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Question 1: Parameter cases (2 × 2) 3 PTS
1
Write the matrix form of $\;x_1 + ax_2 = 4,\; ax_1 + 9x_2 = b$.
1 pt
Worked solution
$$\begin{bmatrix}1 & a \\ a & 9\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}4 \\ b\end{bmatrix}$$
Answer: $A\mathbf x = \mathbf b$ with $A = \begin{bmatrix}1 & a \\ a & 9\end{bmatrix}$
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2
For which values of $a$ is the solution unique?
1 pt
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$-a\cdot\text{Eq}_1 + \text{Eq}_2$: $(9 - a^2)x_2 = b - 4a$.
Unique when $9 - a^2 \neq 0$.
Answer: $a \neq \pm 3$
3
Find the pairs $(a, b)$ giving infinitely many solutions.
1 pt
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Worked solution
Need $9 - a^2 = 0$ and $b - 4a = 0$ together: $b = 4a$.
Answer: $(3, 12)$ and $(-3, -12)$
Question 2: Parameter cases (3 × 3) 2 PTS
1
For which $a$ is the solution unique?$$\begin{cases}x_1 + 2x_2 + x_3 = 3 \\ ax_2 + 5x_3 = 10 \\ 2x_1 + 7x_2 + ax_3 = b\end{cases}$$
1 pt
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Worked solution
$R_3 \leftarrow R_3 - 2R_1$: $(0, 3, a - 2 \mid b - 6)$. Then $R_3 \leftarrow aR_3 - 3R_2$:
$$\left[\begin{array}{ccc|c}1 & 2 & 1 & 3 \\ 0 & a & 5 & 10 \\ 0 & 0 & a^2-2a-15 & ab-6a-30\end{array}\right]$$
$a^2 - 2a - 15 = (a - 5)(a + 3)$. (At $a = 0$ the scaling step is not allowed, but then the matrix still has three pivots.)
Answer: Unique for $a \neq -3$ and $a \neq 5$
2
Find the pairs $(a, b)$ giving infinitely many solutions.
1 pt
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Worked solution
Need $a^2 - 2a - 15 = 0$ and $ab - 6a - 30 = 0$.
$a = 5$: $5b - 30 - 30 = 0 \Rightarrow b = 12$.
$a = -3$: $-3b + 18 - 30 = 0 \Rightarrow -3b = 12 \Rightarrow b = -4$.
Answer: $(5, 12)$ and $(-3, -4)$
Note on the original key: The key writes b = 4 for a = −3 and then lists the pair as (3, −4). Both are slips: the pair is (−3, −4).
Question 3: Inconsistent system 1 PT
1
Solve $$\begin{bmatrix}1 & 2 & -3 \\ 2 & 4 & -2 \\ 3 & 6 & -4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0 \\ 2 \\ 3\end{bmatrix}$$
1 pt
Worked solution
$R_2 - 2R_1$: $(0, 0, 4 \mid 2)$. $R_3 - 3R_1$: $(0, 0, 5 \mid 3)$. $4R_3 - 5R_2$: $(0, 0, 0 \mid 2)$.
That row reads $0 = 2$.
Answer: No solution.
Question 4: Parametric solution 1 PT
1
Solve the system.$$\begin{cases}x_1 + 2x_2 - 3x_3 - 2x_4 + 4x_5 = 1 \\ 2x_1 + 5x_2 - 8x_3 - x_4 + 6x_5 = 4 \\ x_1 + 4x_2 - 7x_3 + 5x_4 + 2x_5 = 8\end{cases}$$
1 pt
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Worked solution
RREF: $$\left[\begin{array}{ccccc|c}1 & 0 & 1 & 0 & 24 & 21 \\ 0 & 1 & -2 & 0 & -8 & -7 \\ 0 & 0 & 0 & 1 & 2 & 3\end{array}\right]$$
Basic $x_1, x_2, x_4$; free $x_3, x_5$: $x_1 = 21 - x_3 - 24x_5$, $x_2 = -7 + 2x_3 + 8x_5$, $x_4 = 3 - 2x_5$.
$$\mathbf x = \begin{bmatrix}21 \\ -7 \\ 0 \\ 3 \\ 0\end{bmatrix} + x_3\begin{bmatrix}-1 \\ 2 \\ 1 \\ 0 \\ 0\end{bmatrix} + x_5\begin{bmatrix}-24 \\ 8 \\ 0 \\ -2 \\ 1\end{bmatrix}$$
Answer: As above, with two free variables.
Note on the original key: The key's final vector drops a row and prints −1 + 2a + 8b for x₂; from its own RREF the constant is −7.
Question 5: Determinant by row reduction 1 PT
1
Find $\det\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$.
1 pt
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Worked solution
$R_2 - 4R_1$, $R_3 - 7R_1$, then $R_3 - 2R_2$ gives a zero row.
Answer: 0
Question 6: 2 × 2 determinant 1 PT
1
Find $\det\begin{bmatrix}a & b \\ c & d\end{bmatrix}$.
1 pt
Worked solution
Main diagonal product minus the other diagonal product.
Answer: $ad - bc$
Note on the original key: The key prints a·b − c·d, which is a typo: the determinant is ad − bc.
Question 7: Determinants with a parameter 2 PTS
a
Find $a$ such that $\det\begin{bmatrix}1 & 2 \\ 3 & a\end{bmatrix} \neq 0$.
1 pt
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Worked solution
$\det = a - 6$.
Answer: $a \neq 6$
b
Write $\det\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$ by cofactor expansion along row 1.
1 pt
Worked solution
$\det = a(ei - fh) - b(di - fg) + c(dh - eg)$.
Answer: $a(ei - fh) - b(di - fg) + c(dh - eg)$
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Question 8: Expansion around a chosen pivot 1 PT
1
Find $\det A$ using the pivot at row 2, column 3, $A = \begin{bmatrix}5 & 4 & 2 & 1 \\ 2 & 3 & 1 & -2 \\ -5 & -7 & -3 & 9 \\ 1 & -2 & -1 & 4\end{bmatrix}$.
1 pt
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Worked solution
$R_1 - 2R_2$, $R_3 + 3R_2$, $R_4 + R_2$ clear column 3; expand along it.
$\det A = (-1)^{5}\begin{vmatrix}1 & -2 & 5 \\ 1 & 2 & 3 \\ 3 & 1 & 2\end{vmatrix} = -(-38) = 38$.
Answer: 38
Question 9: Determinant after row operations 1 PT
1
$A = \begin{bmatrix}1 & -2 & 0 & 5 \\ 2 & 3 & 1 & -2 \\ -5 & -7 & -3 & 9 \\ 1 & -2 & -1 & 4\end{bmatrix}$ has $\det A = 38$. Matrix $B$ is obtained by: swap $R_1, R_3$; $R_2 \leftarrow R_2 + 2R_3$; $R_3 \leftarrow -3R_3$; $R_4 \leftarrow -2R_4$, giving$$B = \begin{bmatrix}-5 & -7 & -3 & 9 \\ 4 & -1 & 1 & 8 \\ -3 & 6 & 0 & -15 \\ -2 & 4 & 2 & -8\end{bmatrix}$$Find $\det B$.
1 pt
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Worked solution
Swap: $-38$. Replacement: still $-38$. Scale by $-3$: $114$. Scale by $-2$: $-228$.
Answer: $\det B = -228$
Note on the original key: On the source sheet, the B printed under "Find the determinant of matrix B" has rows (0, −1, 1, 1) and (3, 6, 0, −15), which do not match its own row operations. This version uses the B those operations produce.
Question 10: 2 × 2 inverse 1 PT
1
Find the inverse of $\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$.
1 pt
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Worked solution
Row reduce $[A \mid I]$: $R_2 - 3R_1$, $R_1 + R_2$, $R_2 / (-2)$.
Or use $\tfrac{1}{\det A}\begin{bmatrix}4 & -2 \\ -3 & 1\end{bmatrix}$ with $\det A = -2$.
Answer: $\begin{bmatrix}-2 & 1 \\ \tfrac32 & -\tfrac12\end{bmatrix}$
Question 11: Singular 3 × 3 1 PT
1
Find the inverse of $\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$.
1 pt
Worked solution
Row reducing $[A \mid I]$ produces a zero row on the left, so $I$ cannot be reached.
Answer: No inverse.
Question 12: Dependent columns 1 PT
1
Find the inverse of $\begin{bmatrix}4 & 2 & -4 & -2 & -6 \\ 2 & 1 & -2 & -1 & -3 \\ -4 & -2 & 4 & 2 & 6 \\ -2 & -1 & 2 & 1 & 3 \\ -6 & -3 & 6 & 3 & 9\end{bmatrix}$.
1 pt
Worked solution
Every row is a multiple of (2, 1, −2, −1, −3): rank 1, so the columns are dependent.
Answer: No inverse.
Question 13: Invertibility independent of a 1 PT
1
For which $a$ is $\begin{bmatrix}-1 & -3 & 1 \\ 1 & a & 2 \\ -2 & 0 & 2\end{bmatrix}$ invertible?
1 pt
Worked solution
$\det = -1(2a) + 3(2 + 4) + 1(0 + 2a) = 12$, whatever $a$ is.
Answer: Every real $a$.
Question 14: Invertibility with a parameter 1 PT
1
For which $a$ is $\begin{bmatrix}a & -3 & 1 \\ 1 & -1 & 2 \\ -2 & 0 & 2\end{bmatrix}$ invertible?
1 pt
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Worked solution
Expand along row 1: $a(-2 - 0) + 3(2 + 4) + 1(0 - 2) = -2a + 18 - 2 = 16 - 2a$.
Answer: $a \neq 8$
Note on the original key: The key gets −2a + 14 and a ≠ 7: it uses +2 for the last cofactor, but 1·0 − (−1)(−2) = −2.
Question 15: Null space membership 3 PTS
1
Write $x_1 - 3x_2 - 2x_3 = 0,\; -5x_1 + 9x_2 + x_3 = 0$ in matrix form.
1 pt
Worked solution
$$\begin{bmatrix}1 & -3 & -2 \\ -5 & 9 & 1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$$
Answer: $A\mathbf x = \mathbf 0$ with $A = \begin{bmatrix}1 & -3 & -2 \\ -5 & 9 & 1\end{bmatrix}$
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2
Is $\mathbf u = (5, 3, -2)$ in Nul $A$?
1 pt
Worked solution
$A\mathbf u = (5 - 9 + 4,\ -25 + 27 - 2) = (0, 0)$.
Answer: Yes.
3
If $\mathbf u, \mathbf w \in \operatorname{Nul}A$, show $\mathbf u + \mathbf w \in \operatorname{Nul}A$.
1 pt
Worked solution
$A(\mathbf u + \mathbf w) = A\mathbf u + A\mathbf w = \mathbf 0 + \mathbf 0 = \mathbf 0$.
Answer: Closed under addition.
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Question 16: Null space and column space 2 PTS
1
Find a spanning set for Nul $A$, $A = \begin{bmatrix}3 & 6 & -1 & 1 & -7 \\ 1 & -2 & 2 & 3 & -1 \\ 2 & -4 & 5 & 8 & -4\end{bmatrix}$, and the nullity.
1 pt
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Worked solution
RREF: $$\begin{bmatrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & \tfrac12 & -\tfrac32 \\ 0 & 0 & 1 & 2 & -2\end{bmatrix}$$
$x_1 = 0$, $x_2 = -\tfrac12x_4 + \tfrac32x_5$, $x_3 = -2x_4 + 2x_5$ with $x_4, x_5$ free:
$$\operatorname{Nul}A = \operatorname{Span}\left\{\begin{bmatrix}0 \\ -\tfrac12 \\ -2 \\ 1 \\ 0\end{bmatrix},\begin{bmatrix}0 \\ \tfrac32 \\ 2 \\ 0 \\ 1\end{bmatrix}\right\}$$
Answer: Two spanning vectors, nullity 2
2
Find a spanning set (basis) for Col $A$ and the rank.
1 pt
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Worked solution
Pivots in columns 1, 2, 3: take those columns of the original $A$.
$$\left\{\begin{bmatrix}3 \\ 1 \\ 2\end{bmatrix},\begin{bmatrix}6 \\ -2 \\ -4\end{bmatrix},\begin{bmatrix}-1 \\ 2 \\ 5\end{bmatrix}\right\}$$
rank $= 5 - 2 = 3$.
Answer: rank 3
Question 17: Row space 1 PT
1
Find a spanning set for Row $A$, $A = \begin{bmatrix}1 & 1 & 0 \\ 2 & 3 & -2 \\ -1 & -4 & 6\end{bmatrix}$.
1 pt
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Worked solution
$R_2 - 2R_1$, $R_3 + R_1$, $R_1 - R_2$, $R_3 + 3R_2$ give $$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}$$
Answer: $\operatorname{Row}A = \operatorname{Span}\{(1, 0, 2), (0, 1, -2)\}$
Note on the original key: The key writes the last step as R3 = R3 + 3R3; the operation that produces its matrix is R3 = R3 + 3R2.
Question 18: Left null space 1 PT
1
Find a spanning set for the left null space of $A = \begin{bmatrix}2 & -1 \\ -6 & 3\end{bmatrix}$.
1 pt
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Worked solution
Solve $A^T\mathbf x = \mathbf 0$: $A^T = \begin{bmatrix}2 & -6 \\ -1 & 3\end{bmatrix}$ reduces to $x_1 = 3x_2$.
Answer: $\operatorname{Span}\{(3, 1)\}$
Question 19: Left null space and Aᵀ 1 PT
1
Show that the left null space of $A$ is the null space of $A^T$.
1 pt
Worked solution
The left null space is $\{\mathbf x : \mathbf x^TA = \mathbf 0^T\}$.
Transpose both sides: $(\mathbf x^TA)^T = A^T\mathbf x = \mathbf 0$.
So $\mathbf x$ is in the left null space exactly when $\mathbf x \in \operatorname{Nul}A^T$.
Answer: $\operatorname{LeftNul}A = \operatorname{Nul}A^T$
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Question 20: Trivial left null space 1 PT
1
Find the left null space of $A = \begin{bmatrix}1 & -2 & -2 \\ 2 & 1 & 3 \\ -1 & 3 & -3\end{bmatrix}$.
1 pt
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Worked solution
$\det A = -32 \neq 0$, so $A^T$ has a pivot in every column and $A^T\mathbf x = \mathbf 0$ forces $\mathbf x = \mathbf 0$.
Answer: $\{\mathbf 0\}$
Question 21: Eigenvalues (2 × 2) 1 PT
1
Find the eigenvalues and eigenvectors of $A = \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$.
1 pt
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Worked solution
$\det(A - \lambda I) = \lambda^2 - 5\lambda - 2 = 0 \Rightarrow \lambda = \tfrac{5 \pm \sqrt{33}}{2}$.
For $\lambda_2 = \tfrac{5 + \sqrt{33}}{2}$: $3v_1 + (4 - \lambda_2)v_2 = 0 \Rightarrow v_1 = \tfrac{\sqrt{33} - 3}{6}v_2$.
For $\lambda_1$: $\mathbf v_1 = \left(\tfrac{-\sqrt{33} - 3}{6},\ 1\right)$.
Answer: $\lambda = \tfrac{5 \pm \sqrt{33}}{2}$
Question 22: Eigenvalues (3 × 3, numerical) 2 PTS
1
Find the eigenvalues of $A = \begin{bmatrix}-1 & 2 & 1 \\ 2 & 1 & -1 \\ 1 & -1 & -2\end{bmatrix}$ (3 decimals).
1 pt
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Worked solution
$\det(A - \lambda I) = -\lambda^3 - 2\lambda^2 + 7\lambda + 6 = 0$ has three real roots (no rational ones), found numerically.
Answer: $\lambda \approx -3.508,\ -0.756,\ 2.264$
2
Find the eigenvector for the largest eigenvalue.
1 pt
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Worked solution
Solve $(A - 2.264I)\mathbf v = \mathbf 0$ with $v_3 = 1$.
Answer: $\mathbf v_3 \approx (-5.957,\ -10.221,\ 1)$
Question 23: Gram-Schmidt 1 PT
1
Apply Gram-Schmidt to the columns of $A = \begin{bmatrix}1 & -2 & 1 \\ 2 & 0 & 1 \\ 3 & -2 & 3\end{bmatrix}$.
1 pt
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Worked solution
$\mathbf u_1 = (1, 2, 3)$, $\|\mathbf u_1\|^2 = 14$.
$\mathbf u_2 = \mathbf c_2 - \tfrac{-8}{14}\mathbf u_1 = \left(-\tfrac{10}{7}, \tfrac87, -\tfrac27\right)$, $\|\mathbf u_2\|^2 = \tfrac{24}{7}$.
$\mathbf u_3 = \mathbf c_3 - \tfrac{12}{14}\mathbf u_1 - \tfrac{-8/7}{24/7}\mathbf u_2 = \left(-\tfrac13, -\tfrac13, \tfrac13\right)$.
Check: $\mathbf u_1\cdot\mathbf u_2 = \mathbf u_1\cdot\mathbf u_3 = \mathbf u_2\cdot\mathbf u_3 = 0$.
Answer: Orthogonal basis $\mathbf u_1, \mathbf u_2, \mathbf u_3$
Question 24: QR decomposition 1 PT
1
Find the QR decomposition of the same $A = \begin{bmatrix}1 & -2 & 1 \\ 2 & 0 & 1 \\ 3 & -2 & 3\end{bmatrix}$.
1 pt
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Worked solution
$Q$ has the normalized $\mathbf u_i$ as columns; $R = Q^TA$, upper triangular.
$$R = \begin{bmatrix}\sqrt{14} & -\tfrac{4\sqrt{14}}{7} & \tfrac{6\sqrt{14}}{7} \\ 0 & \tfrac{2\sqrt{42}}{7} & -\tfrac{2\sqrt{42}}{21} \\ 0 & 0 & \tfrac{\sqrt3}{3}\end{bmatrix}$$
Answer: $A = QR$
Question 25: LU decomposition 1 PT
1
Find the LU decomposition of $A = \begin{bmatrix}1 & -2 & 1 \\ 2 & 0 & 1 \\ 3 & -2 & 3\end{bmatrix}$.
1 pt
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Worked solution
$R_2 - 2R_1$, $R_3 - 3R_1$, then $R_3 - R_2$. The multipliers fill $L$ below the diagonal.
$$L = \begin{bmatrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & 1\end{bmatrix},\quad U = \begin{bmatrix}1 & -2 & 1 \\ 0 & 4 & -1 \\ 0 & 0 & 1\end{bmatrix}$$
Answer: Check: $LU = A$
Result:
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