A=1000110−24, I=100010001 and A−1=16A2+CA+DI . The values of C and D are

6A−1=A2+CA+DI⇒A3+CA2+DA−6I=0C = - trace of A = -6D = sum of minors of a11,a22,a33=11

A=1000110−24, I=100010001 and A−1=16A2+CA+DI . The values of C and D are

6A−1=A2+CA+DI⇒A3+CA2+DA−6I=0C = - trace of A = -6D = sum of minors of a11,a22,a33=11

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$A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{matrix}], I = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ and $A^{- 1} = [\frac{1}{6} (A^{2} + CA + DI)]$ . The values of C and D are

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(-6, 11)

(6, 11)

(-6, -11)

(6, -11)

answer is A.

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$6 A^{- 1} = A^{2} + CA + DI \Rightarrow A^{3} + {CA}^{2} + DA - 6 I = 0$

C = - trace of A = -6
D = sum of minors of $a_{11}, a_{22}, a_{33} = 11$

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A=1000110−24, I=100010001 and A−1=16A2+CA+DI . The values of C and D are

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$A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{matrix}], I = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ and $A^{- 1} = [\frac{1}{6} (A^{2} + CA + DI)]$ . The values of C and D are