13 Problems
Rules, the transpose, and solving the system
The problems follow the sections of the theory. Those that are checked at once get new numbers each time you press New variant: repeat them until the method runs by itself — elimination is learned by doing it, not by watching it. Those with a solution are for paper. Write the whole solution before you open it, then compare line by line. Last come problems in the form of the exam, with points and a marking guide.
Typing an answer. Most answers are numbers; fractions may be typed as -5/4 or as -1.25. One answer is an expression in $t{:}$ type it as 3 - 2t or 3 - 2*t.
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Rules for matrix multiplication, and the identity
Read the theory: 1. Rules for matrix multiplication, and the identity →Let $\mathbf{A}$ and $\mathbf{B}$ be square matrices of the same order $n,$ and $\mathbf{I}=\mathbf{I}_{n}.$
(a) Expand $(\mathbf{A}+2\mathbf{I})(\mathbf{A}-3\mathbf{I})$ and simplify as far as the rules allow.
(b) Expand $(\mathbf{A}-\mathbf{B})^{2}.$ Why can the middle terms be collected in (a) but not in (b)?
(c) Check (a) with $\mathbf{A}=\begin{pmatrix}1&2\\3&0\end{pmatrix}{:}$ compute $\mathbf{A}+2\mathbf{I},$ $\mathbf{A}-3\mathbf{I}$ and their product. Which rule of Watch out 13.5 do you see?
Hint
(a) Four terms first. $$ \begin{aligned} &(\mathbf{A}+2\mathbf{I})(\mathbf{A}-3\mathbf{I})\\ &=\mathbf{A}^{2}-3\mathbf{A}\mathbf{I}\\ &\quad+2\mathbf{I}\mathbf{A}-6\mathbf{I}^{2}\,. \end{aligned} $$
(a) Then the identity. $\mathbf{A}\mathbf{I}=\mathbf{I}\mathbf{A}=\mathbf{A}$ and $\mathbf{I}^{2}=\mathbf{I},$ so the middle terms are $-3\mathbf{A}+2\mathbf{A}=-\mathbf{A}$ and $$ \begin{aligned} &(\mathbf{A}+2\mathbf{I})(\mathbf{A}-3\mathbf{I})\\ &=\mathbf{A}^{2}-\mathbf{A}-6\mathbf{I}\,. \end{aligned} $$
(b) Four terms, and they stay four. $$ \begin{aligned} (\mathbf{A}-\mathbf{B})^{2} &=\mathbf{A}^{2}-\mathbf{A}\mathbf{B}\\ &\quad-\mathbf{B}\mathbf{A}+\mathbf{B}^{2}\,. \end{aligned} $$ In (a) the middle terms are $\mathbf{A}\mathbf{I}$ and $\mathbf{I}\mathbf{A},$ which are equal by Fact 13.4: the identity commutes with every square matrix. In (b) they are $\mathbf{A}\mathbf{B}$ and $\mathbf{B}\mathbf{A},$ which in general differ.
(c) The check. $$ \begin{aligned} \mathbf{A}+2\mathbf{I}&=\begin{pmatrix}3&2\\3&2\end{pmatrix},\\ \mathbf{A}-3\mathbf{I}&=\begin{pmatrix}-2&2\\3&-3\end{pmatrix}, \end{aligned} $$ and their product is $\begin{pmatrix}0&0\\0&0\end{pmatrix}.$ The other side agrees: $\mathbf{A}^{2}=\begin{pmatrix}7&2\\3&6\end{pmatrix},$ so $\mathbf{A}^{2}-\mathbf{A}-6\mathbf{I}=\mathbf{0}.$
(c) What it shows. Neither $\mathbf{A}+2\mathbf{I}$ nor $\mathbf{A}-3\mathbf{I}$ is the zero matrix, yet their product is: failure (2) of Watch out 13.5. For numbers, $(a+2)(a-3)=0$ would force $a=-2$ or $a=3;$ here $\mathbf{A}$ is neither $-2\mathbf{I}$ nor $3\mathbf{I}.$
Answer
(a) $\mathbf{A}^{2}-\mathbf{A}-6\mathbf{I}.$
(b) $\mathbf{A}^{2}-\mathbf{A}\mathbf{B}-\mathbf{B}\mathbf{A}+\mathbf{B}^{2};$ $\mathbf{I}$ commutes with $\mathbf{A},$ while $\mathbf{B}$ need not.
(c) The product is $\mathbf{0},$ with neither factor zero.
How did it go?
Stuck? Read 1. Rules for matrix multiplication, and the identity →
Three rules that do not survive
Read the theory: 2. Three rules that do not survive →Let $$\mathbf{N}=\begin{pmatrix}2&-4\\1&-2\end{pmatrix}.$$
(a) Compute $\mathbf{N}^{2}.$ Which number has the same property among the real numbers?
(b) Find two different $2\times2$ matrices $\mathbf{C}$ and $\mathbf{D}$ with $\mathbf{N}\mathbf{C}=\mathbf{N}\mathbf{D}.$
(c) Which two failures of Watch out 13.5 have you shown?
Hint
For (b), try $\mathbf{C}=\mathbf{I}$ and $\mathbf{D}=\mathbf{I}+\mathbf{N},$ and use (a).
(a) Row by column. $$ \begin{aligned} \mathbf{N}^{2} &=\begin{pmatrix}4-4&-8+8\\2-2&-4+4\end{pmatrix}\\ &=\begin{pmatrix}0&0\\0&0\end{pmatrix}. \end{aligned} $$ Among real numbers only $0$ has $a^{2}=0;$ here $\mathbf{N}\ne\mathbf{0}.$
(b) Two different matrices. With $\mathbf{C}=\mathbf{I}$ and $\mathbf{D}=\mathbf{I}+\mathbf{N}=\begin{pmatrix}3&-4\\1&-1\end{pmatrix},$ $$ \begin{aligned} \mathbf{N}\mathbf{D}&=\mathbf{N}+\mathbf{N}^{2}=\mathbf{N}\,,\\ \mathbf{N}\mathbf{C}&=\mathbf{N}\mathbf{I}=\mathbf{N}\,. \end{aligned} $$ So $\mathbf{N}\mathbf{C}=\mathbf{N}\mathbf{D}$ although $\mathbf{C}\ne\mathbf{D}$ (and $\mathbf{N}\ne\mathbf{0}$).
(c) Failures (2) and (3). $\mathbf{N}\mathbf{N}=\mathbf{0}$ with neither factor zero is failure (2); $\mathbf{N}$ cannot be cancelled from $\mathbf{N}\mathbf{C}=\mathbf{N}\mathbf{D}$ is failure (3). Any $\mathbf{D}=\mathbf{I}+k\mathbf{N}$ works as well.
Answer
(a) $\mathbf{N}^{2}=\mathbf{0},$ although $\mathbf{N}\ne\mathbf{0}.$
(b) For example $\mathbf{C}=\mathbf{I}$ and $\mathbf{D}=\begin{pmatrix}3&-4\\1&-1\end{pmatrix}{:}$ both products are $\mathbf{N}.$
(c) Failure (2) and failure (3).
How did it go?
Stuck? Read 2. Three rules that do not survive →
The transpose
Read the theory: 3. The transpose →(a) Show that $(\mathbf{A}\mathbf{B}\mathbf{C})^{\top}=\mathbf{C}^{\top}\mathbf{B}^{\top}\mathbf{A}^{\top}$ whenever the product $\mathbf{A}\mathbf{B}\mathbf{C}$ exists.
(b) Let $\mathbf{A}$ be $2\times3,$ $\mathbf{B}$ be $3\times4$ and $\mathbf{C}$ be $4\times1.$ Give the orders of $\mathbf{A}\mathbf{B}\mathbf{C},$ of $(\mathbf{A}\mathbf{B}\mathbf{C})^{\top}$ and of the three factors of $\mathbf{C}^{\top}\mathbf{B}^{\top}\mathbf{A}^{\top}.$
(c) Show that $\mathbf{S}=\mathbf{A}+\mathbf{A}^{\top}$ is symmetric for every square $\mathbf{A},$ and compute it for $\mathbf{A}=\begin{pmatrix}1&4\\-2&3\end{pmatrix}.$
Hint
For (a), read $\mathbf{A}\mathbf{B}\mathbf{C}$ as $(\mathbf{A}\mathbf{B})\mathbf{C}$ (associativity) and use $(\mathbf{P}\mathbf{Q})^{\top}=\mathbf{Q}^{\top}\mathbf{P}^{\top}$ twice.
(a) The rule twice. With $\mathbf{P}=\mathbf{A}\mathbf{B}$ and $\mathbf{Q}=\mathbf{C},$ $$ \begin{aligned} (\mathbf{A}\mathbf{B}\mathbf{C})^{\top} &=\bigl((\mathbf{A}\mathbf{B})\mathbf{C}\bigr)^{\top}\\ &=\mathbf{C}^{\top}(\mathbf{A}\mathbf{B})^{\top}\\ &=\mathbf{C}^{\top}\mathbf{B}^{\top}\mathbf{A}^{\top}\,. \end{aligned} $$
(b) The orders. $\mathbf{A}\mathbf{B}\mathbf{C}$ is $(2\times3)(3\times4)(4\times1),$ that is $2\times1,$ and its transpose is $1\times2.$ The factors $\mathbf{C}^{\top},$ $\mathbf{B}^{\top},$ $\mathbf{A}^{\top}$ are $1\times4,$ $4\times3,$ $3\times2{:}$ they meet in the middle, and the ends give $1\times2$ again. The original order, $\mathbf{A}^{\top}\mathbf{B}^{\top}\mathbf{C}^{\top},$ would be $(3\times2)(4\times3)(1\times4),$ which does not exist.
(c) Symmetric. By Fact 13.9, $$ \begin{aligned} \mathbf{S}^{\top}&=\mathbf{A}^{\top}+(\mathbf{A}^{\top})^{\top}\\ &=\mathbf{A}^{\top}+\mathbf{A}=\mathbf{S}\,. \end{aligned} $$ For the given matrix, $$ \begin{aligned} \mathbf{S}&=\begin{pmatrix}1+1&4-2\\-2+4&3+3\end{pmatrix}\\ &=\begin{pmatrix}2&2\\2&6\end{pmatrix}, \end{aligned} $$ which is indeed unchanged by reflection in its main diagonal.
Answer
(a) $(\mathbf{A}\mathbf{B}\mathbf{C})^{\top}=\mathbf{C}^{\top}(\mathbf{A}\mathbf{B})^{\top}=\mathbf{C}^{\top}\mathbf{B}^{\top}\mathbf{A}^{\top}.$
(b) $2\times1$ and $1\times2;$ the factors are $1\times4,$ $4\times3,$ $3\times2.$
(c) $\mathbf{S}^{\top}=\mathbf{S};$ here $\mathbf{S}=\begin{pmatrix}2&2\\2&6\end{pmatrix}.$
How did it go?
Stuck? Read 3. The transpose →
Gaussian elimination
Read the theory: 4. Gaussian elimination →Solve by Gaussian elimination, naming every row operation, and check your answer: $$ \begin{aligned} 2y+3z&=1\\ x+y+z&=2\\ 2x+3y-z&=9\,. \end{aligned} $$
Hint
The first equation has no $x,$ so it cannot give the first leading entry. Start by interchanging the first two equations (operation (1) of Definition 13.13).
Swap, then clear the first column. Interchange rows $1$ and $2;$ then subtract $2\times$ row $1$ from row $3{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c} 0&2&3&1\\ 1&1&1&2\\ 2&3&-1&9 \end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c} 1&1&1&2\\ 0&2&3&1\\ 2&3&-1&9 \end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c} 1&1&1&2\\ 0&2&3&1\\ 0&1&-3&5 \end{array}\right) \end{aligned} $$
Second column. Interchange rows $2$ and $3,$ so that the leading entry is already $1;$ then subtract $2\times$ row $2$ from row $3{:}$ $$ \begin{aligned} \sim{}&\left(\begin{array}{ccc|c} 1&1&1&2\\ 0&1&-3&5\\ 0&2&3&1 \end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c} 1&1&1&2\\ 0&1&-3&5\\ 0&0&9&-9 \end{array}\right) \end{aligned} $$
Back substitution. Divide row $3$ by $9{:}$ $z=-1.$ Row $2{:}$ $y=5+3z=2.$ Row $1{:}$ $x=2-y-z=2-2+1=1.$
Check. $2\cdot2+3\cdot(-1)=1,$ $1+2-1=2,$ $2+6+1=9.$ All three equations hold.
Answer
$(x,y,z)=(1,2,-1),$ the only solution.
How did it go?
Stuck? Read 4. Gaussian elimination →
No solution, one, or infinitely many
Read the theory: 5. No solution, one, or infinitely many →Each array is the finished augmented matrix of a system in $x,$ $y,$ $z.$ How many solutions has each system? Give all of them.
(a) $$\left(\begin{array}{ccc|c}1&0&2&1\\0&1&-1&3\\0&0&0&0\end{array}\right)$$
(b) $$\left(\begin{array}{ccc|c}1&3&0&2\\0&0&1&4\\0&0&0&1\end{array}\right)$$
(c) $$\left(\begin{array}{ccc|c}1&0&0&-1\\0&1&0&0\\0&0&1&5\end{array}\right)$$
Hint
Use Fact 13.20 in its order: first look for a row $(0\;0\;0\mid c)$ with $c\ne0;$ then count the columns with a leading entry.
(a) Infinitely many. The last row is $0=0,$ so there is no contradiction. Leading entries in columns $1$ and $2;$ column $3$ has none, so $z$ is free. With $z=t{:}$ $y=3+t$ and $x=1-2t.$ All solutions: $(x,y,z)=(1-2t,\;3+t,\;t),$ $t\in\mathbb{R}.$
(b) None. The last row reads $0x+0y+0z=1,$ which no numbers satisfy. (The first two rows alone would have left $y$ free; that does not matter once one row is a contradiction.)
(c) Exactly one. No contradiction, and a leading entry in all three columns: $(x,y,z)=(-1,0,5).$
Answer
(a) Infinitely many: $(1-2t,\;3+t,\;t),$ $t\in\mathbb{R}.$
(b) None.
(c) Exactly one: $(-1,0,5).$
How did it go?
Stuck? Read 5. No solution, one, or infinitely many →
The shape of the solution set
Read the theory: 6. The shape of the solution set →Consider the system in $x_{1},x_{2},x_{3},x_{4}{:}$ $$ \begin{aligned} x_{1}+3x_{2}-x_{3}+2x_{4}&=4\\ 2x_{1}+6x_{2}-x_{3}+5x_{4}&=11\,. \end{aligned} $$
(a) Find all solutions by Gaussian elimination. How many degrees of freedom are there?
(b) Write the solution set as $\mathbf{s}+\alpha\mathbf{r}_{1}+\beta\mathbf{r}_{2},$ $\alpha,\beta\in\mathbb{R},$ and check that $\mathbf{s}$ solves the system and that $\mathbf{r}_{1},$ $\mathbf{r}_{2}$ solve the homogeneous system.
Hint
After one row operation the leading entries sit in columns $1$ and $3.$ Two columns are free: call $x_{2}=\alpha$ and $x_{4}=\beta.$
(a) Eliminate. Subtract $2\times$ row $1$ from row $2{:}$ $$ \begin{aligned} &\left(\begin{array}{cccc|c} 1&3&-1&2&4\\ 2&6&-1&5&11 \end{array}\right)\\ \sim{}&\left(\begin{array}{cccc|c} 1&3&-1&2&4\\ 0&0&1&1&3 \end{array}\right)\\ \sim{}&\left(\begin{array}{cccc|c} 1&3&0&3&7\\ 0&0&1&1&3 \end{array}\right) \end{aligned} $$ (the last step adds row $2$ to row $1,$ clearing above the leading entry).
(a) Read off. Leading entries in columns $1$ and $3;$ columns $2$ and $4$ have none, so $x_{2}=\alpha$ and $x_{4}=\beta$ are free: two degrees of freedom. The rows give $$ \begin{aligned} x_{3}&=3-\beta\,,\\ x_{1}&=7-3\alpha-3\beta\,. \end{aligned} $$
(b) The structure. Collect the constants, the $\alpha$ terms and the $\beta$ terms: $$ \begin{aligned} \begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix} &=\begin{pmatrix}7\\0\\3\\0\end{pmatrix} +\alpha\begin{pmatrix}-3\\1\\0\\0\end{pmatrix}\\ &\quad+\beta\begin{pmatrix}-3\\0\\-1\\1\end{pmatrix}. \end{aligned} $$
(b) The check. $\mathbf{s}=(7,0,3,0){:}$ $7-3=4$ and $14-3=11.$ $\mathbf{r}_{1}=(-3,1,0,0){:}$ $-3+3=0$ and $-6+6=0.$ $\mathbf{r}_{2}=(-3,0,-1,1){:}$ $-3+1+2=0$ and $-6+1+5=0.$ By Fact 13.23 the set is exactly $\mathbf{s}$ plus all homogeneous solutions; with two degrees of freedom it is a plane through $\mathbf{s},$ not a line.
Answer
(a) $(x_{1},x_{2},x_{3},x_{4})=(7-3\alpha-3\beta,\;\alpha,\;3-\beta,\;\beta),$ $\alpha,\beta\in\mathbb{R}{:}$ two degrees of freedom.
(b) $\mathbf{s}=(7,0,3,0),$ $\mathbf{r}_{1}=(-3,1,0,0),$ $\mathbf{r}_{2}=(-3,0,-1,1).$
How did it go?
Stuck? Read 6. The shape of the solution set →
The vectors $\mathbf{u}=(1,2,3)$ and $\mathbf{v}=(2,0,4)$ are both solutions of a linear system $\mathbf{A}\mathbf{x}=\mathbf{b}$ in three unknowns.
(a) Find a non-zero solution of $\mathbf{A}\mathbf{x}=\mathbf{0}.$
(b) Show that $\mathbf{u}+\alpha(\mathbf{v}-\mathbf{u})$ solves $\mathbf{A}\mathbf{x}=\mathbf{b}$ for every number $\alpha,$ and give the solution for $\alpha=3.$
(c) Explain why no linear system has exactly two solutions.
Hint
Compute $\mathbf{A}(\mathbf{v}-\mathbf{u})$ with the rules of Fact 13.1. You never need to know $\mathbf{A}.$
(a) The difference. $\mathbf{h}=\mathbf{v}-\mathbf{u}=(1,-2,1),$ and $$ \mathbf{A}\mathbf{h}=\mathbf{A}\mathbf{v}-\mathbf{A}\mathbf{u} =\mathbf{b}-\mathbf{b}=\mathbf{0}\,. $$ It is not the zero vector, since $\mathbf{u}\ne\mathbf{v}.$
(b) A whole line of solutions. $$ \begin{aligned} \mathbf{A}(\mathbf{u}+\alpha\mathbf{h}) &=\mathbf{A}\mathbf{u}+\alpha\mathbf{A}\mathbf{h}\\ &=\mathbf{b}+\alpha\mathbf{0}=\mathbf{b}\,. \end{aligned} $$ For $\alpha=3{:}$ $\mathbf{u}+3\mathbf{h}=(4,-4,6).$
(c) Two give infinitely many. If a system has two different solutions $\mathbf{u}\ne\mathbf{v},$ then (b) gives one solution for every real $\alpha,$ and different $\alpha$ give different vectors because $\mathbf{h}\ne\mathbf{0}.$ So a second solution already forces infinitely many: after "none" and "exactly one", the only possibility left is "infinitely many", as Fact 13.20 says. This argument uses only linearity, not elimination.
Answer
(a) $\mathbf{v}-\mathbf{u}=(1,-2,1).$
(b) $\mathbf{A}(\mathbf{u}+\alpha\mathbf{h})=\mathbf{b}+\alpha\mathbf{0}=\mathbf{b};$ for $\alpha=3$ the solution $(4,-4,6).$
(c) Two different solutions give a whole line of them.
How did it go?
Stuck? Read 6. The shape of the solution set →
Mixed, as on the exam
A workshop makes three products, $x,$ $y$ and $z$ units a week. One unit of the first takes $1$ hour on the cutting machine and $2$ hours on the assembly line; one unit of the second takes $2$ and $1$ hours; one unit of the third takes $3$ hours on each. Both machines are to be used fully: $40$ hours of cutting and $50$ of assembly a week.
(a) Write the conditions as a system $\mathbf{A}\mathbf{x}=\mathbf{b}$ and find all solutions.
(b) Which of the solutions are possible production plans?
(c) The profit per unit is $3,$ $2$ and $6$ hundred kronor, that is $\mathbf{p}^{\top}\mathbf{x}$ with $\mathbf{p}^{\top}=(3\;\;2\;\;6).$ Which possible plan gives the largest profit, and how large is it?
Hint
Two equations in three unknowns: one unknown is free (Fact 13.20). Put $z=t.$ A plan makes sense only if no quantity is negative.
(a) The system. $$ \begin{aligned} x+2y+3z&=40\\ 2x+y+3z&=50 \end{aligned} $$ that is, $$ \mathbf{A}=\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}40\\50\end{pmatrix}. $$
(a) Eliminate. Subtract $2\times$ row $1$ from row $2,$ then divide row $2$ by $-3,$ then subtract $2\times$ row $2$ from row $1{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c}1&2&3&40\\2&1&3&50\end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c}1&2&3&40\\0&-3&-3&-30\end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c}1&0&1&20\\0&1&1&10\end{array}\right) \end{aligned} $$ With $z=t{:}$ $y=10-t,$ $x=20-t.$ All solutions: $(x,y,z)=(20-t,\;10-t,\;t),$ $t\in\mathbb{R}.$ (Check: $20-t+20-2t+3t=40$ and $40-2t+10-t+3t=50.$)
(b) The plans that make sense. We need $x\ge0,$ $y\ge0,$ $z\ge0,$ that is $t\le20,$ $t\le10$ and $t\ge0.$ So the possible plans are $(20-t,\;10-t,\;t)$ with $0\le t\le10.$
(c) The profit. $$ \begin{aligned} \mathbf{p}^{\top}\mathbf{x} &=3(20-t)+2(10-t)\\ &\quad+6t\\ &=80+t\,, \end{aligned} $$ which increases with $t.$ The largest profit on $0\le t\le10$ is at $t=10{:}$ the plan $(10,0,10),$ with profit $90$ hundred kronor. (Each unit of the third product, with what it displaces, adds one hundred kronor.)
Answer
(a) $(x,y,z)=(20-t,\;10-t,\;t),$ $t\in\mathbb{R}.$
(b) Those with $0\le t\le10.$
(c) $(x,y,z)=(10,0,10),$ profit $9000$ kronor.
How did it go?
Stuck? Read 6. The shape of the solution set →
As on the exam
3 problems · 22 pProblems in exam form, with points. Solve them on paper or in the box. Then open the solution one step at a time and mark yourself against the rubric.
Matrix form and Gaussian elimination to the only solution
Consider the linear system $$ \left\{ \begin{aligned} 2x-y+3z&=13\\ x-y+2z&=9\\ 3x-2y+4z&=19\,. \end{aligned} \right. $$
- (a)2 p
Write the system in matrix form $\mathbf{A}\mathbf{v}=\mathbf{b},$ where $\mathbf{A}$ is a $3\times3$ matrix and $\mathbf{v},$ $\mathbf{b}$ are $3\times1$ matrices.
- (b)4 p
Solve the system using Gaussian elimination. Name every row operation, and check your answer.
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(a) Matrix form. With $$ \begin{aligned} \mathbf{A}&=\begin{pmatrix}2&-1&3\\1&-1&2\\3&-2&4\end{pmatrix},\\ \mathbf{v}&=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}13\\9\\19\end{pmatrix}, \end{aligned} $$ the system is $\mathbf{A}\mathbf{v}=\mathbf{b}{:}$ row $i$ of $\mathbf{A}\mathbf{v}$ is the left-hand side of equation $i.$
(b) A leading 1 first. Interchange rows $1$ and $2,$ written $R_{1}\leftrightarrow R_{2},$ so that the first leading entry is $1{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c}2&-1&3&13\\1&-1&2&9\\3&-2&4&19\end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c}1&-1&2&9\\2&-1&3&13\\3&-2&4&19\end{array}\right) \end{aligned} $$
(b) The first column. $R_{2}-2R_{1}$ and $R_{3}-3R_{1}{:}$ $$ \sim\left(\begin{array}{ccc|c}1&-1&2&9\\0&1&-1&-5\\0&1&-2&-8\end{array}\right) $$
(b) The staircase. $R_{3}-R_{2},$ then $(-1)\cdot R_{3}{:}$ $$ \begin{aligned} &\sim\left(\begin{array}{ccc|c}1&-1&2&9\\0&1&-1&-5\\0&0&-1&-3\end{array}\right)\\ &\sim\left(\begin{array}{ccc|c}1&-1&2&9\\0&1&-1&-5\\0&0&1&3\end{array}\right) \end{aligned} $$ A leading entry in every column, so there is exactly one solution (Fact 13.20).
(b) Read it off. Back substitution: $z=3;$ $y=-5+z=-2;$ $x=9+y-2z=9-2-6=1.$ (Clearing upwards instead, $R_{2}+R_{3},$ $R_{1}-2R_{3},$ then $R_{1}+R_{2},$ ends at the rows $(1\;0\;0\mid1),$ $(0\;1\;0\mid-2),$ $(0\;0\;1\mid3).$)
(b) Check. In the original equations: $$ \begin{aligned} 2+2+9&=13\,,\\ 1+2+6&=9\,,\\ 3+4+12&=19\,. \end{aligned} $$
Answer
(a) $\mathbf{A}\mathbf{v}=\mathbf{b}$ with $\mathbf{A}$ the rows $(2,-1,3),$ $(1,-1,2),$ $(3,-2,4)$ and $\mathbf{b}=(13,9,19)^{\top}.$
(b) $(x,y,z)=(1,-2,3).$
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
Your points 0 of 6 p
Common errors
- Starting with $R_{1}/2$ and carrying halves through every row. Allowed, but it is where the arithmetic slips come from; interchanging two rows first costs nothing.
- A sign slip in $R_{3}-3R_{1}{:}$ $-2-(-3)=1,$ not $-5.$ It is found only if the answer is checked in all three original equations.
- Writing a chain of arrays with no operations named. The HT26 solutions name every operation next to its arrow, and a marker cannot give credit for a step that cannot be followed.
A parameter decided by the last row
For a real number $a,$ consider the linear system $$ \left\{ \begin{aligned} x+y+z&=1\\ x+2y+3z&=2\\ 2x+3y+(a^{2}+3)z&=a+2\,. \end{aligned} \right. $$
- (a)3 p
Reduce the augmented matrix of the system to a staircase by Gaussian elimination, naming every row operation.
- (b)3 p
For which values of $a$ does the system have exactly one solution? Find that solution.
- (c)4 p
For each of the remaining values of $a,$ determine whether the system has no solution or infinitely many. Whenever there are infinitely many, give all of them.
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(a) The first column. $R_{2}-R_{1}$ and $R_{3}-2R_{1}{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c}1&1&1&1\\1&2&3&2\\2&3&a^{2}+3&a+2\end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c}1&1&1&1\\0&1&2&1\\0&1&a^{2}+1&a\end{array}\right) \end{aligned} $$
(a) The second column. $R_{3}-R_{2}{:}$ $$ \sim\left(\begin{array}{ccc|c}1&1&1&1\\0&1&2&1\\0&0&a^{2}-1&a-1\end{array}\right) $$ The last row says $(a^{2}-1)z=a-1,$ that is $(a-1)(a+1)z=a-1.$
(b) Exactly one. If $a^{2}-1\ne0,$ that is $a\ne1$ and $a\ne-1,$ the third column has a leading entry, every column does, and there is exactly one solution (Fact 13.20). Only now may we divide by $a^{2}-1{:}$ $$ z=\frac{a-1}{(a-1)(a+1)}=\frac{1}{a+1}\,. $$
(b) Back substitution. Row $2{:}$ $$ \begin{aligned} y&=1-2z=1-\frac{2}{a+1}\\ &=\frac{a-1}{a+1}\,. \end{aligned} $$ Row $1{:}$ $$ \begin{aligned} x&=1-y-z\\ &=\frac{(a+1)-(a-1)-1}{a+1}\\ &=\frac{1}{a+1}\,. \end{aligned} $$
(c) $a=-1.$ The last row is $(0\;0\;0\mid-2){:}$ it says $0=-2,$ which no numbers satisfy. No solution (Watch out 13.19).
(c) $a=1.$ The last row is $(0\;0\;0\mid0),$ that is $0=0{:}$ it is dropped. Leading entries in columns $1$ and $2;$ column $3$ has none, so $z$ is free. Infinitely many solutions.
(c) All of them. Put $z=t.$ Row $2$ gives $y=1-2t,$ and row $1$ gives $x=1-y-z=1-(1-2t)-t=t.$ So $$ \begin{aligned} &(x,y,z)=(t,\;1-2t,\;t)\,,\\ &t\in\mathbb{R}\,. \end{aligned} $$
(c) Check in the original system with $a=1$ (third equation $2x+3y+4z=3$): $$ \begin{aligned} t+(1-2t)+t&=1\,,\\ t+2(1-2t)+3t&=2\,,\\ 2t+3(1-2t)+4t&=3\,. \end{aligned} $$ As Fact 13.23 says, the set is the particular solution $(0,1,0)$ plus all multiples $t\,(1,-2,1)$ of a homogeneous solution: a line.
Answer
(a) The staircase with last row $(0\;0\;a^{2}-1\mid a-1).$
(b) Exactly one solution for $a\ne\pm1{:}$ $(x,y,z)=\Bigl(\dfrac{1}{a+1},\;\dfrac{a-1}{a+1},\;\dfrac{1}{a+1}\Bigr).$
(c) $a=-1{:}$ no solution. $a=1{:}$ infinitely many, $(x,y,z)=(t,\;1-2t,\;t),$ $t\in\mathbb{R}.$
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
Your points 0 of 10 p
Common errors
- Dividing the last row by $a^{2}-1$ (or cancelling $a-1$) at once, and so never seeing that $a=1$ and $a=-1$ behave differently. Split off the zero case before dividing.
- Reading the zero row at $a=1$ as "no solution". A row $(0\;0\;0\mid0)$ is a repeated equation; only $(0\;0\;0\mid c)$ with $c\ne0$ is fatal. Example 13.18 and Watch out 13.19 show the two rows side by side.
- Setting the free unknown to $0$ and giving the single solution $(0,1,0)$ for $a=1.$ A column without a leading entry says nothing about $z;$ the answer is the whole line.
- Answering "infinitely many" for both remaining values because the left-hand side is the same. Whether solutions exist depends on the right-hand side.
Four equations in three unknowns, and two
Find all solutions of each system. Name every row operation.
- (a)3 p$$ \left\{ \begin{aligned} x-y+2z&=3\\ x-z&=1\\ 2x-y+z&=4\\ 3x-4y+9z&=11\,. \end{aligned} \right. $$
- (b)3 p
$$ \left\{ \begin{aligned} x+3y-z&=2\\ 2x+7y-z&=7\,. \end{aligned} \right. $$ Explain also, before solving, why this system cannot have exactly one solution.
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(a) Eliminate. $R_{2}-R_{1},$ $R_{3}-2R_{1},$ $R_{4}-3R_{1}{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c}1&-1&2&3\\1&0&-1&1\\2&-1&1&4\\3&-4&9&11\end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c}1&-1&2&3\\0&1&-3&-2\\0&1&-3&-2\\0&-1&3&2\end{array}\right) \end{aligned} $$ then $R_{3}-R_{2}$ and $R_{4}+R_{2}{:}$ $$ \sim\left(\begin{array}{ccc|c}1&-1&2&3\\0&1&-3&-2\\0&0&0&0\\0&0&0&0\end{array}\right) $$
(a) Read it off. Two rows say $0=0{:}$ no contradiction, and two equations were consequences of the others. Leading entries in columns $1$ and $2;$ $z$ is free. With $z=t{:}$ $y=-2+3t$ and $x=3+y-2z=3-2+3t-2t=1+t.$ $$ \begin{aligned} &(x,y,z)=(1+t,\;-2+3t,\;t)\,,\\ &t\in\mathbb{R}\,. \end{aligned} $$
(b) Before solving. Two equations give at most two leading entries, and there are three unknown columns, so at least one column has none: the system has no solution or infinitely many, never exactly one (Fact 13.20 and the remark after it).
(b) Eliminate. $R_{2}-2R_{1},$ then $R_{1}-3R_{2}{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c}1&3&-1&2\\2&7&-1&7\end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c}1&3&-1&2\\0&1&1&3\end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c}1&0&-4&-7\\0&1&1&3\end{array}\right) \end{aligned} $$ No contradiction; $z$ is free. With $z=t{:}$ $y=3-t$ and $x=-7+4t.$ $$ \begin{aligned} &(x,y,z)=(-7+4t,\;3-t,\;t)\,,\\ &t\in\mathbb{R}\,. \end{aligned} $$
(b) Check. $$ \begin{aligned} (-7+4t)+3(3-t)-t&=2\,,\\ 2(-7+4t)+7(3-t)-t&=7\,. \end{aligned} $$ (In (a) the same check: for instance the fourth equation, $3(1+t)-4(-2+3t)+9t=11.$)
Answer
(a) $(x,y,z)=(1+t,\;-2+3t,\;t),$ $t\in\mathbb{R}.$
(b) Not exactly one: two rows carry at most two leading entries for three columns. $(x,y,z)=(-7+4t,\;3-t,\;t),$ $t\in\mathbb{R}.$
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
Your points 0 of 6 p
Common errors
- In (a): "four equations and three unknowns, so the system is overdetermined and has no solution". Counting equations decides nothing; only a row $0=c$ with $c\ne0$ means no solution.
- In (b): "two equations cannot determine three unknowns, so there is no solution". In the live vote on a finished array of this kind more than a third gave that answer after discussion. Too few equations means too many solutions, or none if a row is contradictory — never because of the count.
- Writing $z=0$ for the free unknown and giving one solution. The answer is every $t,$ and the check must be done with $t$ left general.