Q.If $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, then find the value $\lambda$ and $\mu$ such that $A^2 + \lambda A + \mu I = 0$, where $0$ is zero matrix of order 2. OR Determine the value of $a$ for which the system is consistent. $x + y + z = 1$, $2x + 3y + 2z = 2$, $ax + ay + 2az = 4$.

Computing $A^2$ and matching it against $-\lambda A-\mu I$ entrywise gives $\lambda=-4,\ \mu=1$. (OR: the coefficient determinant of the 3-equation system equals $a$, so the system is consistent — with a unique solution — for every $a\ne0$, and inconsistent at $a=0$.)

Find $\lambda,\mu$ for $A^2+\lambda A+\mu I=0$, $A=\begin{bmatrix}2&3\\1&2\end{bmatrix}$.

$A^2 = \begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix} = \begin{bmatrix}2(2)+3(1) & 2(3)+3(2)\\1(2)+2(1)&1(3)+2(2)\end{bmatrix} = \begin{bmatrix}7&12\\4&7\end{bmatrix}.$

$A^2+\lambda A+\mu I = \begin{bmatrix}7+2\lambda+\mu & 12+3\lambda\\4+\lambda & 7+2\lambda+\mu\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}.$

From the $(1,2)$ entry: $12+3\lambda=0 \Rightarrow \lambda=-4$.

Check $(2,1)$ entry: $4+\lambda=0 \Rightarrow \lambda=-4$ ✓ (consistent).

From the $(1,1)$ entry: $7+2\lambda+\mu=0 \Rightarrow 7-8+\mu=0 \Rightarrow \mu=1$.

$\boxed{\lambda=-4,\ \mu=1}$

OR: Determine $a$ for which the system $x+y+z=1,\ 2x+3y+2z=2,\ ax+ay+2az=4$ is consistent.

The coefficient matrix is $\begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}$. Its determinant:

$\Delta = 1(3\cdot2a-2\cdot a) - 1(2\cdot2a-2\cdot a) + 1(2a-3a) = (6a-2a)-(4a-2a)+(-a) = 4a-2a-a = a.$

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