Instructions: Be sure to put your name on each page of your exam. When in doubt, it is always better to provide more detail than less. And: No calculators or other computational aids allowed.
1. True or False: There exists a system of two linear equations in two unknowns with exactly 17 solutions. What about two equations and three unknowns?
Solution. False. Any system of linear equations either has no solution, a unique solution, or infinitely many solutions.
2. True or False: If \(A\) and \(B\) are invertible \(2\times 2\) matrices, then \(AB\) is invertible. Justify your answer.
Solution. True. \(\det(AB) = \det(A)\cdot\det(B)\not= 0\), since \(\det(A)\not= 0\) and \(\det(B)\not= 0\), so \(AB\) is invertible.
3. True or False: If we remove the line \(y = x\) from \(\mathbb{R}^2\), the remaining subset forms a subspace. Justify your answer.
Solution. False. Any subspace must contain the origin.
4. Define what it means for a matrix to be in reduced row echelon form.
Solution. A matrix is in RREF if:
- (i) The leading entry of each non-zero row is 1.
- (ii) There are zeros above and below each leading entry.
- (iii) The leading entries move from left to right as one moves down the rows.
- (iv) All rows consisting of zeros are at the bottom of the matrix.
5. Define what it means for two vectors \(v_1, v_2\in \mathbb{R}^2\) to span \(\mathbb{R}^2\).
Solution. The vectors \(v_1, v_2\) span \(\mathbb{R}^2\) if, given any vector \(w\in \mathbb{R}^2\), there exist \(\alpha, \beta\in \mathbb{R}\) such that \(w = \alpha v_1+\beta v_2\).
6. We're given a system of equations in \(x, y, z, w\) such that after elementary row reductions, the reduced row echelon form of the augmented matrix has the form
Write the solution set to the original system using set notation as done in class.
Solution. The solution set is \(\{(3-2t_1+7t_2,\ t_1,\ -7-5t_2,\ t_2)\ |\ t_1, t_2\in \mathbb{R}\}\).
1. Suppose the following sequence of elementary row operations was applied to the \(2\times 4\) augmented matrix used to find the inverse of a given \(2\times 2\) matrix: \(R_1\leftrightarrow R_2\); \(-2R_1+R_2\); \(3\cdot R_2\); \(-1\cdot R_2+R_1\). Convert these operations to elementary matrices and take an appropriate product of the elementary matrices to find the inverse of the given matrix.
Solution. Writing the elementary matrices in the order of the given elementary row operations, we have
\[E_1 = \begin{pmatrix} 0 & 1\\1 & 0\end{pmatrix} \quad E_2 = \begin{pmatrix} 1 & 0\\-2 & 1\end{pmatrix} \quad E_3 = \begin{pmatrix} 1 & 0\\0 & 3\end{pmatrix} \quad E_4 = \begin{pmatrix} 1 & -1\\0 & 1\end{pmatrix}.\]The inverse of the matrix is then
\[\begin{aligned} E_4E_3E_2E_1 &= \begin{pmatrix} 1 & -1\\0 & 1\end{pmatrix}\cdot\begin{pmatrix} 1 & 0\\0 & 3\end{pmatrix}\cdot\begin{pmatrix} 1 & 0\\-2 & 1\end{pmatrix}\cdot\begin{pmatrix} 0 & 1\\1 & 0\end{pmatrix}\\ &= \begin{pmatrix} 1 & -3\\0 & 3\end{pmatrix}\cdot\begin{pmatrix} 1 & 0\\-2 & 1\end{pmatrix}\cdot\begin{pmatrix} 0 & 1\\1 & 0\end{pmatrix}\\ &= \begin{pmatrix} 7 & -3\\-6 & 3\end{pmatrix}\cdot\begin{pmatrix} 0 & 1\\1 & 0\end{pmatrix}\\ &= \begin{pmatrix} -3 & 7\\3 & -6\end{pmatrix}. \end{aligned}\]2. Let \(T(x,y) = (3x-y, 2x)\) and \(S(x,y) = (-3y, 5x)\). Let \(E\) be the standard basis for \(\mathbb{R}^2\) and \(\beta\) the basis \(\{(1,1), (-1,0)\}\). Find the matrix of \(T\) with input basis \(\beta\) and output basis \(E\), the matrix of \(S\) with \(E\) as both input and output basis. Then verify the Very Important Formula regarding the matrix of a composition of linear transformations. Be sure to use the relevant notation from class.
Solution. We first calculate \(T\) evaluated on \(\beta\) and \(S\) evaluated on \(E\). \(T(1,1) = (2,2)\), \(T(-1,0) = (-3,-2)\) and \(S(1,0) = (0,5)\), \(S(0,1) = (-3,0)\). When \(E\) is the output basis, the corresponding columns in the matrix are just the values of the linear transformation on the input basis. For example, \(T(-1,0) = (-3,-2) = -3(1,0)+(-2)(0,1)\). Thus, \([T]_{\beta}^E = \begin{pmatrix} 2 & -3\\2 & -2\end{pmatrix}\) and \([S]_E^E = \begin{pmatrix} 0 & -3\\5 & 0\end{pmatrix}\).
By the Very Important Formula, \([ST]_{\beta}^E = [S]_E^E\cdot[T]_{\beta}^E\), so we first calculate \([ST]_{\beta}^E\) directly. We have \(ST(1,1) = S(2,2) = (-6,10)\) and \(ST(-1,0) = S(-3,-2) = (6,-15)\). Thus \([ST]_{\beta}^E = \begin{pmatrix} -6 & 6\\10 & -15\end{pmatrix}\).
To verify the Very Important Formula:
\[[S]_E^E\cdot[T]_{\beta}^E = \begin{pmatrix} 0 & -3\\5 & 0\end{pmatrix}\cdot\begin{pmatrix} 2 & -3\\2 & -2\end{pmatrix} = \begin{pmatrix} -6 & 6\\10 & -15\end{pmatrix} = [ST]_{\beta}^E.\]Given \(v_1 = (a,b)\), \(v_2 = (c,d)\in \mathbb{R}^2\) and \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\), prove that \(v_1, v_2\) are linearly independent if and only if \(\det(A)\not= 0\).
Solution. First suppose that \(v_1, v_2\) are linearly independent. Then we cannot have both \(b = 0\) and \(d = 0\), otherwise \(v_2\) would be a multiple of \(v_1\), contradicting linear independence. If \(\det(A)\) were zero, then \(ad-bc = 0\). This would give
\[dv_1+(-b)v_2 = d(a,b)-b(c,d) = (ad-bc,\ 0) = (0,0),\]which is a contradiction to linear independence, since at least one of \(d\) or \(b\) is non-zero. Therefore \(\det(A)\not= 0\).
Conversely, suppose \(\det(A)\not= 0\). Then \(A\) has an inverse. If we have a dependence relation \(\alpha v_1+\beta v_2 = \vec{0}\), then \(\alpha(a,b)+\beta(c,d) = (0,0)\), i.e.,
\[(0,0) = (\alpha,\beta)\begin{pmatrix} a & b\\c & d\end{pmatrix} = (\alpha,\beta)A.\]Multiplying both sides on the right by \(A^{-1}\) gives \((0,0) = (\alpha,\beta)\), so \(\alpha = 0 = \beta\), showing that \(v_1, v_2\) are linearly independent.
Given the linear transformation \(T(x,y) = (-2y, x+3y)\), show there is a non-zero vector \(v\in \mathbb{R}^2\) and \(\lambda\in \mathbb{R}\) such that \(T(v) = \lambda v\). Hint: You might try setting up a matrix equation. If done correctly you can actually find \(v\) and \(\lambda\), though this is not required.
Solution. We seek a non-zero vector \(v = (a,b)\) and \(\lambda\in \mathbb{R}\) such that \(T(v) = \lambda v\), i.e., \((-2b, a+3b) = (\lambda a, \lambda b)\). This gives two equations \(-2b = \lambda a\) and \(a+3b = \lambda b\). Writing this as a system of homogeneous equations:
\[\begin{aligned} -\lambda a - 2b &= 0\\ a + (3-\lambda)b &= 0. \end{aligned}\]Writing this as a matrix equation, we have \(\begin{pmatrix} -\lambda & -2\\1 & -\lambda+3\end{pmatrix}\cdot\begin{pmatrix} a\\b\end{pmatrix} = \begin{pmatrix} 0\\0\end{pmatrix}\). The system has a non-trivial solution exactly when \(\det\begin{pmatrix} -\lambda & -2\\1 & -\lambda+3\end{pmatrix} = 0\), i.e., when \(\lambda^2-3\lambda+2 = 0\). Thus, when \(\lambda = 1\) or \(\lambda = 2\), the system has a non-trivial solution, and in each case a non-trivial solution gives the vector we seek.
For example, when \(\lambda = 1\), the system becomes \(\begin{pmatrix} -1 & -2\\1 & 2\end{pmatrix}\cdot\begin{pmatrix} a\\b\end{pmatrix} = \begin{pmatrix} 0\\0\end{pmatrix}\), so the vector \(v = (-2,1)\) is a non-trivial solution, and thus satisfies \(T(v) = 1\cdot v\).
When \(\lambda = 2\), the system becomes \(\begin{pmatrix} -2 & -2\\1 & 1\end{pmatrix}\cdot\begin{pmatrix} a\\b\end{pmatrix} = \begin{pmatrix} 0\\0\end{pmatrix}\), so the vector \(v = (-1,1)\) is a non-trivial solution, and thus satisfies \(T(v) = 2\cdot v\).