Title: CJS-dumping
Course: Demonstrating STACK (v4.3.7)

Consider the set of vectors\[ \left\{ [value], [value], [value] \right\}.\]

Find the constants \(a_1\), \(a_2\) and \(a_3\) in the linear combination \[ a_1[value]+a_2[value]+a_3[value]\] which prove the set of vectors are linearly dependent

\(a_1=\)

\(a_2=\)

\(a_3=\)
Feedback: To solve this problem we are essentially trying to solve the equation \[ x[value]+y[value]+z[value] = [value].\] We can write this in matrix form \[ [value][value] = [value].\] Writing the reduced row echelon form gives \[[value]\] We expect to get a dependence of the columns, so this gives us \[ [value] = [value]\] but there are other options.

Use the Gauss-Jordan method to find the inverse of \[ [value] \] over \( \mathbb{Z}_{[value]}\).
Feedback: This question asked you to find the inverse using the Gauss-Jordan method. To use the Gauss-Jordan method set up the problem as \[ \left[ A \big{|} I \right] \] and then use row operations to reduce this to reduced row echelon form. The left hand side will be the identity matrix, and overall we will have \[ \left[ I \big{|} A^{-1} \right]. \] So, using our values we have as \[ [value]\] which has (working over \( \mathbb{Q}\)) reduced row echelon form (working omitted) \[ [value]\] from which we read out the inverse (working over \( \mathbb{Q}\)) as \[ [value].\] Now, in this case we are working over \( \mathbb{Z}_{[value]}\) and so we have to turn any negative numbers and fractions into the equivalent representations over \( \mathbb{Z}_{[value]}\). This gives \[ [value]\] We should, of course, check the answer: \[ [value][value] = [value] \equiv [value] \mbox{ mod } [value].\]

Find the characteristic polynomial of the matrix [value] in terms of the variable \(t\).

\(p(t) = \)

Given that \(t=[value]\) is an eigenvalue, find all the eigenvalues and corresponding eigenvectors of matrix [value].

\(\lambda_1=\)
\(\mathbf{v}_1 =\)

\(\lambda_2=\)
\(\mathbf{v}_2 =\)

\(\lambda_3=\)
\(\mathbf{v}_3 =\)
Feedback: To find the eignenvalues we solve \( \mbox{det}(M-t I)=0\). In our case that is \[ M-\lambda I = [value]\] which has determinant \[ [value].\] We are told \(t=[value]\) is a root allowing us to find the other eigenvalues as roots of \[ [value]=[value].\] Corresponding to each eigenvalue we have \[ t=[value],\ \mathbf{v}_1 = [value].\] \[ t=[value],\ \mathbf{v}_2 = [value].\] \[ t=[value],\ \mathbf{v}_3 = [value].\]

Give an example of a three by three matrix with real entries (i.e. in \(\mathbb{R}^{3\times 3}\)), with exactly one real eigenvalue which is non-zero, and also with two complex eigenvalues.  Repeated eigenvalues will be counted with multiplicity.
Feedback: To build a matrix, we can consider the block structure. To get a real eigen-value we can have a matrix of the form \[ \left[ \begin{array}{ccc} \lambda & 0 & 0 \\ 0 & * & * \\ 0 & * & * \end{array}\right]\] This ensures the characteristic equation has \(\lambda\) as one root. Then we ensure the eigenvalues of the lower-right \(2\times 2\) matrix have complex eigenvalues. A simple example is \[[value].\]

Given than \[ A=[value],\mbox{ and } B=[value]\] solve \[ [value]\] for \(X\)
Feedback: The first task is to solve the equation for \(X\) in the normal way, but avoiding dividing both sides by a matrix! \[ X = [value]\] Given than \[ A=[value],\mbox{ and } B=[value]\] we evaluate this to give \[ [value] = [value]. \]