Exercise 1: Two parameters, three verdicts

For real numbers $k$ and $m$ , consider $\begin{cases} x + y + 2z = 1 \\ 2x + 3y + 3z = 4 \\ x + 2y + kz = m \end{cases}$ .

a) Reduce the augmented matrix to row echelon form, naming every operation.
b) Say, in terms of $k$ and $m$ , when the system has no solution, exactly one solution, infinitely many solutions.
c) In the case of infinitely many solutions, give the general solution as a particular solution plus the solutions of the homogeneous system, and give the rank and the nullity of the coefficient matrix.
d) Solve the system for $k = 3$ and $m = 5$ , and check.

Show the solution

Answers

a) last row $(0, 0, k - 1 \mid m - 3)$
b) exactly one if $k \neq 1$ ; infinitely many if $k = 1$ and $m = 3$ ; none if $k = 1$ and $m \neq 3$
c) $(x, y, z) = (-1, 2, 0) + t(-3, 1, 1)$ ; rank $2$ , nullity $1$
d) $(x, y, z) = (-4, 3, 1)$

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: $-4 + 3 + 2 = 1$ , $-8 + 9 + 3 = 4$ , $-4 + 6 + 3 = 5$ .

Tick the exercises you have done or want to review: a free account, no password, keeps your ticks and your right answers from one visit to the next, tells you which chapter to tackle next and lets you ask for the exercise you are missing. Create your space, an email is enough.

Exercise 2: An inverse, its transpose, and the elementary matrices behind it

Let $A = \begin{pmatrix} 1 & 0 & -1 \\ -2 & 1 & 3 \\ 0 & 2 & 3 \end{pmatrix}$ .

a) Find $A^{-1}$ by reducing $[A \mid I]$ , naming every row operation.
b) Check one row of $A^{-1}A$ .
c) Solve $A^Ty = (1, 0, 0)$ without any new reduction.
d) Give $(3A^T)^{-1}$ .
e) Write $A$ as a product of elementary matrices, using the operations of a).

Show the solution

Answers

a) $A^{-1} = \begin{pmatrix} -3 & -2 & 1 \\ 6 & 3 & -1 \\ -4 & -2 & 1 \end{pmatrix}$
b) row $1$ of $A^{-1}A$ is $(1, 0, 0)$
c) $y = (-3, -2, 1)$ , the first ROW of $A^{-1}$
d) $(3A^T)^{-1} = \frac{1}{3}(A^{-1})^T = \frac{1}{3}\begin{pmatrix} -3 & 6 & -4 \\ -2 & 3 & -2 \\ 1 & -1 & 1 \end{pmatrix}$
e) $A = E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}$ with $E_1^{-1}: R_2 \to R_2 - 2R_1$ , $E_2^{-1}: R_3 \to R_3 + 2R_2$ , $E_3^{-1}: R_2 \to R_2 + R_3$ , $E_4^{-1}: R_1 \to R_1 - R_3$

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.

Exercise 3: Cramer's rule for two unknowns, and one entry of the inverse

Consider the system $Nx = b$ with $N = \begin{pmatrix} 1 & 2 & -1 \\ 3 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}$ , $x = (x_1, x_2, x_3)$ and $b = (3, 7, 5)$ .

a) Compute $\det N$ by expanding along the line of your choice, and justify the choice.
b) Find $x_3$ by Cramer's rule.
c) Find $x_2$ by Cramer's rule, then $x_1$ from one equation of the system, and check the solution.
d) Compute the entry in row $3$ , column $1$ of $N^{-1}$ without computing $N^{-1}$ .
e) Give $\det(\text{adj}\,N)$ and $\det(2N^{-1})$ .

Show the solution

Answers

a) $\det N = -3$
b) $x_3 = \frac{-3}{-3} = 1$
c) $x_2 = 1$ , $x_1 = 2$
d) $(N^{-1})_{31} = \frac{C_{13}}{\det N} = \frac{-2}{-3} = \frac{2}{3}$
e) $\det(\text{adj}\,N) = 9$ and $\det(2N^{-1}) = -\frac{8}{3}$

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 $\det(2N^{-1}) = 2^3 \cdot \frac{1}{\det N} = -\frac{8}{3}$ .

Exercise 4: Two lines that meet, and a parallel that does not

Let $L_1$ be the line through $A(1, -1, 2)$ and $B(3, 0, 1)$ , and $L_2 : (x, y, z) = (3, -3, -6) + s(1, 2, 3)$ .

a) Write a vector equation of $L_1$ . Is the point $C(7, 2, -1)$ on $L_1$ ?
b) Show that $L_1$ and $L_2$ intersect, and find the point of intersection.
c) Let $L_3$ be the line through the origin parallel to $L_1$ . Give the relative position of $L_3$ and $L_2$ .

Show the solution

Answers

a) $L_1 : (x, y, z) = (1, -1, 2) + t(2, 1, -1)$ ; $C$ is on $L_1$ ( $t = 3$ )
b) $t = 2$ , $s = 2$ : they meet at $(5, 1, 0)$
c) $L_3$ and $L_2$ are skew

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.

Exercise 5: A point and a plane: distance, foot, mirror image and side

Let $\Pi$ be the plane $x + 2y + 2z = 10$ and $Q = (3, 5, 3)$ .

The figure below shows the plane edge on, its normal and the point Q.

a) Compute the distance from $Q$ to $\Pi$ by projecting a vector on the normal, starting from a point of $\Pi$ that you choose.
b) Find the point $F$ of $\Pi$ closest to $Q$ , and the mirror image $Q'$ of $Q$ across $\Pi$ .
c) Are $Q$ and the origin on the same side of $\Pi$ ? If not, where does the segment from $O$ to $Q$ cross $\Pi$ ?
d) Find the cosine of the angle between $\Pi$ and the horizontal plane $z = 0$ .

Show the solution

Answers

a) distance $= \frac{9}{3} = 3$
b) $F = (2, 3, 1)$ , $Q' = (1, 1, -1)$
c) opposite sides; the segment crosses $\Pi$ at $\frac{10}{19}Q = (\frac{30}{19}, \frac{50}{19}, \frac{30}{19})$
d) $\cos\theta = \frac{2}{3}$

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.

Exercise 6: A plane through three points, a triangle and a tetrahedron

The points are $A(1, 0, 1)$ , $B(2, 2, 0)$ , $C(0, 1, 3)$ and $D(3, 1, 4)$ , sketched on the figure.

a) Find a normal vector to the plane $ABC$ and its equation in the form $ax + by + cz = d$ . Check it with the three points.
b) Compute the area of triangle $ABC$ .
c) Compute the volume of the tetrahedron $ABCD$ .
d) Deduce the distance from $D$ to the plane $ABC$ , and check it with the distance formula.

Show the solution

Answers

a) $n = (5, -1, 3)$ , plane $5x - y + 3z = 8$
b) $\frac{\sqrt{35}}{2}$
c) $\frac{18}{6} = 3$
d) $\frac{18}{\sqrt{35}}$

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 $\overrightarrow{AD} \cdot n$ .

Exercise 7: Two sets to test, and a family with a parameter

Let $U = \{(a + b,\ a - b,\ 2b,\ a) : a, b \in \mathbb{R}\}$ in $\mathbb{R}^4$ and $W = \{(x, y, z) : x \le y\}$ in $\mathbb{R}^3$ . In $\mathbb{R}^4$ , let $v_1 = (1, 1, 0, 1)$ , $v_2 = (0, 1, 1, 2)$ and $v_3 = (1, 3, k, 5)$ .

a) Show that $U$ is a subspace of $\mathbb{R}^4$ and give a basis and its dimension.
b) Is $W$ a subspace of $\mathbb{R}^3$ ?
c) Find every $k$ for which $\{v_1, v_2, v_3\}$ is linearly dependent, and give the dependency relation.
d) Is $w = (2, 3, 1, 4)$ in $\text{span}\{v_1, v_2\}$ ? Give $\dim \text{span}\{v_1, v_2, v_3\}$ for every $k$ .

Show the solution

Answers

a) $U = \text{span}\{(1, 1, 0, 1), (1, -1, 2, 0)\}$ , basis of $2$ vectors, $\dim U = 2$
b) no: $(0, 1, 0) \in W$ but $-(0, 1, 0) \notin W$
c) dependent exactly when $k = 2$ , and then $v_3 = v_1 + 2v_2$
d) yes, $w = 2v_1 + v_2$ ; the dimension is $2$ for $k = 2$ and $3$ otherwise

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 $\text{span}\{v_1, v_2\}$ .

Exercise 8: One matrix, four subspaces

Let $A = \begin{pmatrix} 1 & 2 & 0 & 1 & 3 \\ 2 & 4 & 1 & 4 & 7 \\ 1 & 2 & 1 & 3 & 4 \end{pmatrix}$ .

a) Reduce $A$ to reduced row echelon form, naming every operation.
b) Give a basis and the dimension of $\text{null}(A)$ , $\text{col}(A)$ and $\text{row}(A)$ , and check the rank theorem.
c) Give $\dim \text{null}(A^T)$ and a non-zero vector of $\text{null}(A^T)$ , read from a relation between the rows of $A$ .
d) Deduce an equation of $\text{col}(A)$ . Are $(1, 1, 1)$ and $(2, 5, 3)$ in $\text{col}(A)$ ?

Type your answers, the page tells you right or wrong 0/14

a)

Entry

(1,4)

of

R

Entry

(1,5)

of

R

Entry

(2,4)

of

R

Entry

(2,5)

of

R

b)

\dim \text{null}(A)

\dim \text{col}(A)

\dim \text{row}(A)

c)

\dim \text{null}(A^{T})

Vector with first component 1: second component

Third component

d)

Equation

y_1 + \beta y_2 + \gamma y_3 = 0

:

\beta

\gamma

Is

(1, 1, 1)

in

\text{col}(A)

?

Is

(2, 5, 3)

in

\text{col}(A)

?

Show the solution

Answers

a) $R = \begin{pmatrix} 1 & 2 & 0 & 1 & 3 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$
b) $\text{null}$ : $(-2, 1, 0, 0, 0)$ , $(-1, 0, -2, 1, 0)$ , $(-3, 0, -1, 0, 1)$ , dim $3$ ; $\text{col}$ : $(1, 2, 1)$ , $(0, 1, 1)$ , dim $2$ ; $\text{row}$ : $(1, 2, 0, 1, 3)$ , $(0, 0, 1, 2, 1)$ , dim $2$ ; $2 + 3 = 5$
c) $\dim \text{null}(A^T) = 1$ , spanned by $(1, -1, 1)$
d) $\text{col}(A)$ is the plane $y_1 - y_2 + y_3 = 0$ : $(1, 1, 1)$ no, $(2, 5, 3)$ yes

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 $\text{null}(A^T)$ .

d)

Every column $c$ of $A$ satisfies $(1, -1, 1) \cdot c = 0$ , since that is exactly the relation $R_1 - R_2 + R_3 = 0$ read column by column. So $\text{col}(A)$ lies in the plane $y_1 - y_2 + y_3 = 0$ , and since both have dimension $2$ , they are equal. $(1, 1, 1)$ gives $1 \neq 0$ : not in $\text{col}(A)$ . $(2, 5, 3)$ gives $0$ : it is, and indeed $2(1, 2, 1) + (0, 1, 1) = (2, 5, 3)$ .

Exercise 9: A linear map known on three vectors, its kernel and a right inverse

$T : \mathbb{R}^3 \to \mathbb{R}^2$ is linear, with $T(1, 0, 1) = (1, 2)$ , $T(0, 1, 1) = (0, 1)$ and $T(1, 1, 1) = (2, 3)$ .

a) Find $T(e_1)$ , $T(e_2)$ , $T(e_3)$ and the standard matrix $A$ of $T$ .
b) Find a basis of $\ker T$ . Is $T$ one-to-one? Is it onto? Give every $x$ with $T(x) = (3, 4)$ .
c) Find a linear map $U : \mathbb{R}^2 \to \mathbb{R}^3$ with $T \circ U = I$ , by giving its matrix. Is it the only one?
d) Explain why no linear map $V : \mathbb{R}^2 \to \mathbb{R}^3$ can satisfy $V \circ T = I$ .

Show the solution

Answers

a) $T(e_1) = (2, 2)$ , $T(e_2) = (1, 1)$ , $T(e_3) = (-1, 0)$ , $A = \begin{pmatrix} 2 & 1 & -1 \\ 2 & 1 & 0 \end{pmatrix}$
b) $\ker T = \text{span}\{(1, -2, 0)\}$ : not one-to-one, onto; $x = (2, 0, 1) + s(1, -2, 0)$
c) $U = \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ -1 & 1 \end{pmatrix}$ ; not unique, any kernel vector may be added to each column
d) $V \circ T$ has rank at most $2$ , and $I_3$ has rank $3$

a)

Write each standard vector with the three given ones: $e_1 = (1, 1, 1) - (0, 1, 1)$ , $e_2 = (1, 1, 1) - (1, 0, 1)$ and $e_3 = (1, 0, 1) + (0, 1, 1) - (1, 1, 1)$ . By linearity $T(e_1) = (2, 3) - (0, 1) = (2, 2)$ , $T(e_2) = (2, 3) - (1, 2) = (1, 1)$ and $T(e_3) = (1, 2) + (0, 1) - (2, 3) = (-1, 0)$ . These are the COLUMNS of $A = \begin{pmatrix} 2 & 1 & -1 \\ 2 & 1 & 0 \end{pmatrix}$ . Check on a given vector: $A(1, 0, 1) = (2 - 1, 2 + 0) = (1, 2)$ .

b)

$R_2 \to R_2 - R_1$ gives $(0, 0, 1)$ , so $Ax = 0$ means $x_3 = 0$ and $2x_1 + x_2 = 0$ : $\ker T = \text{span}\{(1, -2, 0)\}$ . A non-zero kernel means $T$ is not one-to-one. The rank is $2 = \dim \mathbb{R}^2$ , so every $b$ is reached: $T$ is onto, and $2 + 1 = 3$ is the rank theorem. For $(3, 4)$ : $2x_1 + x_2 - x_3 = 3$ and $2x_1 + x_2 = 4$ give $x_3 = 1$ and $x_2 = 4 - 2x_1$ , so $x = (2, 0, 1) + s(1, -2, 0)$ , a particular solution plus the kernel.

c)

$T \circ U = I$ asks $AU = I_2$ , so the columns $u_1$ , $u_2$ of $U$ must satisfy $Au_1 = e_1$ and $Au_2 = e_2$ . For $e_1$ : $2x_1 + x_2 - x_3 = 1$ and $2x_1 + x_2 = 0$ give $x_3 = -1$ ; take $u_1 = (0, 0, -1)$ . For $e_2$ : $2x_1 + x_2 = 1$ and $x_3 = 2x_1 + x_2 = 1$ ; take $u_2 = (0, 1, 1)$ . So $U = \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ -1 & 1 \end{pmatrix}$ , and $AU = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . It is not unique: adding $(1, -2, 0)$ , or any kernel vector, to either column keeps $AU = I$ .

d)

$V \circ T$ has matrix $BA$ for some $3 \times 2$ matrix $B$ , and $\text{rank}(BA) \le \text{rank}(A) = 2$ . The identity of $\mathbb{R}^3$ has rank $3$ . Put differently: $T$ sends $(1, -2, 0)$ and $0$ to the same point, and no map applied afterwards can separate them again. A one-sided inverse exists on the side of the smaller space only.

Exercise 10: A 3 by 3 whose polynomial arrives factored

Let $A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{pmatrix}$ , and use the convention $c_A(x) = \det(xI - A)$ .

a) Compute $c_A(x)$ by cofactor expansion along a line of $xI - A$ that you choose, and give the eigenvalues.
b) Find a basis of each eigenspace, and check each basic eigenvector by one multiplication.
c) Check the eigenvalues against the trace and the determinant of $A$ .
d) Without any new determinant: is $A$ invertible? Give the eigenvalues of $A^{-1}$ and of $A^2 - 2A$ , and say whether $A^2 - 2A$ is invertible.

Show the solution

Answers

a) $c_A(x) = (x - 1)(x - 2)(x - 3)$ : eigenvalues $1$ , $2$ , $3$
b) $E_1 = \text{span}\{(1, 0, -2)\}$ , $E_2 = \text{span}\{(0, 0, 1)\}$ , $E_3 = \text{span}\{(1, 1, -2)\}$
c) $\text{tr}\,A = 6 = 1 + 2 + 3$ and $\det A = 6 = 1 \cdot 2 \cdot 3$
d) invertible; $A^{-1}$ : $1, \frac{1}{2}, \frac{1}{3}$ ; $A^2 - 2A$ : $-1, 0, 3$ , so it is NOT invertible

a)

$xI - A = \begin{pmatrix} x - 1 & -2 & 0 \\ 0 & x - 3 & 0 \\ -2 & 4 & x - 2 \end{pmatrix}$ . Column $3$ has two zeros, and so does row $2$ . Along column $3$ , only the $(3,3)$ entry survives, with sign $+$ : $c_A(x) = (x - 2)\det\begin{pmatrix} x - 1 & -2 \\ 0 & x - 3 \end{pmatrix} = (x - 2)(x - 1)(x - 3)$ . The polynomial comes out FACTORED, which is the reason for the choice: expanding along row $1$ would give a cubic to factor by trial. The eigenvalues are $1$ , $2$ and $3$ .

b)

$\lambda = 1$ : $I - A = \begin{pmatrix} 0 & -2 & 0 \\ 0 & -2 & 0 \\ -2 & 4 & -1 \end{pmatrix}$ . The first row gives $y = 0$ ( $R_2 \to R_2 - R_1$ clears the second), and then the third gives $-2x - z = 0$ : basic eigenvector $(1, 0, -2)$ . $\lambda = 2$ : $2I - A = \begin{pmatrix} 1 & -2 & 0 \\ 0 & -1 & 0 \\ -2 & 4 & 0 \end{pmatrix}$ , so $y = 0$ and then $x = 0$ , with $z$ free: $(0, 0, 1)$ . $\lambda = 3$ : $3I - A = \begin{pmatrix} 2 & -2 & 0 \\ 0 & 0 & 0 \\ -2 & 4 & 1 \end{pmatrix}$ , so $x = y$ , and $-2x + 4x + z = 0$ gives $z = -2x$ : $(1, 1, -2)$ . Checks: $A(1, 0, -2) = (1, 0, 2 - 4) = 1 \cdot (1, 0, -2)$ , $A(0, 0, 1) = (0, 0, 2) = 2(0, 0, 1)$ , $A(1, 1, -2) = (3, 3, -6) = 3(1, 1, -2)$ .

c)

$\text{tr}\,A = 1 + 3 + 2 = 6 = 1 + 2 + 3$ . $\det A$ along column $3$ : $2\det\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} = 6 = 1 \cdot 2 \cdot 3$ . Both agree.

d)

$0$ is not an eigenvalue, so $A$ is invertible. From $Ax = \lambda x$ : $A^{-1}x = \frac{1}{\lambda}x$ , giving $1$ , $\frac{1}{2}$ , $\frac{1}{3}$ with the same eigenvectors; and $(A^2 - 2A)x = (\lambda^2 - 2\lambda)x$ , giving $-1$ , $0$ , $3$ . Since $0$ is among them, $A^2 - 2A$ is NOT invertible, and $(0, 0, 1)$ is in its null space. Three distinct eigenvalues also make $A$ diagonalizable, with no rank test needed.

Exercise 11: A repeated eigenvalue that passes the test, and a power without P inverse

Let $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ .

a) Compute $c_A(x)$ , factored, and give the eigenvalues with their algebraic multiplicities.
b) Find a basis of each eigenspace, and decide whether $A$ is diagonalizable.
c) Give an invertible $P$ and a diagonal $D$ with $P^{-1}AP = D$ , and check the answer without computing $P^{-1}$ .
d) Let $N = A - I$ . Show that $N^2 = 2N$ , and deduce that $A^n = I + \frac{3^n - 1}{2}N$ for every $n \ge 1$ . Check the formula at $n = 2$ .

Show the solution

Answers

a) $c_A(x) = (x - 1)^2(x - 3)$ : $1$ (double) and $3$ (simple)
b) $E_1 = \text{span}\{(-1, 1, 0), (-1, 0, 1)\}$ , $E_3 = \text{span}\{(1, 1, 0)\}$ : diagonalizable
c) $P = \begin{pmatrix} -1 & -1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$ , $D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}$
d) $A^n = I + \frac{3^n - 1}{2}\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ ; $A^2 = \begin{pmatrix} 5 & 4 & 4 \\ 4 & 5 & 4 \\ 0 & 0 & 1 \end{pmatrix}$

a)

Row $3$ of $xI - A = \begin{pmatrix} x - 2 & -1 & -1 \\ -1 & x - 2 & -1 \\ 0 & 0 & x - 1 \end{pmatrix}$ has two zeros. Along it, with sign $+$ at $(3,3)$ : $c_A(x) = (x - 1)\left[(x - 2)^2 - 1\right] = (x - 1)(x - 1)(x - 3) = (x - 1)^2(x - 3)$ . The factor $(x - 2)^2 - 1$ is a difference of squares, $(x - 3)(x - 1)$ : no quadratic formula needed. Eigenvalues: $1$ with algebraic multiplicity $2$ , and $3$ .

b)

$\lambda = 1$ : $I - A = \begin{pmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ 0 & 0 & 0 \end{pmatrix}$ ; $R_2 \to R_2 - R_1$ leaves the single equation $x + y + z = 0$ , rank $1$ , so $\dim E_1 = 3 - 1 = 2$ , with basis $(-1, 1, 0)$ and $(-1, 0, 1)$ . $\lambda = 3$ : $3I - A = \begin{pmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ 0 & 0 & 2 \end{pmatrix}$ gives $z = 0$ and $x = y$ : basis $(1, 1, 0)$ . Each eigenspace has the dimension of its algebraic multiplicity, so $A$ has three independent eigenvectors and IS diagonalizable. The simple eigenvalue never needs the test; the double one is the only place the count could fall short, and it does not.

c)

$P = \begin{pmatrix} -1 & -1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$ , the eigenvectors as COLUMNS, and $D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ in the same order. $P$ is invertible: along row $3$ , $\det P = -\det\begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix} = 2$ . Check $AP = PD$ column by column, which is the statement $Av = \lambda v$ for each column: $A(-1, 1, 0) = (-1, 1, 0)$ , $A(-1, 0, 1) = (-1, 0, 1)$ , $A(1, 1, 0) = (3, 3, 0)$ .

d)

$N = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ and $N^2 = \begin{pmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 0 & 0 & 0 \end{pmatrix} = 2N$ . By induction $N^k = 2^{k-1}N$ for $k \ge 1$ . Since $I$ and $N$ commute, the binomial expansion is legal: $A^n = (I + N)^n = I + \sum_{k=1}^{n}\binom{n}{k}2^{k-1}N = I + \frac{1}{2}\left[(1 + 2)^n - 1\right]N = I + \frac{3^n - 1}{2}N$ . At $n = 2$ : $I + 4N = \begin{pmatrix} 5 & 4 & 4 \\ 4 & 5 & 4 \\ 0 & 0 & 1 \end{pmatrix}$ , and $A^2$ computed directly gives the same matrix (row $1$ of $A$ against its columns: $4 + 1 + 0 = 5$ , $2 + 2 + 0 = 4$ , $2 + 1 + 1 = 4$ ).

The formula is diagonalization in disguise: on $E_1$ , $N$ acts as $0$ and $A^n = I$ ; on $E_3$ , $N$ acts as $2$ and $A^n$ multiplies by $1 + (3^n - 1) = 3^n$ . Two eigenvalues, so a polynomial of degree one in $A$ is enough, and $P^{-1}$ never had to be computed.

Exercise 12: Two lakes joined by a channel, and the fishing quota

Two lakes, $A$ and $B$ , are joined by a channel. Let $a_k$ and $b_k$ be their trout populations in year $k$ , in hundreds. Each year reproduction, mortality and migration through the channel combine into $a_{k+1} = 0.8a_k + 0.4b_k$ and $b_{k+1} = 0.4a_k + 0.8b_k$ , so $x_{k+1} = Mx_k$ with $x_k = (a_k, b_k)$ and $M = \begin{pmatrix} 0.8 & 0.4 \\ 0.4 & 0.8 \end{pmatrix}$ . At the start, $x_0 = (5, 1)$ : $500$ trout in lake $A$ and $100$ in lake $B$ . The figure plots the first five years.

a) Diagonalize $M$ .
b) Write $x_0$ as a combination of the eigenvectors and give a closed formula for $x_k$ . Check it against year $1$ on the figure.
c) Describe the long run: the yearly growth rate and the eventual ratio of the two populations. From which year on does lake $B$ hold at least $90$ per cent of the population of lake $A$ ?
d) A fishing quota removes each year, on top of the update, a number of trout equal to the same fraction $h$ of each lake's population at the start of that year, so that $x_{k+1} = Mx_k - hx_k$ . For which $h$ do the populations neither grow nor die out in the long run, and where do they settle?

Type your answers, the page tells you right or wrong 0/13

a)

Larger eigenvalue

Its eigenvector with first component 1: second component

Smaller eigenvalue

Its eigenvector with first component 1: second component

b)

x_0 = \alpha(1, 1) + \beta(1, -1)

:

\alpha

\beta

a_1

b_1

c)

Growth factor per year

Limit of

b_k / a_k

First year with lake B at 90 per cent of lake A

d)

Settling population of lake A (hundreds)

Settling population of lake B (hundreds)

Show the solution

Answers

a) $P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ , $D = \begin{pmatrix} 1.2 & 0 \\ 0 & 0.4 \end{pmatrix}$
b) $x_0 = 3(1, 1) + 2(1, -1)$ and $x_k = 3(1.2)^k(1, 1) + 2(0.4)^k(1, -1)$ ; $x_1 = (4.4, 2.8)$
c) growth $20$ per cent a year, ratio tending to $1$ ; $\frac{b_k}{a_k} = \frac{3 \cdot 3^k - 2}{3 \cdot 3^k + 2}$ reaches $0.9$ from year $3$ on
d) $h = 0.2$ ; the populations settle at $(3, 3)$ , $300$ trout in each lake

a)

$c_M(x) = (x - 0.8)^2 - 0.16 = (x - 0.8 - 0.4)(x - 0.8 + 0.4) = (x - 1.2)(x - 0.4)$ , a difference of squares again. For $1.2$ : $1.2I - M = \begin{pmatrix} 0.4 & -0.4 \\ -0.4 & 0.4 \end{pmatrix}$ , so $a = b$ and $(1, 1)$ . For $0.4$ : $0.4I - M = \begin{pmatrix} -0.4 & -0.4 \\ -0.4 & -0.4 \end{pmatrix}$ , so $a = -b$ and $(1, -1)$ . Hence $P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ and $D = \begin{pmatrix} 1.2 & 0 \\ 0 & 0.4 \end{pmatrix}$ with $P^{-1}MP = D$ . Check: $M(1, 1) = (1.2, 1.2)$ and $M(1, -1) = (0.4, -0.4)$ .

b)

$(5, 1) = c_1(1, 1) + c_2(1, -1)$ gives $c_1 + c_2 = 5$ and $c_1 - c_2 = 1$ , so $c_1 = 3$ and $c_2 = 2$ . Each eigenvector is simply multiplied by its eigenvalue every year, so $x_k = 3(1.2)^k(1, 1) + 2(0.4)^k(1, -1)$ . Year $1$ : $3(1.2)(1, 1) + 2(0.4)(1, -1) = (3.6 + 0.8, 3.6 - 0.8) = (4.4, 2.8)$ , the point of the figure. Year $2$ gives $(4.32 + 0.32, 4.32 - 0.32) = (4.64, 4)$ , also on the figure.

c)

Factor out the dominant term: $x_k = (1.2)^k\left[3(1, 1) + 2\left(\frac{1}{3}\right)^k(1, -1)\right]$ , since $\frac{0.4}{1.2} = \frac{1}{3}$ . The bracket tends to $3(1, 1)$ : in the long run both lakes hold the same number of trout, and each population grows by $20$ per cent a year. The corrected figure shows the points approaching the eigenline $y = x$ . The ratio is $\frac{b_k}{a_k} = \frac{3(1.2)^k - 2(0.4)^k}{3(1.2)^k + 2(0.4)^k} = \frac{3 \cdot 3^k - 2}{3 \cdot 3^k + 2}$ , dividing top and bottom by $(0.4)^k$ . It is at least $0.9$ when $3 \cdot 3^k - 2 \ge 2.7 \cdot 3^k + 1.8$ , that is $0.3 \cdot 3^k \ge 3.8$ , or $3^k \ge \frac{38}{3}$ . Since $3^2 = 9$ and $3^3 = 27$ , this first happens in year $3$ : $\frac{79}{83} \approx 0.95$ , while year $2$ gives $\frac{25}{29} \approx 0.86$ .

d)

With the quota, $x_{k+1} = (M - hI)x_k$ . The matrix $M - hI$ has the SAME eigenvectors, with eigenvalues $1.2 - h$ and $0.4 - h$ , because $(M - hI)v = (\lambda - h)v$ . The populations neither explode nor vanish when the dominant eigenvalue is exactly $1$ : $h = 0.2$ , a fifth of each lake caught every year. Then $x_k = 3(1, 1) + 2(0.2)^k(1, -1) \to (3, 3)$ : $300$ trout in each lake. For $h$ slightly above $0.2$ both populations die out; slightly below, they still grow. The limit $(3, 3)$ comes from the coefficient $c_1 = \frac{a_0 + b_0}{2}$ , so it depends on the starting TOTAL, not on how it was split.

MATH 133 practice final exam with full solutions (McGill)

Part A: systems, matrices and determinants (/33)

Exercise 1: Two parameters, three verdicts

Exercise 2: An inverse, its transpose, and the elementary matrices behind it

Exercise 3: Cramer's rule for two unknowns, and one entry of the inverse

Part B: vectors, lines and planes (/22)

Exercise 4: Two lines that meet, and a parallel that does not

Exercise 5: A point and a plane: distance, foot, mirror image and side

Exercise 6: A plane through three points, a triangle and a tetrahedron

Part C: subspaces, bases and linear maps (/22)

Exercise 7: Two sets to test, and a family with a parameter

Exercise 8: One matrix, four subspaces

Exercise 9: A linear map known on three vectors, its kernel and a right inverse

Part D: eigenvalues and diagonalization (/23)

Exercise 10: A 3 by 3 whose polynomial arrives factored

Exercise 11: A repeated eigenvalue that passes the test, and a power without P inverse

Exercise 12: Two lakes joined by a channel, and the fishing quota

See also

Looking for a tutor in Montreal?