91Ó°ÊÓ

Since we have given a matrix A such that$A=\left[\begin{array}{cc}1& 10\\ -3& 12\end{array}\right]$.

Then

$$\begin{gathered} A-\mathrm{λ}{l}_2=\left[\begin{array}{cc}1& 10\\ -3& 12\end{array}\right]-\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ ⇒A-\mathrm{λ}{l}_2=\left[\begin{array}{cc}1-\mathrm{λ}& 10\\ -3& 12-\mathrm{λ}\end{array}\right] \end{gathered}$$

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

$$\begin{gathered} \mathrm{A}-{\mathrm{λ\pm ô}}_2=0 \\ \left[\begin{array}{cc}1-\mathrm{λ}& 10\\ -3& 12-\mathrm{λ}\end{array}\right]=0 \\ \left(1-\mathrm{λ}\right)\left(12-\mathrm{λ}\right)+30=0 \\ {\mathrm{λ}}^2-13\mathrm{λ}+42=0 \\ \mathrm{λ}=7\mathrm{or}\mathrm{λ}=6 \end{gathered}$$

$$\begin{gathered} \mathrm{From}\mathrm{part}\left(\mathrm{a}\right)\mathrm{we}\mathrm{have}\mathrm{λ}=6\mathrm{or}\mathrm{λ}=7.\mathrm{Let}\mathrm{us}\mathrm{choose}\mathrm{λ}=6,\mathrm{then}\mathrm{we}\mathrm{have} \\ \left(\mathrm{A}-{\mathrm{λ\pm ô}}_2\right)\stackrel{â‡\P Ä}{\mathrm{x}}=\stackrel{â‡\P Ä}{0} \\ \left(\mathrm{A}-6{\mathrm{l}}_2\right)\stackrel{â‡\P Ä}{\mathrm{x}}=\stackrel{â‡\P Ä}{0} \\ \left[\begin{array}{cc}1& 10\\ -3& 12\end{array}\right]-\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ \left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right] \\ \stackrel{â‡\P Ä}{\mathrm{x}}=\left[\begin{array}{c}2\\ 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{We}\mathrm{have}\mathrm{λ}=6\mathrm{from}\mathrm{part}\left(\mathrm{a}\right)\mathrm{and}\stackrel{â‡\P Ä}{\mathrm{x}}=\left[\begin{array}{c}2\\ 1\end{array}\right]\mathrm{from}\mathrm{part}\left(\mathrm{b}\right). \\ \mathrm{Then}\mathrm{A}\stackrel{â‡\P Ä}{\mathrm{x}}=\left[\begin{array}{cc}1& 10\\ -3& 12\end{array}\right]\left[\begin{array}{c}2\\ 1\end{array}\right]=\left[\begin{array}{c}2+10\\ -6+12\end{array}\right]=\left[\begin{array}{c}12\\ 6\end{array}\right] \\ \mathrm{And} \\ \mathrm{λ}\stackrel{â‡\P Ä}{\mathrm{x}}=6\left[\begin{array}{c}2\\ 1\end{array}\right]=\left[\begin{array}{c}12\\ 6\end{array}\right] \\ \mathrm{This}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{equation}\mathrm{A}\stackrel{â‡\P Ä}{\mathrm{x}}=\mathrm{λ}\stackrel{â‡\P Ä}{\mathrm{x}}\mathrm{holds}\mathrm{for}\mathrm{our}\mathrm{λ}\mathrm{from}\mathrm{part}\left(\mathrm{a}\right)\mathrm{and}\mathrm{our}\stackrel{â‡\P Ä}{\mathrm{x}}\mathrm{from}\mathrm{part}\left(\mathrm{b}\right). \end{gathered}$$

a. The scalar$\mathrm{λ}$ such that the matrix $A-\mathrm{λ}{l}_2$fails to be invertible is $\mathrm{λ}=6or\mathrm{λ}=7$.

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

c. The equation $A\stackrel{â‡\P Ä}{x}=\mathrm{λ}\stackrel{â‡\P Ä}{x}$holds for our $\mathrm{λ}=6$from part (a) and our $\stackrel{â‡\P Ä}{x}=\left[\begin{array}{c}2\\ 1\end{array}\right]$from part (b).