Question

If A is nonsingular matrix such that $(A-2I)(A-4I)=0$ then $A+8A^{-1}=$

• A. I
• B. 0
• C. 3I
• D. 6I

Answer 1

Answer: (D)

6I

Answer 2

The given equation is $(A-2I)(A-4I)=0$.
First, expand the product:
$A \cdot A - A \cdot 4I - 2I \cdot A + 2I \cdot 4I = 0$
$A^2 - 4A - 2A + 8I = 0$
$A^2 - 6A + 8I = 0$

We are given that A is a nonsingular matrix, which means its inverse $A^{-1}$ exists.
Multiply the entire equation by $A^{-1}$ from the right (or left):
$(A^2 - 6A + 8I)A^{-1} = 0 \cdot A^{-1}$
$A^2 A^{-1} - 6A A^{-1} + 8I A^{-1} = 0$

Using the properties of matrix multiplication ($A^2 A^{-1} = A$, $A A^{-1} = I$, and $I A^{-1} = A^{-1}$):
$A - 6I + 8A^{-1} = 0$

Now, we need to find the value of $A+8A^{-1}$.
Rearrange the equation obtained:
$A + 8A^{-1} = 6I$

Thus, $A+8A^{-1} = 6I$.