Matrices and Determinants 3 Question 10
12. If $A=$
$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & -2 & 4 \end{bmatrix}$, $6 A^{-1}=A^{2}+c A+d I$, then $(c, d)$ is
(2005, 1M)
(a) $(-6,11)$
(b) $(-11,6)$
(c) $(11,6)$
(d) $(6,11)$
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Answer:
Correct Answer: 12. (a)
Solution:
- Every square matrix satisfied its characteristic equation,
$ \begin{aligned} & \text { i.e. }|A-\lambda I|=0 \Rightarrow\left|\begin{array}{ccc} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 1 \\ 0 & -2 & 4-\lambda \end{array}\right|=0 \\ & \Rightarrow \quad(1-\lambda){(1-\lambda)(4-\lambda)+2}=0 \\ & \Rightarrow \quad \lambda^{3}-6 \lambda^{2}+11 \lambda-6=0 \\ & \Rightarrow \quad 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 \text { and } d=11 $
