Question

Let $\mathit{A}\mathbf{=}\left[\begin{array}{cc}3& 1\\ 3& 5\end{array}\right]$in all parts of this problem.

1. Find the scalar$\mathit{\lambda }$such that the matrix$\mathit{A}\mathbf{-}\mathit{\lambda }{\mathit{l}}_{\mathbf{2}}$fails to be invertible. There are two solutions; choose one and use it in parts (b) and (c).

2. For the$\mathit{\lambda }$you choose in part (a), find a non-zero vector$\overrightarrow{\mathbf{x}}$such thatrole="math" localid="1659859315863" $\mathbf{(}\mathit{A}\mathbf{-}\mathit{\lambda }{\mathit{l}}_{\mathbf{2}}\mathbf{)}\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{0}}$

3. Note that the equation $\mathbf{(}\mathit{A}\mathbf{-}\mathit{\lambda }{\mathit{l}}_{\mathbf{2}}\mathbf{)}\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{0}}$can be written as$\mathbf{A}\overrightarrow{\mathbf{x}}\mathbf{-}\mathbf{\lambda }\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{0}}\mathbf{}\mathbf{or}\mathbf{}\mathbf{A}\overrightarrow{\mathbf{x}}\mathbf{=}\mathbf{\lambda }\overrightarrow{\mathbf{x}}$. Check that the equation $\mathbf{A}\overrightarrow{\mathbf{x}}\mathbf{-}\mathbf{\lambda }\overrightarrow{\mathbf{x}}$holds for your $\mathbf{\lambda }$from part (a) and your $\overrightarrow{\mathbf{x}}$from part (b).

Short answer

1. The scalar $\lambda$such that the matrix $A-\lambda l_2$fails to be invertible is .

2. For$\mathrm{\lambda }=6\mathrm{or}\mathrm{\lambda }=7$we choose from part (a), we have find a non-zero vector$\overrightarrow{x}=\left[\begin{array}{c}1\\ 3\end{array}\right]$such thatlocalid="1659861307439" $(A-\lambda l_2)\overrightarrow{x}=\overrightarrow{0}$

3. The equation $A\overrightarrow{x}=\lambda \overrightarrow{x}$holds for our $\lambda =6$from part (a) and our$\overrightarrow{x}=\left[\begin{array}{c}2\\ 1\end{array}\right]$from part (b).

Step-by-step solution

Finding the value of λ

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

Then

$$\begin{gathered} A-\lambda l_2=\left[\begin{array}{cc}3& 1\\ 3& 5\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow A-\lambda l_2=\left[\begin{array}{cc}3-\lambda & 1\\ 3& 5-\lambda \end{array}\right] \end{gathered}$$

It is given that$A-\lambda l_2$is not invertible.

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

Finding a non-zero vector x→ such that (A-λl2)x→=0→

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

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

To check that the equation Ax→=Ax→ holds for our λ

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

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

And

$\lambda \overrightarrow{x}=6\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}6\\ 18\end{array}\right]$

This shows that the equation$A\overrightarrow{x}=A\overrightarrow{x}$holds for our$\lambda$from part (a) and our$\overrightarrow{x}$from part (b).

Final Answer

1. The scalar$\lambda$such that the matrix$A-\lambda l_2$fails to be invertible is$\lambda =6or\lambda =2$.

2. For$\lambda =6$we choose from part (a), we have find a non-zero vector$\overrightarrow{x}=\left[\begin{array}{c}1\\ 3\end{array}\right]$such that $(A-\lambda l_2)\overrightarrow{x}=\overrightarrow{0}$

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