SATHEE: Matrices and Determinants 3 Question 10

Matrices and Determinants 3 Question 10

12. If $A =$

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

(2005, 1M)

(a) $(- 6, 11)$

(b) $(- 11, 6)$

(d) $(6, 11)$

Show Answer

Answer:

Correct Answer: 12. (a)

Solution:

Every square matrix satisfied its characteristic equation,

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

Given, $6 A^{- 1} = A^{2} + c A + d I$ , multiplying both sides by $A$ , we get

$6 I = A^{3} + c A^{2} + d A \Rightarrow A^{3} + c A^{2} + d A - 6 I = O$

On comparing Eqs. (i) and (ii), we get

$c = - 6 and d = 11$

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Matrices and Determinants 3 Question 10

12. If A=

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12. If $A =$