91Ó°ÊÓ

Since we have given a matrix A such that$A=\left[\begin{array}{cc}3& 1\\ 3& 5\end{array}\right]$.

Then

$$\begin{gathered} A-\mathrm{λ}{l}_2=\left[\begin{array}{cc}3& 1\\ 3& 5\end{array}\right]-\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ ⇒A-\mathrm{λ}{l}_2=\left[\begin{array}{cc}3-\mathrm{λ}& 1\\ 3& 5-\mathrm{λ}\end{array}\right] \end{gathered}$$

It is given that$A-\mathrm{λ}{l}_2$is not invertible.

$$\begin{gathered} \mathrm{A}-{\mathrm{λ\pm ô}}_2=0 \\ \left[\begin{array}{cc}3-\mathrm{λ}& 1\\ 3& 5-\mathrm{λ}\end{array}\right]=0 \\ \left(3-\mathrm{λ}\right)\left(5-\mathrm{λ}\right)-3=0 \\ {\mathrm{λ}}^2-8\mathrm{λ}+12=0 \\ \mathrm{λ}=2\mathrm{or}\mathrm{λ}=6 \end{gathered}$$

From part (a) we have $\mathrm{λ}=6or\mathrm{λ}=2$. Let us choose$\mathrm{λ}=6$, then we have:

$$\begin{gathered} (\mathrm{A}-{\mathrm{λ\pm ô}}_2)\stackrel{→}{\mathrm{x}}=\stackrel{→}{0} \\ (\mathrm{A}-6{\mathrm{l}}_2)\stackrel{→}{\mathrm{x}}=\stackrel{→}{0} \\ \left[\begin{array}{cc}3& 1\\ 3& 5\end{array}\right]-\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ \left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=\left[\begin{array}{c}1\\ 3\end{array}\right] \\ \stackrel{→}{\mathrm{x}}=\left[\begin{array}{c}1\\ 3\end{array}\right] \end{gathered}$$

We have$\mathrm{λ}=6$from part (a) and$\stackrel{→}{x}=\left[\begin{array}{c}1\\ 3\end{array}\right]$from part (b).

Then$A\stackrel{→}{x}=\left[\begin{array}{cc}3& 1\\ 3& 5\end{array}\right]\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}3+3\\ 3+15\end{array}\right]=\left[\begin{array}{c}6\\ 18\end{array}\right]$

And

$\mathrm{λ}\stackrel{→}{x}=6\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}6\\ 18\end{array}\right]$

This shows that the equation$A\stackrel{→}{x}=A\stackrel{→}{x}$holds for our$\mathrm{λ}$from part (a) and our$\stackrel{→}{x}$from part (b).

1. The scalar$\mathrm{λ}$such that the matrix$A-\mathrm{λ}{l}_2$fails to be invertible is$\mathrm{λ}=6or\mathrm{λ}=2$.

2. For$\mathrm{λ}=6$we choose from part (a), we have find a non-zero vector$\stackrel{→}{x}=\left[\begin{array}{c}1\\ 3\end{array}\right]$such that $(A-\mathrm{λ}{l}_2)\stackrel{→}{x}=\stackrel{→}{0}$

3. The equation$A\stackrel{→}{x}=A\stackrel{→}{x}$holds for our$\mathrm{λ}=6$from part (a) and our$\stackrel{→}{x}=\left[\begin{array}{c}1\\ 3\end{array}\right]$from part (b).