CS223 // Major Exam 2 // Term 251
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Major 2 (Term 251)
Determinants by cofactors, row reduction and Cramer's rule, then null space, column space, bases and rank.
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Question 1: Determinants 11 PTS
$$A = \begin{bmatrix}0 & 2 & 1 \\ 3 & 1 & 4 \\ 2 & 0 & 5\end{bmatrix}$$
a
Compute $\det A$ by cofactor expansion along the first row.
2 pts
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$\det A = 0\cdot C_{11} + 2\cdot C_{12} + 1\cdot C_{13}$.
$C_{12} = -\begin{vmatrix}3 & 4 \\ 2 & 5\end{vmatrix} = -(15 - 8) = -7$ and $C_{13} = \begin{vmatrix}3 & 1 \\ 2 & 0\end{vmatrix} = 0 - 2 = -2$.
$\det A = 2(-7) + 1(-2) = -16$.
Answer: $\det A = -16$
b
Compute $\det A$ by row reduction.
2 pts
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Swap $R_1 \leftrightarrow R_2$ (sign flips): $$\begin{bmatrix}3 & 1 & 4 \\ 0 & 2 & 1 \\ 2 & 0 & 5\end{bmatrix}$$
$R_3 \leftarrow R_3 - \tfrac23 R_1$: $(0, -\tfrac23, \tfrac73)$. Then $R_3 \leftarrow R_3 + \tfrac13 R_2$: $(0, 0, \tfrac83)$.
Triangular: product of the diagonal $= 3 \cdot 2 \cdot \tfrac83 = 16$. One swap, so $\det A = -16$.
Answer: $\det A = -16$
c
Using your determinant, is $A$ invertible?
1 pt
Worked solution
$\det A = -16 \neq 0$, and a square matrix is invertible exactly when its determinant is nonzero.
Answer: Yes, $A$ is invertible.
d
Using Cramer's rule and $\det A$, solve$$\begin{cases}2y + z = 0 \\ 3x + y + 4z = 0 \\ 2x + 5z = 1\end{cases}$$
3 pts
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Worked solution
The coefficient matrix is $A$ and $\mathbf b = (0, 0, 1)$. Replace one column of $A$ by $\mathbf b$ each time.
$\det A_1(\mathbf b) = \begin{vmatrix}0 & 2 & 1 \\ 0 & 1 & 4 \\ 1 & 0 & 5\end{vmatrix} = 1\cdot(8 - 1) = 7$
$\det A_2(\mathbf b) = \begin{vmatrix}0 & 0 & 1 \\ 3 & 0 & 4 \\ 2 & 1 & 5\end{vmatrix} = 1\cdot(3 - 0) = 3$
$\det A_3(\mathbf b) = \begin{vmatrix}0 & 2 & 0 \\ 3 & 1 & 0 \\ 2 & 0 & 1\end{vmatrix} = 1\cdot(0 - 6) = -6$
$x = \tfrac{7}{-16}$, $y = \tfrac{3}{-16}$, $z = \tfrac{-6}{-16}$.
Answer: $x = -\tfrac{7}{16},\ y = -\tfrac{3}{16},\ z = \tfrac38$
e-i
Using $\det A$, find $\begin{vmatrix}0 & 4 & 2 \\ 4 & 0 & 10 \\ 6 & 2 & 8\end{vmatrix}$. Justify.
1 pt
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Worked solution
Row 1 is $2\times$ row 1 of $A$, row 2 is $2\times$ row 3 of $A$, row 3 is $2\times$ row 2 of $A$.
Factor 2 out of each row: $2^3 = 8$. The rows are $A$'s with rows 2 and 3 swapped: factor $-1$.
$8 \cdot (-1) \cdot (-16) = 128$.
Answer: $128$
e-ii
Using $\det A$, find $\begin{vmatrix}0 & 2 & 1 \\ 1 & 1 & -1 \\ 2 & 0 & 5\end{vmatrix}$. Justify.
1 pt
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Row 2 is $(3, 1, 4) - (2, 0, 5)$: $A$ with $R_2 \leftarrow R_2 - R_3$.
Adding a multiple of one row to another does not change the determinant.
Answer: $-16$
e-iii
Using $\det A$, find $\det(-A^2)$. Justify.
1 pt
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Worked solution
$\det(-A^2) = (-1)^3\det(A^2) = -(\det A)^2$ for a $3\times 3$ matrix.
$-(-16)^2 = -256$.
Answer: $-256$
Question 2: Vector spaces 9 PTS
a
Let $A = \begin{bmatrix}-6 & 12 \\ -3 & 6\end{bmatrix}$ and $\mathbf w = \begin{bmatrix}2 \\ 1\end{bmatrix}$. Is $\mathbf w$ in $\operatorname{Col} A$? Is it in $\operatorname{Nul} A$?
2 pts
Worked solution
$A\mathbf w = (-12 + 12,\ -6 + 6) = (0, 0)$, so $\mathbf w \in \operatorname{Nul} A$.
$\operatorname{Col} A = \operatorname{Span}\{(-6, -3)\} = \operatorname{Span}\{(2, 1)\}$ because column 2 is $-2\times$ column 1.
$\mathbf w = -\tfrac13(-6, -3)$, so $\mathbf w \in \operatorname{Col} A$ as well.
Answer: $\mathbf w$ is in both spaces.
b
Let $\mathbf v_1 = \begin{bmatrix}7 \\ 4 \\ -9 \\ -5\end{bmatrix}$, $\mathbf v_2 = \begin{bmatrix}4 \\ -7 \\ 2 \\ 5\end{bmatrix}$, $\mathbf v_3 = \begin{bmatrix}1 \\ -5 \\ 3 \\ 4\end{bmatrix}$ with $\mathbf v_1 - 3\mathbf v_2 + 5\mathbf v_3 = \mathbf 0$. Find a basis for $H = \operatorname{Span}\{\mathbf v_1, \mathbf v_2, \mathbf v_3\}$.
2 pts
Worked solution
The relation gives $\mathbf v_3 = \tfrac15(3\mathbf v_2 - \mathbf v_1)$, so $\mathbf v_3$ adds nothing to the span.
$\mathbf v_1$ and $\mathbf v_2$ are not multiples of each other, so they are independent.
By the Spanning Set Theorem, the two remaining vectors form a basis (any two of the three would work).
Answer: Basis: $\{\mathbf v_1, \mathbf v_2\}$, so $\dim H = 2$.
c
Assume $A$ is row equivalent to $B$. Find bases for Nul $A$, Col $A$ and Row $A$, and find rank $A$, dim Nul $A$, dim Col $A$ and dim Row $A$.$$A = \begin{bmatrix}-2 & 4 & -2 & -4 \\ 2 & -6 & -3 & 1 \\ -3 & 8 & 2 & -3\end{bmatrix},\qquad B = \begin{bmatrix}1 & 0 & 6 & 5 \\ 0 & 2 & 5 & 3 \\ 0 & 0 & 0 & 0\end{bmatrix}$$
4 pts
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Worked solution
$B$ has pivots in columns 1 and 2, so rank $A = 2$ and $\dim\operatorname{Nul}A = 4 - 2 = 2$.
Nul $A$: from $B$, $x_1 = -6x_3 - 5x_4$ and $2x_2 = -5x_3 - 3x_4$. Basis: $$\left\{\begin{bmatrix}-6 \\ -\tfrac52 \\ 1 \\ 0\end{bmatrix},\begin{bmatrix}-5 \\ -\tfrac32 \\ 0 \\ 1\end{bmatrix}\right\}$$
Col $A$: the pivot columns of the original $A$: $$\left\{\begin{bmatrix}-2 \\ 2 \\ -3\end{bmatrix},\begin{bmatrix}4 \\ -6 \\ 8\end{bmatrix}\right\}$$
Row $A$: the nonzero rows of the echelon form $B$: $\{(1, 0, 6, 5),\ (0, 2, 5, 3)\}$.
$\dim\operatorname{Col}A = \dim\operatorname{Row}A = \operatorname{rank}A = 2$.
Answer: rank = 2, dim Nul = 2, dim Col = 2, dim Row = 2
d
A $4\times 7$ matrix $A$ has rank 4. Find nullity $A$ and rank $A^T$.
1 pt
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Worked solution
Rank theorem: rank $A$ + nullity $A = n = 7$, so nullity $= 3$.
rank $A^T$ = rank $A = 4$ (row rank equals column rank).
Answer: nullity $A = 3$, rank $A^T = 4$
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