If the matrix A=02K−1 satisfies AA3+3I=2I, then the value of K is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$\frac{1}{2}$
B
$- \frac{1}{2}$
C
$- 1$
D
1

Solution:

Given matrix,

$A = [\begin{array}{cc} 0 & 2 \\ K & - 1 \end{array}]$

Characteristic equation of A is

$|A - λl| = 0 \begin{array}{l} \Rightarrow |\begin{array}{cc} - λ & 2 \\ K & - 1 - λ \end{array}| = 0 \\ \Rightarrow λ (1 + λ) - 2 k = 0 \Rightarrow λ^{2} + λ - 2 k = 0 \end{array}$

$∵$ Every square matrix satisfied its own characteristic equation.

$\begin{array}{l} ∴ A^{2} + A - 2 KI = 0 \\ \Rightarrow A^{2} = 2 KI - A \\ \Rightarrow A^{4} = 4 K^{2} I + A^{2} - 2 (2 KI) (A) \\ \Rightarrow A^{4} = 4 K^{2} I + 2 KI - A - 4 KA \\ \Rightarrow A^{4} = (4 K^{2} + 2 K) I - (I + 4 K) A . . . (i) \end{array}$

$\begin{array}{l} Now, A (A^{3} + 3 l) = 2 I \\ \Rightarrow A^{4} = 2 I - 3 A . . . (ii) \end{array}$

Comparing Eqs. (i) and (ii), we get

$1 + 4 K = 3 \Rightarrow K = \frac{1}{2}$

If the matrix $A = (\begin{array}{cc} 0 & 2 \\ K & - 1 \end{array})$ satisfies $A (A^{3} + 3 I) = 2 I$ , then the value of K is

Solution:

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