Question

Let A be a $3\times 3$ matrix such that ${{A}^{2}}-5A+7I=0$
Statement-I: ${{A}^{-1}}=\dfrac{1}{7}\left( 5I-A \right)$
Statement-II: The polynomial ${{A}^{3}}-2{{A}^{2}}-3A+I=0$ can be reduced to $5\left( A-4I \right)$.
Then,
(A) Both the statements are true
(B) Both the statements are false
(C) Statement-I is true, but Statement-II is false
(D) Statement-I is false, but Statement-II is true

Answer 1

Hint: We solve this question by first multiplying the given equation, ${{A}^{2}}-5A+7I=0$ with the matrix ${{A}^{-1}}$. Then using the property \[{{A}^{-1}}A=A{{A}^{-1}}=I\] and simplifying it we get the value of the inverse of A, that is ${{A}^{-1}}$. Next, we consider the polynomial given in statement-II and substitute the value of ${{A}^{2}}$ from the given equation, ${{A}^{2}}-5A+7I=0$. Then we simplify the equation and find the reduced form of the polynomial and find whether the given statements are correct or wrong and mark the answer accordingly.

Complete step-by-step solution:
We are given that A is a $3\times 3$ matrix such that ${{A}^{2}}-5A+7I=0$.
Now let us multiply it with ${{A}^{-1}}$. Then we can write it as,
\[\begin{align}
  & \Rightarrow {{A}^{-1}}\left( {{A}^{2}}-5A+7I \right)={{A}^{-1}}\left( 0 \right) \\
 & \Rightarrow {{A}^{-1}}{{A}^{2}}-5{{A}^{-1}}A+7{{A}^{-1}}I=0 \\
\end{align}\]
As we can write ${{A}^{2}}=A\times A$, we can write the above equation as,
\[\Rightarrow {{A}^{-1}}A\times A-5{{A}^{-1}}A+7{{A}^{-1}}I=0\]
Now let us consider the property of inverse of A. For any matrix A, we can write its inverse as, \[{{A}^{-1}}A=A{{A}^{-1}}=I\]
So, we can write the above equation as,
\[\begin{align}
  & \Rightarrow I\times A-5I+7{{A}^{-1}}=0 \\
 & \Rightarrow A-5I+7{{A}^{-1}}=0 \\
 & \Rightarrow 7{{A}^{-1}}=5I-A \\
 & \Rightarrow {{A}^{-1}}=\dfrac{1}{7}\left( 5I-A \right) \\
\end{align}\]
So, we get the inverse of A as, \[{{A}^{-1}}=\dfrac{1}{7}\left( 5I-A \right)\].
So, we get that Statement-I is true.
Now let us consider the polynomial given in Statement-II, that is ${{A}^{3}}-2{{A}^{2}}-3A+I=0$.
As we can write ${{A}^{3}}$ as $A\times {{A}^{2}}$, we can write the polynomial given as,
$\Rightarrow A\times {{A}^{2}}-2{{A}^{2}}-3A+I=0$
Now let us substitute the value of ${{A}^{2}}$ given, that is ${{A}^{2}}=5A-7I$. Then we get,
\[\begin{align}
  & \Rightarrow A\left( 5A-7I \right)-2\left( 5A-7I \right)-3A+I=0 \\
 & \Rightarrow 5{{A}^{2}}-7A-10A+14I-3A+I=0 \\
 & \Rightarrow 5{{A}^{2}}-20A+15I=0 \\
\end{align}\]
Now let us substitute the value of ${{A}^{2}}$ again. Then we get,
\[\begin{align}
  & \Rightarrow 5\left( 5A-7I \right)-20A+15I=0 \\
 & \Rightarrow 25A-35I-20A+15I=0 \\
 & \Rightarrow 5A-20I=0 \\
 & \Rightarrow 5\left( A-4I \right)=0 \\
\end{align}\]
So, we can say that the polynomial ${{A}^{3}}-2{{A}^{2}}-3A+I=0$ can be reduced to $5\left( A-4I \right)$.
So, we get that the Statement-II is true.
So, we get that both the equations are correct. Hence the answer is Option A.

Note: We can also simplify the polynomial given in statement-II by multiplying it with ${{A}^{-1}}$ and then simplifying it and then substituting the value of ${{A}^{-1}}$ obtained before in statement-I and then substituting the value of ${{A}^{2}}$ and then by simplifying it.