adjA \right) \right) \right|={{\left| A \right|}^{{{\left( n-1 \right)}^{3}}}} \\
\end{align}$
Therefore, we can see that the assertion is true.
Hence, we can say that the assertion is correct but the reason is incorrect.
Note: Whenever solving a problem based on assertion and reason, we have to be careful when both the assertion and reason are correct. In such a case, the reason should be a proper explanation of the assertion. If in our problem, the reason would have been correct, then it would have been a correct explanation of our assertion.
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