If A=1 0 00 1 10 −2 4,6A−1=A2+cA+dI, then (c,d)=

A
(- 6, 11)
B
(-11, 6)
C
(11, 6)
D
(6, 11)

Solution:

Every square matrix A satisfies its characteristic equation i.e. $| A - λ I | = 0$

Here, $| A - λ I | = 0$

$\begin{array}{l} \Rightarrow |\begin{array}{ccc} 1 - λ & 0 & 0 \\ 0 & 1 - λ & 1 \\ 0 & - 2 & 4 - λ \end{array}| = 0 \\ \Rightarrow (1 - λ) {(1 - λ) (4 - λ) + 2} = 0 \\ \Rightarrow λ^{3} - 6 λ^{2} + 11 λ - 6 = 0 \\ \Rightarrow A^{3} - 6 A^{2} + 11 A - 6 I = 0 \\ \Rightarrow 6 I = A^{3} - 6 A^{2} + 11 A \end{array}$

$\Rightarrow 6 A^{- 1} = A^{2} - 6 A + 11 I$ [Multiplying both sides by $A^{- 1}$ ]

$∴ 6 A^{- 1} = A^{2} + c A + d I \Rightarrow c = - 6$ and d=11

If $A = [\begin{array}{rrr} 1 0 0 \\ 0 1 1 \\ 0 - 2 4 \end{array}], 6 A^{- 1} = A^{2} + c A + d I$ , then $(c, d) =$

Solution:

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