If $$A=1223$$ and $$A2-$$λ$$A-I2=0$$, then λ is equal to

we are $$givenA=1223And$$ $$A2-$$λ$$A-I2=0$$→(1) so $$A2=A$$.$$A=1223$$ $$1223$$ $$=56613From$$ $$eq(1)A2-$$λ$$A+I2=58813-$$λ$$1223-1001=0$$ ⇒$$5-$$λ$$-18-2$$λ$$-08-2$$λ$$-013-3$$λ$$-1$$ $$=0000$$ ⇒$$4-$$λ$$8-2$$λ$$8-2$$λ$$12-3$$λ$$=0000$$ ⇒$$4-$$λ$$=0$$ λ$$=4Hence$$ option (d) is correct

Questions

If $A = [\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}]$ and $A^{2} - λ A - I_{2} = 0$ , then $λ$ is equal to

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detailed solution

Correct option is D

we are given

$A = [\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}]$

And $A^{2} - λ A - I_{2} = 0 \to (1)$

so $A^{2} = A . A = [\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}] [\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}] = [\begin{matrix} 5 & 6 \\ 6 & 13 \end{matrix}]$

From eq(1)

$A^{2} - λ A + I_{2} = [\begin{matrix} 5 & 8 \\ 8 & 13 \end{matrix}] - λ [\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}] - [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 5 - λ - 1 & 8 - 2 λ - 0 \\ 8 - 2 λ - 0 & 13 - 3 λ - 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} 4 - λ & 8 - 2 λ \\ 8 - 2 λ & 12 - 3 λ \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow 4 - λ = 0 λ = 4$