Question

Let $A^{-1}$ be a $3\times 3$ matrix such that $A^2-5A+7I=0.$
Statement-I :-$A^{-1}=\frac{1}{7}(5I-A).$
Statement-II :-The polynomial $A^3-2A^2-3A+I$ can be reduced to $5(A-4I)$.Then

• A. Both the statement are true.
• B. Statement-I is true, but Statement-II is false.
• C. Statement-I is false, but Statement-II is true.
• D. Both the statement are false.

Answer 1

The correct option is A Both the statement are true.

Multiply the equation with $A^{-1}$. Now we have

$A^{-1}{A}^2-5A^{-1}A+7I{A}^{-1}=0.A-5I+7A^{-1}=0$
$A^{-1}=\frac{1}{7}(5I-A).$
Now for second statement
The polynomial $A^3-2A^2-3A+I$ can be reduced to $5(A-4I)$
$A^3–2A^2–3A+I=A\left(A^2\right)–2A^2–3A+I=A(5A–7I)–2A^2–3A+I=3A^2–10A+I=3(5A–7I)–10A+I=5A–20I=5(A–4I)$
So both statements are true