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Q.

If $A = [\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}]$ then find the value of $(A - 2 I) (A - 3 I)$

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a

0

b

1

c

2

d

3

answer is A.

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Detailed Solution

Given,

A = [\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}]

We know that,

I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

Here,

\begin{matrix} A - 2 I = [\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}] - 2 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ = [\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}] - [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}] \\ = [\begin{matrix} 4 - 2 & 2 - 0 \\ - 1 - 0 & 1 - 2 \end{matrix}] = [\begin{matrix} 2 & 2 \\ - 1 & - 1 \end{matrix}] \dots . . (1) \end{matrix}

Similarly,

\begin{matrix} A - 3 I = [\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}] - 3 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ = [\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}] - [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}] \\ = [\begin{matrix} 4 + 3 & 2 - 0 \\ - 1 - 0 & 1 - 3 \end{matrix}] = [\begin{matrix} 1 & 2 \\ - 1 & - 2 \end{matrix}] . \dots . . (2) \end{matrix}

Now, add equation (1) and (2), to get,

\begin{matrix} (A - 2 I) (A - 3 I) = [\begin{matrix} 2 & 2 \\ - 1 & - 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ - 1 & - 2 \end{matrix}] \\ = [\begin{matrix} 2 \times 1 + 2 \times (- 1) & 2 \times 2 + 2 \times (- 2) \\ - 1 \times 1 + (- 1) \times (- 1) & - 1 \times 2 + (- 1) \times (- 2) \end{matrix}] \\ = [\begin{matrix} 2 - 2 & 4 - 4 \\ - 1 + 1 & - 2 + 2 \end{matrix}] \\ = [\begin{matrix} 0 0 \\ 0 0 \end{matrix}] \end{matrix}

= 0

Thus,

(A - 2 I) (A - 3 I) = 0

Option 1 is Correct.

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