If A=[array|c|c| 1 & 5 & 10 array],A^-1= A+ I and + =-2, then 4 ^2+ ^2+ ^2 is equal to :

Question

If $A=[\begin{array}{|c|c|} 1 & 5 \ \lambda  & 10 \end{array}],{A}^{-1}=\alpha A+\beta I$ and $\alpha +\beta =-2$, then $4{\alpha }^{2}+{\beta }^{2}+{\lambda }^{2}$ is equal to :

Accepted Answer

$$
\begin{aligned}
& \mathrm{A}=\left[\[\begin{array}{cc} 1 & 5 \ \lambda & 10 \end{array}\end{aligned}$$\right] \\
\& \Rightarrow|\mathrm{A}-x \mathrm{I}|=0 \\
\& \Rightarrow\left|$$\begin{array}{cc} 1-x & 5 \ \lambda & 10-x \end{array}$$\right|=0 \\
\& \Rightarrow(1-x)(10-x)-5 \lambda=0 \\
\& \Rightarrow 10-11 x+x\^2-5 \lambda=0

\]

Also, $\Rightarrow \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$

$$
\Rightarrow \alpha A^2+\beta A-I=0
$$

and $A^2-11 A+(10-5 \lambda) I=0$

On solving, we get

$$
\alpha=\frac{1}{5}, \beta=-\frac{11}{5}
$$

$$
\begin{aligned}
& \text { So, } 5 \lambda-10=5 \Rightarrow \lambda=3 \\
& \therefore 4 \alpha^2+\beta^2+\lambda^2 \\
& =4\left(\frac{1}{25}\right)+\left(\frac{121}{25}\right)+9 \\
& =\frac{125}{25}+9=14
\end{aligned}
$$

Solutions · 1