15 Problems
One array, four questions
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. Those with a solution are for paper. Write the whole solution before you open it — every row operation named next to its arrow — 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 here are numbers; fractions may be typed as -5/4 or as -1.25. A matrix is typed row by row, with a semicolon between the rows, for example 1 2; 3 4.
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The method, and where the marks go
Read the theory: 1. The method, and where the marks go →A student solves $$ \begin{aligned} x + 2y + z &= 8\\ 2x + y + z &= 7\\ x + y + 2z &= 9 \end{aligned} $$ and writes three steps:
(i) $R_2\to R_2-2R_1$;
(ii) "interchange columns $2$ and $3,$ so that the second row starts with $-1$";
(iii) $R_3\to 0\cdot R_3,$ "since the third equation is not needed".
(a) Which of the three steps are elementary row operations? For each one that is not, say what it does to the system.
(b) Solve the system correctly, naming every row operation, and check your answer.
Hint
The three operations are: interchange two rows, multiply a row by a number $\alpha\ne0,$ add a multiple of one row to a different row. Ask of each step whether it acts on rows and whether it can be undone.
(a) Step (i). Adding $-2$ times row $1$ to row $2$ is the third operation. It is allowed, and gives the row $(0\;\;{-3}\;\;{-1}\mid{-9}).$
(a) Step (ii). Not a row operation. Columns $2$ and $3$ hold the coefficients of $y$ and $z,$ so swapping them renames the unknowns: the new array stands for a system whose second unknown is $z$ and third is $y.$ Every answer read off afterwards is attached to the wrong letter unless it is swapped back (Watch out 15.1).
(a) Step (iii). Not allowed: multiplying by $\alpha=0$ is the one scaling that cannot be undone. It deletes the third equation, and two equations in three unknowns have infinitely many solutions — the solution set has grown.
(b) The first column. $R_2-2R_1$ and $R_3-R_1{:}$ $$ \left(\begin{array}{ccc|c} 1 & 2 & 1 & 8\\ 0 & -3 & -1 & -9\\ 0 & -1 & 1 & 1 \end{array}\right) $$
(b) The second column. $R_2\leftrightarrow R_3,$ so that the leading entry is $-1,$ then $R_3-3R_2{:}$ $$ \left(\begin{array}{ccc|c} 1 & 2 & 1 & 8\\ 0 & -1 & 1 & 1\\ 0 & 0 & -4 & -12 \end{array}\right) $$ The row swap does what step (ii) wanted, legally.
(b) Back substitution. $-4z=-12$ gives $z=3;$ $-y+z=1$ gives $y=2;$ $x=8-2y-z=1.$
Check. $1+4+3=8,$ $2+2+3=7,$ $1+2+6=9.$
Answer
(a) Only (i). Step (ii) renames $y$ and $z;$ step (iii) deletes an equation and cannot be undone.
(b) $(x,y,z)=(1,2,3).$
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Stuck? Read 1. The method, and where the marks go →
Reading the finished array
Read the theory: 2. Reading the finished array →Each array below is the end of an elimination. Give the verdict (none, exactly one, infinitely many) and, where there are solutions, all of them.
(a) Unknowns $x_1,x_2,x_3,x_4{:}$ $$ \left(\begin{array}{cccc|c} 1 & 2 & 0 & -1 & 3\\ 0 & 0 & 1 & 2 & 1\\ 0 & 0 & 0 & 0 & 0 \end{array}\right) $$
(b) Unknowns $x_1,x_2,x_3{:}$ $$ \left(\begin{array}{ccc|c} 1 & 0 & 1 & 2\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 4 \end{array}\right) $$
(c) Unknowns $x_1,x_2,x_3{:}$ $$ \left(\begin{array}{ccc|c} 1 & 0 & 0 & 2\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{array}\right) $$
Hint
First look for a row $(0\;\cdots\;0\mid c)$ with $c\ne0.$ Then mark the columns of the unknowns that carry a leading entry. The column after the rule is not an unknown.
(a) The verdict. No contradictory row. Leading entries in columns $1$ and $3;$ columns $2$ and $4$ have none, so $x_2$ and $x_4$ are free: infinitely many solutions, two degrees of freedom.
(a) All solutions. Put $x_2=s$ and $x_4=t.$ The rows say $x_1+2x_2-x_4=3$ and $x_3+2x_4=1,$ so $$ \begin{aligned} x_1 &= 3-2s+t\,,\\ x_3 &= 1-2t\,. \end{aligned} $$ Check with $s=1,$ $t=1{:}$ $(2,1,-1,1)$ gives $2+2-1=3$ and $-1+2=1.$
(b) The last row reads $0=4.$ No solutions, whatever the other rows say.
(c) No contradictory row, and a leading entry in each of the three columns of unknowns: exactly one solution, $(x_1,x_2,x_3)=(2,-1,0).$ The fourth row $0=0$ only repeats information (four equations, three unknowns). And $x_3=0$ here is a value, written by a leading entry — not a blank column.
Answer
(a) Infinitely many: $(x_1,x_2,x_3,x_4)=(3-2s+t,\,s,\,1-2t,\,t),$ $s,t\in\mathbb{R}.$
(b) None.
(c) Exactly one: $(2,-1,0).$
How did it go?
Stuck? Read 2. Reading the finished array →
Finding the inverse by row operations
Read the theory: 3. Finding the inverse by row operations →Let $$ \mathbf{A}=\begin{pmatrix}2&1&0\\1&1&1\\0&1&3\end{pmatrix}. $$ (a) Find $\mathbf{A}^{-1}$ with Method 15.3, naming every row operation, and check your answer with one product.
(b) Use it to solve $\mathbf{A}\mathbf{x}=(3,2,4).$
Hint
A leading entry $1$ is easier than $2:$ interchange rows $1$ and $2$ first. Then clear below, scale, and clear above until the left block is $\mathbf{I}.$
(a) A swap, then the first column. $R_1\leftrightarrow R_2,$ then $R_2-2R_1{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 1 & 0\\ 0 & -1 & -2 & 1 & -2 & 0\\ 0 & 1 & 3 & 0 & 0 & 1 \end{array}\right) $$
(a) The staircase. $(-1)\cdot R_2,$ then $R_3-R_2{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 1 & 0\\ 0 & 1 & 2 & -1 & 2 & 0\\ 0 & 0 & 1 & 1 & -2 & 1 \end{array}\right) $$
(a) Clear above. $R_2-2R_3$ and $R_1-R_3,$ then $R_1-R_2{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 2 & -3 & 1\\ 0 & 1 & 0 & -3 & 6 & -2\\ 0 & 0 & 1 & 1 & -2 & 1 \end{array}\right) $$
(a) Check. In $\mathbf{A}\mathbf{A}^{-1},$ row $1$ of $\mathbf{A}$ times the columns of $\mathbf{A}^{-1}$ gives $4-3=1,$ $-6+6=0,$ $2-2=0;$ rows $2$ and $3$ give $(0,1,0)$ and $(0,0,1)$ in the same way. One side is enough (Fact 14.21).
(b) $\mathbf{x}=\mathbf{A}^{-1}(3,2,4){:}$ $$ \begin{aligned} x_1&=6-6+4=4\,,\\ x_2&=-9+12-8=-5\,,\\ x_3&=3-4+4=3\,. \end{aligned} $$ Check: $8-5=3,$ $4-5+3=2,$ $-5+9=4.$
Answer
(a) $\mathbf{A}^{-1}=\begin{pmatrix}2&-3&1\\-3&6&-2\\1&-2&1\end{pmatrix}.$
(b) $\mathbf{x}=(4,-5,3).$
How did it go?
Stuck? Read 3. Finding the inverse by row operations →
Let $$ \mathbf{B}=\begin{pmatrix}1&2&0\\2&5&1\\1&3&k\end{pmatrix}. $$ (a) Reduce $(\mathbf{B}\mid\mathbf{I})$ as far as you can without knowing $k.$ For which $k$ does a row of the left block become zero?
(b) Finish the reduction for $k=2$ and give $\mathbf{B}^{-1}.$
(c) Explain, without computing a determinant, why a zero row in the left block means $|\mathbf{B}|=0.$ Then confirm by computing $|\mathbf{B}|.$
Hint
Two operations clear the first column; one more clears the second. For (c): a zero row leaves a column of the left block without a leading entry. What does that say about the system $\mathbf{B}\mathbf{x}=\mathbf{0}$?
(a) Reduce. $R_2-2R_1$ and $R_3-R_1,$ then $R_3-R_2{:}$ $$ \small \left(\begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & -2 & 1 & 0\\ 0 & 0 & k-1 & 1 & -1 & 1 \end{array}\right) $$ The third row of the left block is zero exactly when $k=1.$ Then the method stops: $\mathbf{B}$ has no inverse.
(b) $k=2.$ The third row is already $(0\;\;0\;\;1\mid1\;\;{-1}\;\;1).$ $R_2-R_3,$ then $R_1-2R_2{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 7 & -4 & 2\\ 0 & 1 & 0 & -3 & 2 & -1\\ 0 & 0 & 1 & 1 & -1 & 1 \end{array}\right) $$ Check: row $1$ of $\mathbf{B}$ times column $1$ is $7-6=1,$ times column $2$ is $-4+4=0.$
(c) The argument. The row operations act on the systems $\mathbf{B}\mathbf{x}=\mathbf{0}$ as well. A zero row in the left block leaves at most two leading entries for three columns, so $\mathbf{B}\mathbf{x}=\mathbf{0}$ has a free unknown and infinitely many solutions. It has $\mathbf{x}=\mathbf{0}$ and others, so not exactly one; by Fact 14.26 that is $|\mathbf{B}|=0.$
(c) The determinant. Expanding along the first row, $$ \begin{aligned} |\mathbf{B}| &= 1\cdot(5k-3)\\ &\quad-2\cdot(2k-1)+0\\ &= k-1\,, \end{aligned} $$ which is $0$ exactly for $k=1$ — the value found in (a).
Answer
(a) A zero row exactly for $k=1.$
(b) $\mathbf{B}^{-1}=\begin{pmatrix}7&-4&2\\-3&2&-1\\1&-1&1\end{pmatrix}.$
(c) A zero row means $\mathbf{B}\mathbf{x}=\mathbf{0}$ has a free unknown, so not exactly one solution, so $|\mathbf{B}|=0;$ indeed $|\mathbf{B}|=k-1.$
How did it go?
Stuck? Read 3. Finding the inverse by row operations →
Many right-hand sides at once
Read the theory: 4. Many right-hand sides at once →Let $$ \mathbf{A}=\begin{pmatrix}1&2&1\\0&1&1\\1&2&2\end{pmatrix}. $$ (a) Solve $\mathbf{A}\mathbf{x}=\mathbf{b}$ for the three right-hand sides $(4,2,5),$ $(3,1,4)$ and $(0,0,1)$ in one reduction.
(b) One of your three answers is a column of $\mathbf{A}^{-1}.$ Which one, and which column?
Hint
Write all three right-hand sides after the rule, in a $3\times6$ array, and reduce once. The operations are chosen by the left block alone.
(a) One array. $R_3-R_1$ clears the first column at once: $$ \left(\begin{array}{ccc|ccc} 1 & 2 & 1 & 4 & 3 & 0\\ 0 & 1 & 1 & 2 & 1 & 0\\ 0 & 0 & 1 & 1 & 1 & 1 \end{array}\right) $$ A leading entry in every column of the left block: each system has exactly one solution.
(a) Clear above. $R_2-R_3$ and $R_1-R_3,$ then $R_1-2R_2{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 2 & 1\\ 0 & 1 & 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 1 & 1 & 1 \end{array}\right) $$
(a) Read off by columns. $(1,1,1),$ $(2,0,1)$ and $(1,-1,1).$ Check the second: $2+0+1=3,$ $0+1=1,$ $2+0+2=4.$
(b) The third right-hand side is $\mathbf{e}_3,$ the third column of $\mathbf{I}.$ Its solution is the third column of $\mathbf{A}^{-1}$ (Method 15.3 is this reduction with $\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$ as right-hand sides).
Answer
(a) $(1,1,1),$ $(2,0,1)$ and $(1,-1,1).$
(b) $(1,-1,1),$ the solution for $\mathbf{e}_3,$ is the third column of $\mathbf{A}^{-1}.$
How did it go?
Stuck? Read 4. Many right-hand sides at once →
Which tool for which question
Read the theory: 5. Which tool for which question →Consider $$ \begin{aligned} x + y + z &= 2\\ x + 2y + 3z &= 3\\ 2x + 3y + az &= c \end{aligned} $$ (a) For which $a$ has the system exactly one solution, whatever $c$ is?
(b) Let $a$ be the remaining value. For which $c$ are there solutions? Give all of them.
Hint
(a) is a question for the determinant (Fact 15.7). In (b) the determinant has said all it can: eliminate, and read the last row as a function of $c.$
(a) Predict. Expanding along the first row, $$ \begin{aligned} |\mathbf{A}|&=(2a-9)-(a-6)\\ &\quad+(3-4)\\ &=a-4\,. \end{aligned} $$ Exactly one solution for every right-hand side when $a\ne4.$
(b) Decide. For $a=4,$ $R_2-R_1$ and $R_3-2R_1,$ then $R_3-R_2{:}$ $$ \left(\begin{array}{ccc|c} 1 & 1 & 1 & 2\\ 0 & 1 & 2 & 1\\ 0 & 0 & 0 & c-5 \end{array}\right) $$ For $c\ne5$ the last row is a contradiction: no solutions. For $c=5$ it is $0=0.$
(b) All solutions for $c=5.$ $R_1-R_2$ gives the row $(1\;\;0\;\;{-1}\mid1).$ Put $z=t;$ then $y=1-2t$ and $x=1+t.$ Check in the third equation: $2(1+t)+3(1-2t)+4t=5.$
The lesson. $|\mathbf{A}|=0$ did not mean "no solutions": for one value of $c$ there are infinitely many (Watch out 15.8).
Answer
(a) $a\ne4.$
(b) $a=4{:}$ solutions only for $c=5,$ namely $(x,y,z)=(1+t,\,1-2t,\,t),$ $t\in\mathbb{R};$ none for $c\ne5.$
How did it go?
Stuck? Read 5. Which tool for which question →
Mixed, as on the exam
A workshop makes three goods from three inputs. One unit of good $j$ uses $a_{ij}$ units of input $i,$ where $$ \mathbf{A}=(a_{ij})=\begin{pmatrix}1&1&1\\1&2&1\\1&1&2\end{pmatrix}. $$ In a given week the amounts $\mathbf{b}=(b_1,b_2,b_3)$ of the inputs are available, and the plan $\mathbf{x}$ (units of each good) must use them exactly: $\mathbf{A}\mathbf{x}=\mathbf{b}.$
(a) Find $\mathbf{A}^{-1}$ by row operations.
(b) Find the plan for the three weeks with $\mathbf{b}=(10,13,14),$ $(9,12,11)$ and $(6,9,10).$
(c) One of the three plans cannot be carried out. Which, and why?
(d) Why is computing $\mathbf{A}^{-1}$ worth it here, and when would one elimination of $(\mathbf{A}\mid\mathbf{b})$ be better?
Hint
Three operations bring $(\mathbf{A}\mid\mathbf{I})$ to $(\mathbf{I}\mid\mathbf{A}^{-1})$: clear the first column, then clear above. A plan is a list of quantities produced.
(a) The inverse. $R_2-R_1$ and $R_3-R_1$ give the rows $(0\;\;1\;\;0\mid{-1}\;\;1\;\;0)$ and $(0\;\;0\;\;1\mid{-1}\;\;0\;\;1);$ then $R_1-R_2-R_3{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 3 & -1 & -1\\ 0 & 1 & 0 & -1 & 1 & 0\\ 0 & 0 & 1 & -1 & 0 & 1 \end{array}\right) $$ Check: row $1$ of $\mathbf{A}$ times column $1$ is $3-1-1=1,$ row $2$ times column $1$ is $3-2-1=0.$
(b) The plans. $\mathbf{x}=\mathbf{A}^{-1}\mathbf{b},$ that is $x_1=3b_1-b_2-b_3,$ $x_2=b_2-b_1,$ $x_3=b_3-b_1{:}$ $$ \begin{aligned} (10,13,14)&\mapsto(3,3,4)\,,\\ (9,12,11)&\mapsto(4,3,2)\,,\\ (6,9,10)&\mapsto(-1,3,4)\,. \end{aligned} $$
(c) The third week. Its plan has $x_1=-1{:}$ a negative quantity of the first good. The system has exactly one solution (the inverse exists), so no plan with all quantities $\ge0$ uses the inputs exactly; the mathematics is fine, the plan is not feasible. In general $x_1\ge0$ needs $3b_1\ge b_2+b_3.$
(d) Why the inverse. The row operations depend on $\mathbf{A}$ only. Once $\mathbf{A}^{-1}$ is known, every week costs one product, and the formulas in (b) even say in advance which input combinations are feasible. For a single week one elimination of $(\mathbf{A}\mid\mathbf{b})$ is less work; the inverse pays when the same $\mathbf{A}$ is used again.
Answer
(a) $\mathbf{A}^{-1}=\begin{pmatrix}3&-1&-1\\-1&1&0\\-1&0&1\end{pmatrix}.$
(b) $(3,3,4),$ $(4,3,2),$ $(-1,3,4).$
(c) The third: it needs $-1$ units of the first good.
(d) One inverse serves every week; for one week, eliminate.
How did it go?
Stuck? Read 4. Many right-hand sides at once →
For a real number $a,$ consider $$ \begin{aligned} x + ay + z &= 1\\ x + 3y + az &= 1\\ ax + 3y + z &= 2 \end{aligned} $$ (a) Compute the determinant of the coefficient matrix and factor it completely. For which $a$ is there exactly one solution?
(b) For each remaining value of $a,$ decide whether there are no solutions or infinitely many, and give them all when there are infinitely many.
Hint
The determinant is a cubic in $a.$ An integer root divides the constant term (Lecture 1); try $a=1$ first. For each root, eliminate and read the last row.
(a) The determinant. Expanding along the first row, $$ \begin{aligned} |\mathbf{A}| &= 1\cdot(3-3a)-a\cdot(1-a^{2})\\ &\quad+1\cdot(3-3a)\\ &= a^{3}-7a+6\,. \end{aligned} $$
(a) Factor. $a=1$ is a root: $1-7+6=0.$ Dividing by $a-1$ (Polynomials of Lecture 1) gives $$ \begin{aligned} &a^{3}-7a+6\\ &=(a-1)(a^{2}+a-6)\\ &=(a-1)(a-2)(a+3)\,. \end{aligned} $$ Exactly one solution for $a\ne1,$ $a\ne2,$ $a\ne-3.$
(b) $a=1.$ The second and third equations read $x+3y+z=1$ and $x+3y+z=2.$ They contradict each other: no solutions.
(b) $a=2.$ $R_2-R_1$ and $R_3-2R_1,$ then $R_3+R_2{:}$ $$ \left(\begin{array}{ccc|c} 1 & 2 & 1 & 1\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 \end{array}\right) $$ The last row is $0=0.$ Put $z=t;$ then $y=-t$ and $x=1-2y-z=1+t.$ Check in the third equation: $2(1+t)-3t+t=2.$
(b) $a=-3.$ $R_2-R_1$ and $R_3+3R_1,$ then $R_3+R_2{:}$ $$ \left(\begin{array}{ccc|c} 1 & -3 & 1 & 1\\ 0 & 6 & -4 & 0\\ 0 & 0 & 0 & 5 \end{array}\right) $$ The last row is $0=5{:}$ no solutions.
Answer
(a) $|\mathbf{A}|=(a-1)(a-2)(a+3);$ exactly one solution for $a\notin\{1,2,-3\}.$
(b) $a=1{:}$ none. $a=2{:}$ infinitely many, $(x,y,z)=(1+t,\,-t,\,t),$ $t\in\mathbb{R}.$ $a=-3{:}$ none.
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As on the exam
3 problems · 24 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.
A determinant with a parameter, and every value decided
For a real number $a,$ consider the linear system $$ \begin{aligned} x - 2y + az &= -3\\ ax + y \phantom{{}+az} &= 0\\ x + y + 2z &= 6 \end{aligned} $$ and let $\mathbf{A}$ be its coefficient matrix.
- (a)3 p
Compute $|\mathbf{A}|$ and determine for which values of $a$ the system has exactly one solution.
- (b)4 p
For each of the remaining values of $a,$ determine whether the system has no solution or infinitely many solutions. Whenever there are infinitely many, give all of them.
- (c)3 p
Solve the system for $a=1$ using Gaussian elimination.
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(a) The determinant. With $$ \mathbf{A}=\begin{pmatrix}1&-2&a\\a&1&0\\1&1&2\end{pmatrix}, $$ expansion along the first row (Method 14.7) gives $$ \begin{aligned} |\mathbf{A}| &= 1\cdot(2-0)\\ &\quad-(-2)\cdot(2a-0)\\ &\quad+a\cdot(a-1)\\ &= 2+4a+a^{2}-a\\ &= a^{2}+3a+2\,. \end{aligned} $$
(a) Factor and conclude. $a^{2}+3a+2=(a+1)(a+2),$ which is zero exactly for $a=-1$ and $a=-2.$ The system has exactly one solution precisely when $|\mathbf{A}|\ne0$ (Fact 14.26), that is for $a\ne-1$ and $a\ne-2.$
(b) $a=-2.$ The determinant only says not exactly one (Watch out 15.8); eliminate. $R_2+2R_1$ and $R_3-R_1{:}$ $$ \left(\begin{array}{ccc|c} 1 & -2 & -2 & -3\\ 0 & -3 & -4 & -6\\ 0 & 3 & 4 & 9 \end{array}\right) $$ then $R_3+R_2$ gives the row $(0\;\;0\;\;0\mid3),$ that is $0=3.$ No solution.
(b) $a=-1.$ $R_2+R_1$ and $R_3-R_1{:}$ $$ \left(\begin{array}{ccc|c} 1 & -2 & -1 & -3\\ 0 & -1 & -1 & -3\\ 0 & 3 & 3 & 9 \end{array}\right) $$ then $R_3+3R_2$ gives $(0\;\;0\;\;0\mid0).$ The system is consistent and the third column has no leading entry: infinitely many solutions.
(b) All of them. $(-1)\cdot R_2$ gives $y+z=3,$ and $R_1+2R_2$ then gives $x+z=3.$ Put $z=t$ (free): $$ \begin{aligned} &(x,y,z)=(3-t,\ 3-t,\ t)\,,\\ &t\in\mathbb{R}\,. \end{aligned} $$ Check in the original system with $a=-1{:}$ $(3-t)-2(3-t)-t=-3,$ $-(3-t)+(3-t)=0,$ $(3-t)+(3-t)+2t=6.$
(c) $a=1.$ $R_2-R_1$ and $R_3-R_1,$ then $R_3-R_2{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c} 1 & -2 & 1 & -3\\ 1 & 1 & 0 & 0\\ 1 & 1 & 2 & 6 \end{array}\right)\\ &\sim \left(\begin{array}{ccc|c} 1 & -2 & 1 & -3\\ 0 & 3 & -1 & 3\\ 0 & 3 & 1 & 9 \end{array}\right)\\ &\sim \left(\begin{array}{ccc|c} 1 & -2 & 1 & -3\\ 0 & 3 & -1 & 3\\ 0 & 0 & 2 & 6 \end{array}\right) \end{aligned} $$
(c) Clear above. $\tfrac12R_3,$ then $R_2+R_3$ and $\tfrac13R_2,$ then $R_1+2R_2-R_3{:}$ $$ \left(\begin{array}{ccc|c} 1 & 0 & 0 & -2\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 3 \end{array}\right) $$ so $(x,y,z)=(-2,\,2,\,3).$ Check: $-2-4+3=-3,$ $-2+2=0,$ $-2+2+6=6.$
Answer
(a) $|\mathbf{A}|=a^{2}+3a+2=(a+1)(a+2);$ exactly one solution for $a\ne-1$ and $a\ne-2.$
(b) $a=-2{:}$ no solution. $a=-1{:}$ infinitely many, $(x,y,z)=(3-t,\,3-t,\,t),$ $t\in\mathbb{R}.$
(c) $(x,y,z)=(-2,\,2,\,3).$
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 10 p
Common errors
- Reading $|\mathbf{A}|=0$ as "no solution" for both values. At $a=-1$ there are infinitely many: the determinant says only not exactly one (Watch out 15.8).
- Setting the free unknown to zero and answering $(3,3,0)$ for $a=-1.$ That is one solution, not all of them (Watch out 15.2).
- A sign slip in the middle term of the expansion: the cofactor of $-2$ enters with the sign $-,$ so $-(-2)(2a)=+4a.$
- Elimination without the row operations written next to the arrows, or with a column swapped to get a nicer pivot (Watch out 15.1).
An inverse by row operations, used twice, and a matrix without one
Let $$ \begin{aligned} &\mathbf{A}=\begin{pmatrix}1&2&1\\1&3&1\\2&4&3\end{pmatrix},\\ &\mathbf{C}=\begin{pmatrix}1&2&1\\1&3&1\\3&7&3\end{pmatrix}. \end{aligned} $$
- (a)3 p
Find $\mathbf{A}^{-1}$ by elementary row operations on $(\mathbf{A}\mid\mathbf{I}),$ and check your answer.
- (b)2 p
Use $\mathbf{A}^{-1}$ to solve $\mathbf{A}\mathbf{x}=\mathbf{b}$ for $\mathbf{b}=(4,5,9)$ and for $\mathbf{b}=(1,0,1).$
- (c)1 p
Apply the same method to $\mathbf{C}.$ What happens, and what does it tell you about $\mathbf{C}$?
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(a) The first column. $R_2-R_1$ and $R_3-2R_1{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0\\ 0 & 1 & 0 & -1 & 1 & 0\\ 0 & 0 & 1 & -2 & 0 & 1 \end{array}\right) $$ The left block is a staircase, but not yet $\mathbf{I}.$
(a) Clear above (Watch out 15.6). $R_1-2R_2,$ then $R_1-R_3{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 5 & -2 & -1\\ 0 & 1 & 0 & -1 & 1 & 0\\ 0 & 0 & 1 & -2 & 0 & 1 \end{array}\right) $$ so $$ \mathbf{A}^{-1}=\begin{pmatrix}5&-2&-1\\-1&1&0\\-2&0&1\end{pmatrix}. $$
(a) Check. In $\mathbf{A}\mathbf{A}^{-1}$ the first row is $(5-2-2,\ -2+2,\ -1+1)=(1,0,0),$ the second $(5-3-2,\ -2+3,\ -1+1)=(0,1,0),$ the third $(10-4-6,\ -4+4,\ -2+3)=(0,0,1).$ So $\mathbf{A}\mathbf{A}^{-1}=\mathbf{I},$ and one side is enough (Fact 14.21).
(b) Two products. $\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}$ (Fact 14.24), with $\mathbf{A}^{-1}$ on the left: $$ \begin{aligned} &\mathbf{A}^{-1}(4,5,9)\\ &=(20-10-9,\\ &\qquad -4+5,\ -8+9)\\ &=(1,1,1)\,,\\[4pt] &\mathbf{A}^{-1}(1,0,1)\\ &=(5-1,\ -1,\ -2+1)\\ &=(4,-1,-1)\,. \end{aligned} $$ Check the second: $4-2-1=1,$ $4-3-1=0,$ $8-4-3=1.$
(c) $\mathbf{C}.$ $R_2-R_1$ and $R_3-3R_1,$ then $R_3-R_2{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0\\ 0 & 1 & 0 & -1 & 1 & 0\\ 0 & 0 & 0 & -2 & -1 & 1 \end{array}\right) $$ A row of the left block is zero, so it can never become $\mathbf{I}{:}$ $\mathbf{C}$ is singular, it has no inverse, and $|\mathbf{C}|=0$ (Method 15.3, step 4; its third row is twice the first plus the second). The right block means nothing (Example 15.5).
Answer
(a) $\mathbf{A}^{-1}=\begin{pmatrix}5&-2&-1\\-1&1&0\\-2&0&1\end{pmatrix}.$
(b) $(1,1,1)$ and $(4,-1,-1).$
(c) The left block gets a zero row: $\mathbf{C}$ has no inverse ($|\mathbf{C}|=0$).
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
- Stopping at the staircase after the first two operations and giving the right block $\begin{pmatrix}1&0&0\\-1&1&0\\-2&0&1\end{pmatrix}$ as the inverse (Watch out 15.6). The check $\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}$ catches it at once.
- Writing $\mathbf{x}=\mathbf{b}\mathbf{A}^{-1}:$ a $1\times3$ row times a $3\times3$ matrix is defined but is not the solution. The inverse multiplies from the left.
- For $\mathbf{C},$ reading the numbers in the right block as an inverse, or redoing the arithmetic in search of a slip. The verdict is in the left block alone (Example 15.5).
One matrix, four questions
For a real number $k,$ let $$ \mathbf{A}=\begin{pmatrix}1&1&2\\2&3&k\\1&2&3\end{pmatrix}. $$
- (a)2 p
Compute $|\mathbf{A}|.$ For which $k$ has $\mathbf{A}\mathbf{x}=\mathbf{b}$ exactly one solution for every $\mathbf{b}$?
- (b)3 p
Let $k=5.$ In one reduction, solve $\mathbf{A}\mathbf{x}=\mathbf{b}^{1}$ and $\mathbf{A}\mathbf{x}=\mathbf{b}^{2},$ where $\mathbf{b}^{1}=(1,3,2)$ and $\mathbf{b}^{2}=(0,1,0).$ For each, say how many solutions there are and give them all.
- (c)3 p
Let $k=4.$ Find $\mathbf{A}^{-1}$ by row operations, and check it.
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(a) Predict. Expanding along the first row, $$ \begin{aligned} |\mathbf{A}| &= 1\cdot(9-2k)-1\cdot(6-k)\\ &\quad+2\cdot(4-3)\\ &= 5-k\,. \end{aligned} $$ By Fact 14.26 there is exactly one solution for every $\mathbf{b}$ exactly when $k\ne5.$ For $k=5$ there is exactly one for no $\mathbf{b}$ — and the determinant says nothing more (Fact 15.7).
(b) Decide, both at once. For $k=5,$ reduce $(\mathbf{A}\mid\mathbf{b}^{1}\;\mathbf{b}^{2})$ with $R_2-2R_1$ and $R_3-R_1,$ then $R_3-R_2{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|cc} 1 & 1 & 2 & 1 & 0\\ 2 & 3 & 5 & 3 & 1\\ 1 & 2 & 3 & 2 & 0 \end{array}\right)\\ &\sim \left(\begin{array}{ccc|cc} 1 & 1 & 2 & 1 & 0\\ 0 & 1 & 1 & 1 & 1\\ 0 & 0 & 0 & 0 & -1 \end{array}\right) \end{aligned} $$
(b) Read the last row. For $\mathbf{b}^{2}$ it says $0=-1{:}$ no solution. For $\mathbf{b}^{1}$ it says $0=0,$ and the third column has no leading entry: infinitely many. $R_1-R_2$ gives the row $(1\;\;0\;\;1\mid0).$ Put $x_3=t{:}$ $$ \begin{aligned} &(x_1,x_2,x_3)=(-t,\ 1-t,\ t)\,,\\ &t\in\mathbb{R}\,. \end{aligned} $$ Check: $-t+(1-t)+2t=1,$ $-2t+3(1-t)+5t=3,$ $-t+2(1-t)+3t=2.$
(c) The inverse. For $k=4,$ $R_2-2R_1$ and $R_3-R_1,$ then $R_3-R_2{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 1 & 2 & 1 & 0 & 0\\ 0 & 1 & 0 & -2 & 1 & 0\\ 0 & 0 & 1 & 1 & -1 & 1 \end{array}\right) $$
(c) Clear above. $R_1-R_2,$ then $R_1-2R_3{:}$ $$ \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 1 & -2\\ 0 & 1 & 0 & -2 & 1 & 0\\ 0 & 0 & 1 & 1 & -1 & 1 \end{array}\right) $$ Check: the first row of $\mathbf{A}$ against the three columns gives $1-2+2=1,$ $1+1-2=0,$ $-2+0+2=0;$ the second $2-6+4=0,$ $2+3-4=1,$ $-4+0+4=0;$ the third $1-4+3=0,$ $1+2-3=0,$ $-2+0+3=1.$
Answer
(a) $|\mathbf{A}|=5-k;$ exactly one solution for every $\mathbf{b}$ when $k\ne5.$
(b) $\mathbf{b}^{1}{:}$ infinitely many, $(x_1,x_2,x_3)=(-t,\,1-t,\,t),$ $t\in\mathbb{R}.$ $\mathbf{b}^{2}{:}$ none.
(c) $\mathbf{A}^{-1}=\begin{pmatrix}1&1&-2\\-2&1&0\\1&-1&1\end{pmatrix}.$
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 8 p
Common errors
- In (b), "$|\mathbf{A}|=0,$ so neither system has a solution". The determinant decides nothing between none and infinitely many; $\mathbf{b}^{1}$ has infinitely many (Watch out 15.8).
- In (b), the free unknown read as zero, so that $(0,1,0)$ is given as the solution for $\mathbf{b}^{1}$ (Watch out 15.2).
- Two separate eliminations with different operations, and an arithmetic slip in one of them; the operations depend on $\mathbf{A}$ only, so one array is less work and less risk.
- In (c), the right block of the staircase given as $\mathbf{A}^{-1}$ (Watch out 15.6).