So on my last exam on linear algebra there was this question:
Exam question
$\text{Let the matrix A be equal to:}$ \begin{pmatrix}1&-1&2&1\\2&1&1&1\\ -2&2&a&2\\3&-3&6&a\end{pmatrix}
$\text{1.Find the values of $\alpha$ so the matrix A is inversable}$
$\text{2.For the values of $\alpha$ that matrix A is inversable what should be the values of}$ $\text{eigenvalues of}$
$\text{matrix A so the matrix T is also inversable?}$ $\text{Consider T be: } T\equiv A^4-A^2-1$
My solution to the first part is this:
By apllying elemenetary row operations we have:
$\left\{\begin{array}{cc} row_2\Rightarrow row_2-2row_1\\row_3 \Rightarrow row_3+2row_1\\row_4 \Rightarrow row_4-3row_1\end{array}\right\}$.
Thus the matrix is equal to:\begin{pmatrix}1&-1&2&1\\0&3&-3&-1\\0&0& a+4&4\\0&0&0&a-3\end{pmatrix}
$\text{So the determinat is equal to}:det(\alpha)=3 \cdot (\alpha+4) \cdot (\alpha-3) \text{ (1)} $
$\text{And } det(\alpha)=0 \Rightarrow \alpha=-4 \text{ or } \alpha=3.$
$\text{So the final answer is the matrix is inversable for: } \boxed{\alpha \neq{-4,3} } $
My solution to the second part is this:
$\text{Let matrix C be equal to:}
$\begin{pmatrix}1&-1\\0&3\end{pmatrix}
$\text{Let matrix D be equal to:}
$\begin{pmatrix}a+4&4\\0&a-3\end{pmatrix}
Thus the matrix from the first question can be wriiten:
$\begin{pmatrix}C&*\\0&D\end{pmatrix}$
$\text{So:}$
$x_A$($\lambda)=x_C$($\lambda)\cdot x_D$($\lambda$) (2)
$x_C(\lambda)=det(C)=(\lambda-1)\cdot (\lambda-3)$
$x_D(\lambda)=det(D)=(\lambda -a-4)\cdot(\lambda-a+3)$
$\text{So the equation (2) can be wriiten as: }(\lambda-1)\cdot (\lambda-3)\cdot (\lambda -a-4)\cdot(\lambda-a+3)$
$\text{For the eigenvalues of matrix A we demand that equation (2) is equal to 0.}$
$\text{The solutions are:}$
$$\begin{cases}
\displaystyle \lambda=a+4 \\
\displaystyle \lambda=a+3\\
\displaystyle \lambda=1\\
\displaystyle \lambda=3
\end{cases}$$
$\text{So the eigenvalues of matrix A are: }
\boxed{\lambda \epsilon \left\{\begin{array}{cc} a+3,a+4 ,a\epsilon \mathbb{R}-\{-4,3\}\end{array}\right\}\cup \{ 1,3 \}}$
$\text{Now by solving the equation:}
\lambda^4-\lambda^2-1=0$
$\text{We found out that the only solution that we can accept is:}
\lambda=\pm \sqrt{\displaystyle\frac {1+\sqrt{5}} 2}$
$\text{Note:We are limited to search for eigenvalues in real numbers}$
Final answer
$\text{In order for T to be inversable: }$
$\boxed{\lambda \epsilon\left\{\begin{array}{cc}a+3,a+4 ,a\epsilon \mathbb{R}-\{-4,3\}\displaystyle\end{array}\right\}- \left\{\begin{array}{cc}\pm \sqrt{\displaystyle\frac {1+\sqrt{5}} 2}\end{array}\right\}} $
Is the second solution correct?Is there any easier way?i failed to give a full answer to this question so i was wondering if someone could give a check to my solution.
Thanks in advance
Best Answer
So after a long time i can verify that the solution on this question is correct.Thanks everyone for the help.