MATH 133 practice final exam with full solutions (McGill)

a) Start from \left(\begin{array}{ccc|c} 1 & 1 & 2 & 1 \\ 2 & 3 & 3 & 4 \\ 1 & 2 & k & m \end{array}\right). $R_2 \to R_2 - 2R_1$ gives $(0, 1, -1 \mid 2)$ and $R_3 \to R_3 - R_1$ gives $(0, 1, k - 2 \mid m - 1)$. Then $R_3 \to R_3 - R_2$ gives $(0, 0, k - 1 \mid m - 3)$. The parameters have been gathered in the last row without a single division.

b) The last row reads $(k - 1)z = m - 3$. If $k \neq 1$, it has a non-zero pivot: three pivots, exactly one solution, whatever $m$ is. If $k = 1$, the coefficient vanishes and the row reads $0 = m - 3$: consistent only when $m = 3$, and then rank $2$ for $3$ unknowns gives infinitely many solutions; when $m \neq 3$ it is a contradiction and there is no solution. With two parameters the verdict is a statement about the PAIR $(k, m)$, and $m$ only matters on the line $k = 1$. Answering “no solution for $k = 1$” without looking at $m$ is the typical half-mark answer.

c) For $k = 1$ and $m = 3$ the system is $x + y + 2z = 1$ and $y - z = 2$. With $z = t$: $y = 2 + t$ and $x = 1 - y - 2z = -1 - 3t$. So $(x, y, z) = (-1, 2, 0) + t(-3, 1, 1)$: $(-1, 2, 0)$ is a particular solution, and $t(-3, 1, 1)$ is the general solution of the homogeneous system, whose basic solution is $(-3, 1, 1)$. The coefficient matrix has rank $2$ and nullity $3 - 2 = 1$, the number of parameters. Check in the third equation with $k = 1$: $(-1 - 3t) + 2(2 + t) + t = 3$.

d) For $k = 3$ and $m = 5$, the last row reads $2z = 2$, so $z = 1$. Then $y - z = 2$ gives $y = 3$, and $x = 1 - y - 2z = -4$. Check: $-4 + 3 + 2 = 1$, $-8 + 9 + 3 = 4$, $-4 + 6 + 3 = 5$.

a) Start from \left(\begin{array}{ccc|ccc} 1 & 0 & -1 & 1 & 0 & 0 \\ -2 & 1 & 3 & 0 & 1 & 0 \\ 0 & 2 & 3 & 0 & 0 & 1 \end{array}\right). $R_2 \to R_2 + 2R_1$ gives $(0, 1, 1 \mid 2, 1, 0)$. $R_3 \to R_3 - 2R_2$ gives $(0, 0, 1 \mid -4, -2, 1)$. The left block is now triangular with pivots $1, 1, 1$. Upwards: $R_2 \to R_2 - R_3$ gives $(0, 1, 0 \mid 6, 3, -1)$ and $R_1 \to R_1 + R_3$ gives $(1, 0, 0 \mid -3, -2, 1)$. So $A^{-1} = \begin{pmatrix} -3 & -2 & 1 \\ 6 & 3 & -1 \\ -4 & -2 & 1 \end{pmatrix}$.

b) Row $1$ of $A^{-1}$, $(-3, -2, 1)$, against the columns of $A$: $-3 + 4 + 0 = 1$, $0 - 2 + 2 = 0$, $3 - 6 + 3 = 0$. That is row $1$ of $I$.

c) $A^T$ is invertible with $(A^T)^{-1} = (A^{-1})^T$, so $y = (A^{-1})^Te_1$, the first column of $(A^{-1})^T$, which is the first ROW of $A^{-1}$: $y = (-3, -2, 1)$. Check: $A^T = \begin{pmatrix} 1 & -2 & 0 \\ 0 & 1 & 2 \\ -1 & 3 & 3 \end{pmatrix}$ and $A^Ty = (-3 + 4, -2 + 2, 3 - 6 + 3) = (1, 0, 0)$. Taking the first COLUMN of $A^{-1}$, $(-3, 6, -4)$, solves $Ax = e_1$, a different system.

d) $(3A^T)^{-1} = \frac{1}{3}(A^T)^{-1} = \frac{1}{3}(A^{-1})^T = \frac{1}{3}\begin{pmatrix} -3 & 6 & -4 \\ -2 & 3 & -2 \\ 1 & -1 & 1 \end{pmatrix}$. The scalar inverts to $\frac{1}{3}$, the transpose stays a transpose.

e) The four operations of a) are left multiplications by $E_1$ ($R_2 \to R_2 + 2R_1$), $E_2$ ($R_3 \to R_3 - 2R_2$), $E_3$ ($R_2 \to R_2 - R_3$) and $E_4$ ($R_1 \to R_1 + R_3$), and $E_4E_3E_2E_1A = I$. Multiply on the left by the inverses, the last one first: $A = E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}$, where each inverse undoes its operation, $E_1^{-1}$ is $R_2 \to R_2 - 2R_1$, $E_2^{-1}$ is $R_3 \to R_3 + 2R_2$, $E_3^{-1}$ is $R_2 \to R_2 + R_3$ and $E_4^{-1}$ is $R_1 \to R_1 - R_3$. As matrices, \begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $= A$. The order is the trap: the operation performed FIRST has its inverse written FIRST, on the far left.

a) Row $3$ holds a zero, and so do row $2$ and column $2$ and column $3$. Along row $3$: $\det N = 2C_{31} + 0 \cdot C_{32} + 1 \cdot C_{33}$ with $C_{31} = +\det\begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} = 1$ and $C_{33} = +\det\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} = -5$. So $\det N = 2 - 5 = -3$. Both surviving positions, $(3,1)$ and $(3,3)$, carry the sign $+$.

b) Replace column $3$ by $b$: $N_3 = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 7 \\ 2 & 0 & 5 \end{pmatrix}$. Along row $3$: $2\det\begin{pmatrix} 2 & 3 \\ 1 & 7 \end{pmatrix} + 5\det\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} = 2(11) + 5(-5) = -3$. So $x_3 = \frac{-3}{-3} = 1$.

c) Replace column $2$ by $b$: $N_2 = \begin{pmatrix} 1 & 3 & -1 \\ 3 & 7 & 0 \\ 2 & 5 & 1 \end{pmatrix}$. Along column $3$: $(-1)C_{13} + 0 + 1 \cdot C_{33} = -\det\begin{pmatrix} 3 & 7 \\ 2 & 5 \end{pmatrix} + \det\begin{pmatrix} 1 & 3 \\ 3 & 7 \end{pmatrix} = -1 - 2 = -3$, so $x_2 = 1$. The second equation, $3x_1 + x_2 = 7$, gives $x_1 = 2$ at once, cheaper than a third determinant. Check: $2 + 2 - 1 = 3$, $6 + 1 = 7$, $4 + 1 = 5$.

d) $N^{-1} = \frac{1}{\det N}\text{adj}(N)$ and $\text{adj}(N)$ is the TRANSPOSE of the cofactor matrix, so the entry in row $3$, column $1$ of $N^{-1}$ is $\frac{C_{13}}{\det N}$, indices swapped. $C_{13} = +\det\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix} = -2$, hence $(N^{-1})_{31} = \frac{-2}{-3} = \frac{2}{3}$. Using $C_{31} = 1$ instead would give $-\frac{1}{3}$, the entry in row $1$, column $3$.

e) From $N\,\text{adj}(N) = (\det N)I$ and $n = 3$: $\det N \cdot \det(\text{adj}\,N) = (\det N)^3$, so $\det(\text{adj}\,N) = (\det N)^2 = 9$. And $\det(2N^{-1}) = 2^3 \cdot \frac{1}{\det N} = -\frac{8}{3}$.

a) Direction $\overrightarrow{AB} = (2, 1, -1)$, so $L_1 : (x, y, z) = (1, -1, 2) + t(2, 1, -1)$. For $C$: the first coordinate gives $1 + 2t = 7$, so $t = 3$, and then $(1 + 6, -1 + 3, 2 - 3) = (7, 2, -1)$. All three coordinates agree with ONE value of $t$, so $C$ is on $L_1$. One coordinate alone proves nothing.

b) The directions $(2, 1, -1)$ and $(1, 2, 3)$ are not proportional, so the lines meet or are skew. Use two DIFFERENT parameters: $1 + 2t = 3 + s$, $-1 + t = -3 + 2s$, $2 - t = -6 + 3s$. The first gives $s = 2t - 2$; in the second, $-1 + t = -3 + 4t - 4$, so $t = 2$ and $s = 2$. The third equation is the test: $2 - 2 = 0$ and $-6 + 6 = 0$. It holds, so the lines meet, at $t = 2$ on $L_1$: $(5, 1, 0)$, which is also $s = 2$ on $L_2$.

c) $L_3 : (x, y, z) = u(2, 1, -1)$. Its direction is not proportional to $(1, 2, 3)$, so $L_3$ and $L_2$ are not parallel. Solve $2u = 3 + s$, $u = -3 + 2s$, $-u = -6 + 3s$. The last two give $-3 + 2s = 6 - 3s$, so $s = \frac{9}{5}$ and $u = \frac{3}{5}$. The first equation then reads $\frac{6}{5} = \frac{24}{5}$, false. No common point and not parallel: $L_3$ and $L_2$ are skew. Parallel to a line that meets $L_2$ does not mean meeting $L_2$: moving $L_1$ to the origin moved it out of the plane that $L_1$ and $L_2$ span.

a) $P_0 = (10, 0, 0)$ is on $\Pi$, and $\overrightarrow{P_0Q} = (-7, 5, 3)$. With $n = (1, 2, 2)$ and $\|n\| = 3$, the scalar component of $\overrightarrow{P_0Q}$ along $n$ is $\frac{\overrightarrow{P_0Q} \cdot n}{\|n\|} = \frac{-7 + 10 + 6}{3} = 3$. The distance is $3$. The formula gives the same thing: $\frac{|3 + 10 + 6 - 10|}{3} = 3$, and the choice of $P_0$ cannot matter, since two points of $\Pi$ differ by a vector orthogonal to $n$.

b) From $Q$, walk back along the normal by the projection of $\overrightarrow{P_0Q}$ on $n$, which is $\frac{9}{9}n = n$: $F = Q - n = (2, 3, 1)$. Check: $2 + 6 + 2 = 10$, so $F$ is on $\Pi$. The mirror image is as far beyond $F$ as $Q$ is before it: $Q' = 2F - Q = (1, 1, -1)$. Check: $1 + 2 - 2 = 1$, so its distance is $\frac{|1 - 10|}{3} = 3$, on the other side. The corrected figure shows the cross-section along the normal.

c) Evaluate $x + 2y + 2z - 10$: at $Q$ it is $19 - 10 = 9 > 0$, at $O$ it is $-10 < 0$. The signs differ: opposite sides. The segment is $sQ$ for $0 \le s \le 1$, and it meets $\Pi$ when $s(3 + 10 + 6) = 10$, so $s = \frac{10}{19}$ and the crossing point is $(\frac{30}{19}, \frac{50}{19}, \frac{30}{19})$.

d) The angle between two planes is the angle between their normals: $\cos\theta = \frac{(1, 2, 2) \cdot (0, 0, 1)}{3 \times 1} = \frac{2}{3}$, an acute angle, so no absolute value is needed.

a) Edges from the same vertex: $\overrightarrow{AB} = (1, 2, -1)$ and $\overrightarrow{AC} = (-1, 1, 2)$. Their cross product is $n = (2 \cdot 2 - (-1) \cdot 1,\ (-1)(-1) - 1 \cdot 2,\ 1 \cdot 1 - 2(-1)) = (5, -1, 3)$. Check: $n \cdot \overrightarrow{AB} = 5 - 2 - 3 = 0$ and $n \cdot \overrightarrow{AC} = -5 - 1 + 6 = 0$. The plane is $5x - y + 3z = d$ with $d = 5(1) - 0 + 3(1) = 8$. Check: $B$ gives $10 - 2 + 0 = 8$, $C$ gives $0 - 1 + 9 = 8$.

b) Area $= \frac{1}{2}\|\overrightarrow{AB} \times \overrightarrow{AC}\| = \frac{1}{2}\sqrt{25 + 1 + 9} = \frac{\sqrt{35}}{2}$. The cross product was already there: the normal of a) gives the area for free.

c) $\overrightarrow{AD} = (2, 1, 3)$ and the triple product is $\overrightarrow{AD} \cdot (\overrightarrow{AB} \times \overrightarrow{AC}) = 10 - 1 + 9 = 18$. The parallelepiped on the three edges has volume $18$, and the tetrahedron is one sixth of it: $3$. One sixth, not one third: a tetrahedron is one third of a PRISM on the triangle, and the triangle is half the parallelogram.

d) Volume $= \frac{1}{3}(\text{base area})(\text{height})$, so $h = \frac{3 \times 3}{\sqrt{35}/2} = \frac{18}{\sqrt{35}}$. Check with the formula: $\frac{|5(3) - 1 + 3(4) - 8|}{\sqrt{35}} = \frac{18}{\sqrt{35}}$. The two agree, and the same $18$ appears in both, which is no accident: it is $\overrightarrow{AD} \cdot n$.

a) Split by parameter: $(a + b, a - b, 2b, a) = a(1, 1, 0, 1) + b(1, -1, 2, 0)$. So $U = \text{span}\{(1, 1, 0, 1), (1, -1, 2, 0)\}$, and a span is always a subspace: it contains $0$ ($a = b = 0$), and sums and multiples of combinations are combinations. The two spanning vectors are not multiples of each other (the third coordinates are $0$ and $2$ while the fourth are $1$ and $0$), so they are independent: they form a basis and $\dim U = 2$.

b) No. $W$ contains $0$ and is closed under addition, since $x_1 \le y_1$ and $x_2 \le y_2$ give $x_1 + x_2 \le y_1 + y_2$. But $(0, 1, 0) \in W$ while $(-1)(0, 1, 0) = (0, -1, 0)$ is not, since $0 \le -1$ is false. One failing condition is enough, and it must be shown with numbers.

c) Put the vectors as columns and reduce: rows $(1, 0, 1)$, $(1, 1, 3)$, $(0, 1, k)$, $(1, 2, 5)$. $R_2 \to R_2 - R_1$ gives $(0, 1, 2)$, $R_4 \to R_4 - R_1$ gives $(0, 2, 4)$; then $R_3 \to R_3 - R_2$ gives $(0, 0, k - 2)$ and $R_4 \to R_4 - 2R_2$ gives a zero row. There are three pivots, hence independence, exactly when $k \neq 2$. For $k = 2$ the third column is free, and the reduced form reads $c_1 = -c_3$ and $c_2 = -2c_3$: with $c_3 = -1$, $v_1 + 2v_2 - v_3 = 0$, that is $v_3 = v_1 + 2v_2$. Check: $v_1 + 2v_2 = (1, 1, 0, 1) + (0, 2, 2, 4) = (1, 3, 2, 5)$, which is $v_3$ for $k = 2$.

d) $2v_1 + v_2 = (2, 2, 0, 2) + (0, 1, 1, 2) = (2, 3, 1, 4) = w$: yes. The span of the three vectors has dimension $3$ when they are independent, $k \neq 2$, and $2$ when $k = 2$, since $v_3$ then adds nothing to $\text{span}\{v_1, v_2\}$.

a) $R_2 \to R_2 - 2R_1$ gives $(0, 0, 1, 2, 1)$ and $R_3 \to R_3 - R_1$ gives $(0, 0, 1, 2, 1)$ as well; $R_3 \to R_3 - R_2$ gives a zero row. The matrix \begin{pmatrix} 1 & 2 & 0 & 1 & 3 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} is reduced: pivots in columns $1$ and $3$, with a $0$ above the second pivot.

b) Free variables $x_2 = s$, $x_4 = t$, $x_5 = u$; the rows give $x_3 = -2t - u$ and $x_1 = -2s - t - 3u$. So $\text{null}(A)$ has basis $(-2, 1, 0, 0, 0)$, $(-1, 0, -2, 1, 0)$, $(-3, 0, -1, 0, 1)$, dimension $3$. $\text{col}(A)$ takes the columns of the ORIGINAL $A$ in the pivot positions $1$ and $3$: $(1, 2, 1)$ and $(0, 1, 1)$, dimension $2$. The pivot columns of $R$, $(1, 0, 0)$ and $(0, 1, 0)$, would be wrong: row operations change the column space. $\text{row}(A)$ takes the non-zero rows of $R$: $(1, 2, 0, 1, 3)$ and $(0, 0, 1, 2, 1)$, dimension $2$. Rank theorem: $2 + 3 = 5$, the number of columns.

c) $\dim \text{null}(A^T) = 3 - 2 = 1$, the number of rows minus the rank. The reduction showed that row $3$ minus row $1$ equals row $2$ minus $2$ row $1$, that is $R_1 - R_2 + R_3 = 0$. Check: $(1 - 2 + 1, 2 - 4 + 2, 0 - 1 + 1, 1 - 4 + 3, 3 - 7 + 4) = 0$. So $(1, -1, 1)^TA = 0$, and $(1, -1, 1)$ spans $\text{null}(A^T)$.