Find the following matrix product, being sure to show your work:
\[\begin{pmatrix} 2 & 0 & -1 & 0 & 4 \\ 1 & 1 & 3 & 0 & 2 \\ -1 & 2 & 0 & -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ -1 & 1 \\ 0 & 3 \\ 2 & 4 \\ 1 & 0 \end{pmatrix}\]You are welcome to use a calculator or computer program to check your answer, but I strongly advise you to first compute the result by hand. Remember, you won’t have a calculator on our upcoming test.
\[\begin{pmatrix} 6 & -1 \\ 2 & 11 \\ -4 & -3 \end{pmatrix}\]
Express the following system of linear equations as a matrix equation \(A\vec x = b\). You do not have to solve the system.
\[\begin{aligned} 2a + 2b + 2c - d &= 0 \\ -2a + 5b + 2c + d&= 1 \\ 8a + b + 4c + 2d &= -1 \end{aligned}\]\[\begin{pmatrix} 2 & 2 & 2 & -1 \\ -2 & 5 & 2 & 1 \\ 8 & 1 & 4 & 2 \end{pmatrix} \vec x = \begin{pmatrix}0\\1\\-1\end{pmatrix}\]
This question has two parts.
Use Gauss-Jordan elimination to put the following matrix in reduced row echelon form.
\[\begin{pmatrix} 2 & 1 & 2 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{pmatrix}\]\[\begin{pmatrix} 1 & 0 & 0 & 0 & -1 & 1 \\ 0 & 1 & 0 & -1 & 0 & 2 \\ 0 & 0 & 1 & 1 & 1 & -2 \end{pmatrix}\]
Use your answer to part (a) to find the inverse \(A^{-1}\) of the matrix
\[A = \begin{pmatrix} 2 & 1 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}\]Part (a) has you reduce a matrix of the form \((A|I)\). So the RREF matrix you found looks like \((I|A^{-1})\). Hence to find the answer here we just copy the right hand side to get,
\[A^{-1} = \begin{pmatrix} 0 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & 1 & -2 \end{pmatrix}\]
Find all values of \(t\) that satisfy the equation,
\[\begin{pmatrix}2&2&t\end{pmatrix} \begin{pmatrix} 1&2&0\\ 2&0&3\\ 0&3&1 \end{pmatrix} \begin{pmatrix}2\\2\\t\end{pmatrix} = 0\]This one might seem a little scary, but part 1 is just matrix multiplication. We have here three matrices being multiplied together, like \(ABC\). But by associativity we know that \(ABC = (AB)C = A(BC)\) so we can pick whichever pair we want first to multiply, as long as the order doesn’t change. We’ll just pick the right two here, but it doesn’t matter. We get,
\[\begin{pmatrix}2&2&t\end{pmatrix} \begin{pmatrix} 6 \\ 3t+4 \\ t+6 \end{pmatrix} = 0\]Then we multiply the two we have left, to get…
\[\begin{pmatrix}t^2 + 12t + 20\end{pmatrix} = 0\]This is just a quadratic equation! So, we use the quadratic formula to find,
\[t = \frac{-12 \pm \sqrt{144 - 80}}{2} = -10, -2.\]
Use the definition of a linear function to explain whether the following functions are linear or not. Answers that don’t use the definition cannot receive full credit.
\(f\left(\begin{pmatrix}x\\y\end{pmatrix}\right) = \begin{pmatrix}2x + y \\ y\end{pmatrix}\)
This one looks linear (it is!), so we have to show that. We do that by picking two arbitrary vectors (variables! It’s not enough to just pick numbers for this one) and check our two properties.
Addition.
\[\begin{aligned} f\begin{pmatrix}x\\y\end{pmatrix} + f\begin{pmatrix}a\\b\end{pmatrix} &= \begin{pmatrix}2x + y\\y\end{pmatrix} + \begin{pmatrix}2a + b\\b\end{pmatrix} \\ &= \begin{pmatrix}2x + y + 2a + b\\y + b\end{pmatrix} \\ &= \begin{pmatrix}2(x+a) + (y+b)\\y + b\end{pmatrix} \\ &= f\begin{pmatrix}x+a\\y+b\end{pmatrix} \end{aligned}\]Scalar multiplication.
\[f\left(\lambda \begin{pmatrix}x \\ y\end{pmatrix}\right) = f\begin{pmatrix}\lambda x \\ \lambda y\end{pmatrix} = \begin{pmatrix} 2\lambda x + \lambda y \\ \lambda y\end{pmatrix} = \lambda \begin{pmatrix}2x + y \\ y\end{pmatrix} = \lambda f \begin{pmatrix}x\\y\end{pmatrix}\]
\(f\left(\begin{pmatrix}x\\y\\z\end{pmatrix}\right) = \begin{pmatrix}x + y + z \\ y + \frac 2z\end{pmatrix}\)
This one doesn’t look linear (that \(2/z\) is suspicious), and it isn’t. To check that something isn’t linear, we can just show that one thing doesn’t work. This means it’s okay to pick some numbers. Since \(z\) is in that troubling term, we’ve got to make sure to pick some numbers where \(z\) shows up. We’ll see that scalar multiplication doesn’t work:
\[f\begin{pmatrix}0\\0\\2\end{pmatrix} = \begin{pmatrix}2\\1\end{pmatrix}\]On the other hand,
\[2f\begin{pmatrix}0\\0\\1\end{pmatrix} = 2\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}2\\4\end{pmatrix}\]These should be the same if \(f\) were linear, but they’re not. So \(f\) isn’t linear.