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Q.

Let $A = (\begin{matrix} 6 & 2 & 2 \\ - 2 & - 3 & - 1 \\ 2 & - 1 & 3 \end{matrix})$ and $f (x) = |xI - A|,$ then $f (A)$ equals.

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a

0

b

A

c

2 A

d

- A

answer is A.

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Detailed Solution

Given,

A = (\begin{matrix} 6 & 2 & 2 \\ - 2 & - 3 & - 1 \\ 2 & - 1 & 3 \end{matrix})

f (x) = |xI - A|, . . . . . . (1)

Substitute the values in equation

(1)

we get,

\Rightarrow |xI - A| = |x (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) - (\begin{matrix} 6 & 2 & 2 \\ - 2 & - 3 & - 1 \\ 2 & - 1 & 3 \end{matrix})|

\Rightarrow |xI - A| = (\begin{matrix} x - 6 & - 2 & - 2 \\ 2 & x + 3 & 1 \\ - 2 & 1 & x - 3 \end{matrix})

Apply row operation

R_{3} \to R_{2} + R_{3},

\Rightarrow |xI - A| = (\begin{matrix} x - 6 & - 2 & - 2 \\ 2 & x + 3 & 1 \\ 0 & x + 4 & x - 2 \end{matrix})

\Rightarrow f (x) = (x - 6) [(x - 2) (x + 3) - 1 (x + 4)] - 2 [- 2 (x - 2) + 2 (x + 4)]

\Rightarrow f (x) = (x - 6) ((x^{2} + 3 x - 2 x - 6) - x - 4) - 2 (- 2 x + 4 + 2 x + 8)

\Rightarrow f (x) = (x - 6) (x^{2} - 10) - 2 (12)

\Rightarrow f (x) = x^{3} - 10 x - 6 x^{2} + 60 - 24

\Rightarrow f (x) = x^{3} - 10 x - 6 x^{2} + 36

\Rightarrow f (x) = x^{3} - 10 x - 6 x^{2} + 36

Replace A from x,

f (A) = A^{3} - 6 A^{2} - 10 A + 36 I

= (\begin{matrix} 240 & 44 & 116 \\ 68 & - 30 & - 34 \\ 140 & 14 & 78 \end{matrix}) - 6 (\begin{matrix} 36 & 4 & 16 \\ - 8 & 6 & - 4 \\ 20 & 4 & 14 \end{matrix}) - 10 (\begin{matrix} 6 & 2 & 2 \\ - 2 & - 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}) + 36 (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})

= (\begin{matrix} 240 & 44 & 116 \\ 68 & - 30 & - 34 \\ 140 & 14 & 78 \end{matrix}) + (\begin{matrix} - 216 & 44 & 116 \\ 48 & - 30 & - 34 \\ - 120 & 14 & 78 \end{matrix}) + (\begin{matrix} - 60 & - 20 & - 20 \\ 20 & 30 & 10 \\ - 20 & 10 & - 30 \end{matrix}) + (\begin{matrix} 36 & 0 & 0 \\ 0 & 36 & 0 \\ 0 & 0 & 36 \end{matrix})

= (\begin{matrix} 240 & 44 & 116 \\ 68 & - 30 & - 34 \\ 140 & 14 & 78 \end{matrix}) + (\begin{matrix} - 216 & 44 & 116 \\ 48 & - 30 & - 34 \\ - 120 & 14 & 78 \end{matrix}) + (\begin{matrix} - 60 + 36 & - 20 & - 20 \\ 20 & 30 + 36 & 10 \\ - 20 & 10 & - 30 + 36 \end{matrix})

= (\begin{matrix} 240 & 44 & 116 \\ 68 & - 30 & - 34 \\ 140 & 14 & 78 \end{matrix}) + (\begin{matrix} - 24 - 216 & - 20 - 44 & - 20 - 96 \\ 20 + 48 & 66 - 36 & 10 + 24 \\ - 20 - 120 & 10 - 24 & 6 - 84 \end{matrix})

= (\begin{matrix} 240 & 44 & 116 \\ 68 & - 30 & - 34 \\ 140 & 14 & 78 \end{matrix}) + (\begin{matrix} - 240 & - 44 & - 116 \\ - 68 & 30 & 34 \\ - 140 & - 14 & - 78 \end{matrix})

= (\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

= 0

Hence the correct option is 1.

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