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

A
-4
B
-2
C
2
D
4

Solution:

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)

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

Hence option (d) is correct

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

Solution:

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