If A=3-342-340-11, then A−1=

|A|=3(−3+4)+3(2−0)+4(−2)=3+6−8=1A−1=Adj Adet A=111−10−23−4−23−3A2=3-342-340-11 3-342-340-11=9−6+0−9+9−412−12+46−6+0−6+9−48−12+40−2+00+3−10−4+1=3-440-10-22-3A3=A2.A=3−440−10−22−3 3−342−340−11=9−8+0−9+12−412−16+40−2+00+3+00−4+0−6+4+06−6+3−8+8−3=1−10−23−4−23−3=A−1∴A−1=A3

If A=3-342-340-11, then A−1=

|A|=3(−3+4)+3(2−0)+4(−2)=3+6−8=1A−1=Adj Adet A=111−10−23−4−23−3A2=3-342-340-11 3-342-340-11=9−6+0−9+9−412−12+46−6+0−6+9−48−12+40−2+00+3−10−4+1=3-440-10-22-3A3=A2.A=3−440−10−22−3 3−342−340−11=9−8+0−9+12−412−16+40−2+00+3+00−4+0−6+4+06−6+3−8+8−3=1−10−23−4−23−3=A−1∴A−1=A3

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If $A = [\begin{matrix} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{matrix}]$ , then $A^{- 1} =$

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$- A$

$A^{3}$

$A$

$A^{2}$

answer is A.

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Detailed Solution

$| A | = 3 (- 3 + 4) + 3 (2 - 0) + 4 (- 2)$

$= 3 + 6 - 8$

$= 1$

$A^{- 1} = \frac{Adj A}{\det A} = \frac{1}{1} [\begin{matrix} 1 & - 1 & 0 \\ - 2 & 3 & - 4 \\ - 2 & 3 & - 3 \end{matrix}]$

$A^{2} = [\begin{matrix} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{matrix}] [\begin{matrix} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{matrix}]$

$= [\begin{matrix} 9 - 6 + 0 & - 9 + 9 - 4 & 12 - 12 + 4 \\ 6 - 6 + 0 & - 6 + 9 - 4 & 8 - 12 + 4 \\ 0 - 2 + 0 & 0 + 3 - 1 & 0 - 4 + 1 \end{matrix}]$

$= [\begin{matrix} 3 & - 4 & 4 \\ 0 & - 1 & 0 \\ - 2 & 2 & - 3 \end{matrix}]$

$A^{3} = A^{2} . A = [\begin{matrix} 3 & - 4 & 4 \\ 0 & - 1 & 0 \\ - 2 & 2 & - 3 \end{matrix}] [\begin{matrix} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{matrix}]$

$= [\begin{matrix} 9 - 8 + 0 & - 9 + 12 - 4 & 12 - 16 + 4 \\ 0 - 2 + 0 & 0 + 3 + 0 & 0 - 4 + 0 \\ - 6 + 4 + 0 & 6 - 6 + 3 & - 8 + 8 - 3 \end{matrix}]$