14 Problems
Determinants and the inverse
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Determinants of order 2
Read the theory: 1. Determinants of order 2 →Let $\mathbf{A}=\begin{pmatrix}4&1\\1&3\end{pmatrix}.$
(a) Compute $|\mathbf{A}|,$ and give the area of the parallelogram spanned by the rows $(4,1)$ and $(1,3).$
(b) What is the determinant when the two rows are interchanged? What happens to the area?
(c) For which $c$ do the rows $(4,1)$ and $(c,2)$ lie on one line? Check your answer.
Hint
The area is the absolute value of the determinant; the sign records the order of the rows. For (c), set the determinant equal to $0.$
(a) $|\mathbf{A}|=4\cdot3-1\cdot1=11.$ The parallelogram has area $|11|=11.$
(b) $\begin{vmatrix}1&3\\4&1\end{vmatrix}=1-12=-11.$ The parallelogram is the same set of points, so the area is still $11;$ only the sign of the determinant changes.
(c) $\begin{vmatrix}4&1\\c&2\end{vmatrix}=8-c,$ which is $0$ exactly when $c=8.$ Check: $(8,2)=2\cdot(4,1),$ so the second row lies on the line through the first: the parallelogram is flat.
Answer
(a) $|\mathbf{A}|=11,$ area $11.$ (b) $-11;$ the area is unchanged. (c) $c=8.$
How did it go?
Stuck? Read 1. Determinants of order 2 →
Consider the system $$ \begin{aligned} (k+1)x+2y&=3\\ 4x+(k-1)y&=-6\,. \end{aligned} $$
(a) For which $k$ does it have exactly one solution?
(b) For those $k,$ find the solution by Cramer's rule, and simplify.
(c) Decide what happens at each of the remaining values of $k.$
(d) The simplified formula of (b) makes sense at one of the remaining values. Does it give "the" solution there?
Hint
$|\mathbf{A}|=(k+1)(k-1)-8.$ Factor it, and factor the numerators of Cramer's rule too. In (c), write out the two equations for each value and compare them.
(a) The determinant. $$ \begin{aligned} |\mathbf{A}|&=(k+1)(k-1)-2\cdot4\\ &=k^{2}-9=(k-3)(k+3)\,. \end{aligned} $$ Exactly one solution if and only if $|\mathbf{A}|\ne0,$ that is for $k\ne3$ and $k\ne-3.$
(b) Cramer. Replace a column by $\mathbf{b}=(3,-6){:}$ $$ \begin{aligned} |\mathbf{A}_{1}|&=3(k-1)-2(-6)\\ &=3k+9\\ &=3(k+3)\,, \end{aligned} $$ $$ \begin{aligned} |\mathbf{A}_{2}|&=(k+1)(-6)-4\cdot3\\ &=-6k-18\\ &=-6(k+3)\,. \end{aligned} $$ So $$ x=\frac{3}{k-3}\,,\qquad y=\frac{-6}{k-3}\,. $$
(c) $k=3.$ The equations are $4x+2y=3$ and $4x+2y=-6.$ The same left-hand side cannot equal both $3$ and $-6{:}$ no solution.
(c) $k=-3.$ The equations are $-2x+2y=3$ and $4x-4y=-6.$ The second is $-2$ times the first, so only one equation is left, $y=x+\tfrac32.$ Infinitely many solutions: $(x,y)=(t,\,t+\tfrac32),$ $t\in\mathbb{R}.$
(d) A formula that hides the breakdown. At $k=-3$ the formula gives $x=-\tfrac12,$ $y=1.$ That pair is one of the infinitely many solutions, the one with $t=-\tfrac12,$ not the solution. Cancelling $k+3$ in (b) divided by $0$ at $k=-3$ without saying so. Cramer's rule needs $|\mathbf{A}|\ne0,$ and the cases with $|\mathbf{A}|=0$ must be decided by looking at the equations.
Answer
(a) $k\ne\pm3.$ (b) $x=3/(k-3),$ $y=-6/(k-3).$ (c) $k=3{:}$ no solution; $k=-3{:}$ infinitely many, $(t,\,t+\tfrac32).$ (d) No: it gives one of infinitely many.
How did it go?
Stuck? Read 1. Determinants of order 2 →
Determinants of order 3
Read the theory: 2. Determinants of order 3 →Let $$ \mathbf{A}=\begin{pmatrix}2&0&1\\ p&3&4\\ 1&0&5\end{pmatrix}, $$ where $p$ is a number.
(a) Which row or column is the cheapest to expand along, and what is the sign of the one term that survives?
(b) Compute $|\mathbf{A}|.$ Does it depend on $p?$
(c) Check the value by expanding along the first row.
Hint
Look for the row or column with the most zeros. The sign of the $(i,j)$ term is $(-1)^{i+j}.$
(a) Column 2. It is $(0,3,0),$ so only the $3$ at $(2,2)$ survives, with the sign $(-1)^{2+2}=+1.$
(b) The value. Deleting row $2$ and column $2$ leaves $$ \begin{aligned} |\mathbf{A}|&=+3\begin{vmatrix}2&1\\1&5\end{vmatrix}\\ &=3\cdot(10-1)=27\,. \end{aligned} $$ The entry $p$ is in the deleted row, so $|\mathbf{A}|=27$ for every $p.$
(c) The check. Along the first row, $$ \begin{aligned} |\mathbf{A}|&=2\begin{vmatrix}3&4\\0&5\end{vmatrix} -0+1\begin{vmatrix}p&3\\1&0\end{vmatrix}\\ &=2\cdot15+(0-3)=27\,. \end{aligned} $$ The same number, as it must be.
Answer
Column $2,$ sign $+;$ $|\mathbf{A}|=27,$ whatever $p$ is.
How did it go?
Stuck? Read 2. Determinants of order 3 →
Cramer's rule of order 3, and the determinant of a product
Read the theory: 3. Cramer's rule of order 3, and the determinant of a product →Find $z$ in the solution of $$ \begin{aligned} 2x+y&=1\\ x+3y+z&=0\\ y+2z&=3 \end{aligned} $$ by Cramer's rule, without finding $x$ and $y.$ Then find $y$ the same way and check both in an equation.
Hint
$z=|\mathbf{A}_{3}|/|\mathbf{A}|,$ where $\mathbf{A}_{3}$ is $\mathbf{A}$ with its third column replaced by the right-hand side. Write the missing coefficients as $0.$
The determinant. $\mathbf{A}=\begin{pmatrix}2&1&0\\1&3&1\\0&1&2\end{pmatrix},$ and along the first row $$ \begin{aligned} |\mathbf{A}|&=2(6-1)-1\cdot(2-0)+0\\ &=8\ne0\,, \end{aligned} $$ so Fact 14.11 applies.
z. Third column replaced by $(1,0,3){:}$ $$ \begin{aligned} |\mathbf{A}_{3}|&=\begin{vmatrix}2&1&1\\1&3&0\\0&1&3\end{vmatrix}\\ &=2\cdot9-1\cdot3\\ &\quad+1\cdot1\\ &=16\,, \end{aligned} $$ so $z=16/8=2.$
y. Second column replaced: $$ \begin{aligned} |\mathbf{A}_{2}|&=\begin{vmatrix}2&1&0\\1&0&1\\0&3&2\end{vmatrix}\\ &=2(0-3)\\ &\quad-1\cdot(2-0)+0\\ &=-8\,, \end{aligned} $$ so $y=-8/8=-1.$
Check. The third equation: $y+2z=-1+4=3.$ The first then gives $x=(1-y)/2=1,$ and the second $1-3+2=0.$
Answer
$z=2;$ also $y=-1$ and $x=1.$
How did it go?
Stuck? Read 3. Cramer's rule of order 3, and the determinant of a product →
$\mathbf{A}$ and $\mathbf{B}$ are of order $3,$ with $|\mathbf{A}|=2$ and $|\mathbf{B}|=-3.$ Compute
(a) $|\mathbf{A}\mathbf{B}^{\top}|,$
(b) $|-\mathbf{A}|,$
(c) $|\mathbf{A}^{2}\mathbf{B}|,$
(d) $|2\mathbf{B}|.$
(e) Explain, with an example, why $|\mathbf{A}+\mathbf{B}|$ cannot be computed from the given information.
Hint
Fact 14.13 for products and transposes, Watch out 14.14 for a scalar: the scalar is raised to the power $3.$
(a) $|\mathbf{A}\mathbf{B}^{\top}|=|\mathbf{A}|\cdot|\mathbf{B}^{\top}|=|\mathbf{A}|\cdot|\mathbf{B}|=2\cdot(-3)=-6.$
(b) $-\mathbf{A}=(-1)\mathbf{A},$ so $|-\mathbf{A}|=(-1)^{3}\cdot2=-2.$ (For a matrix of order $2$ the sign would not change.)
(c) $|\mathbf{A}^{2}\mathbf{B}|=|\mathbf{A}|^{2}|\mathbf{B}|=4\cdot(-3)=-12.$
(d) $|2\mathbf{B}|=2^{3}\cdot(-3)=-24.$
(e) The determinant does not respect sums. With $\mathbf{A}=\operatorname{diag}(2,1,1)$ and $\mathbf{B}=\operatorname{diag}(-3,1,1)$ (diagonal matrices) the sum is $\operatorname{diag}(-1,2,2),$ with determinant $-4;$ with $\mathbf{B}=\operatorname{diag}(1,1,-3)$ instead it is $\operatorname{diag}(3,2,-2),$ with determinant $-12.$ The same $|\mathbf{A}|$ and $|\mathbf{B}|,$ two different answers.
Answer
(a) $-6.$ (b) $-2.$ (c) $-12.$ (d) $-24.$ (e) Matrices with the same determinants can give different $|\mathbf{A}+\mathbf{B}|,$ for example $-4$ and $-12.$
How did it go?
Stuck? Read 3. Cramer's rule of order 3, and the determinant of a product →
The inverse, and when it exists
Read the theory: 4. The inverse, and when it exists →Let $\mathbf{A}$ be a square matrix.
(a) Show in two ways that if $\mathbf{A}^{2}=\mathbf{0}$ then $\mathbf{A}$ has no inverse: once with determinants, once by multiplying by $\mathbf{A}^{-1}.$
(b) Give a matrix $\mathbf{A}\ne\mathbf{0}$ of order $2$ with $\mathbf{A}^{2}=\mathbf{0}.$
(c) Show that if $\mathbf{A}$ is invertible and $\mathbf{A}^{2}=\mathbf{A},$ then $\mathbf{A}=\mathbf{I}.$
Hint
For (a), suppose $\mathbf{A}^{-1}$ exists and reach a contradiction. For (c), multiply $\mathbf{A}\mathbf{A}=\mathbf{A}$ by $\mathbf{A}^{-1}$ on one side.
(a) With determinants. By Fact 14.13 $|\mathbf{A}|^{2}=|\mathbf{A}^{2}|=|\mathbf{0}|=0,$ so $|\mathbf{A}|=0,$ and by Fact 14.18 $\mathbf{A}$ has no inverse.
(a) With the inverse. Suppose $\mathbf{A}^{-1}$ existed. Multiplying $\mathbf{A}\mathbf{A}=\mathbf{0}$ on the left by $\mathbf{A}^{-1}$ gives $$ \begin{aligned} \mathbf{A}&=\mathbf{A}^{-1}\mathbf{A}\mathbf{A}\\ &=\mathbf{A}^{-1}\mathbf{0}=\mathbf{0}\,, \end{aligned} $$ and the zero matrix has no inverse, since $\mathbf{0}\mathbf{X}=\mathbf{0}\ne\mathbf{I}$ for every $\mathbf{X}.$ A contradiction.
(b) An example. $\mathbf{A}=\begin{pmatrix}2&-4\\1&-2\end{pmatrix}$ has $$ \mathbf{A}^{2}=\begin{pmatrix}4-4&-8+8\\2-2&-4+4\end{pmatrix}=\mathbf{0}\,, $$ and $|\mathbf{A}|=-4+4=0,$ as (a) says it must. Another one is $\begin{pmatrix}0&1\\0&0\end{pmatrix}.$
(c) Cancel with the inverse. Multiply $\mathbf{A}\mathbf{A}=\mathbf{A}$ on the left by $\mathbf{A}^{-1}{:}$ $\mathbf{I}\mathbf{A}=\mathbf{I},$ that is $\mathbf{A}=\mathbf{I}.$ Without the inverse there is no such cancellation: $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ also satisfies $\mathbf{A}^{2}=\mathbf{A},$ and it is not $\mathbf{I}$ — it is singular.
Answer
(a) $|\mathbf{A}|^{2}=0;$ or $\mathbf{A}=\mathbf{A}^{-1}\mathbf{A}^{2}=\mathbf{0},$ a contradiction. (b) e.g. $\begin{pmatrix}2&-4\\1&-2\end{pmatrix}.$ (c) $\mathbf{A}=\mathbf{A}^{-1}\mathbf{A}^{2}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}.$
How did it go?
Stuck? Read 4. The inverse, and when it exists →
The inverse of order 2, and its rules
Read the theory: 5. The inverse of order 2, and its rules →Let $\mathbf{A}=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ and $\mathbf{B}=\begin{pmatrix}1&2\\0&1\end{pmatrix}.$
(a) Compute $\mathbf{A}\mathbf{B}$ and invert it with Fact 14.20.
(b) Compute $\mathbf{A}^{-1}$ and $\mathbf{B}^{-1},$ and then $\mathbf{B}^{-1}\mathbf{A}^{-1}$ and $\mathbf{A}^{-1}\mathbf{B}^{-1}.$ Which of the two is $(\mathbf{A}\mathbf{B})^{-1}?$
Hint
All three determinants are $1,$ so no fractions appear.
(a) $$ \mathbf{A}\mathbf{B}=\begin{pmatrix}2&5\\1&3\end{pmatrix}, $$ with determinant $6-5=1,$ so $$ (\mathbf{A}\mathbf{B})^{-1}=\begin{pmatrix}3&-5\\-1&2\end{pmatrix}. $$
(b) The two inverses. $$ \begin{aligned} \mathbf{A}^{-1}&=\begin{pmatrix}1&-1\\-1&2\end{pmatrix},\\[4pt] \mathbf{B}^{-1}&=\begin{pmatrix}1&-2\\0&1\end{pmatrix}. \end{aligned} $$
(b) The two products. $$ \begin{aligned} \mathbf{B}^{-1}\mathbf{A}^{-1}&=\begin{pmatrix}3&-5\\-1&2\end{pmatrix},\\[4pt] \mathbf{A}^{-1}\mathbf{B}^{-1}&=\begin{pmatrix}1&-3\\-1&4\end{pmatrix}. \end{aligned} $$ The first is $(\mathbf{A}\mathbf{B})^{-1},$ as Fact 14.22 says; the second is not. It is $(\mathbf{B}\mathbf{A})^{-1}{:}$ the order reverses.
Answer
$(\mathbf{A}\mathbf{B})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}=\begin{pmatrix}3&-5\\-1&2\end{pmatrix};$ $\mathbf{A}^{-1}\mathbf{B}^{-1}=\begin{pmatrix}1&-3\\-1&4\end{pmatrix}$ is different.
How did it go?
Stuck? Read 5. The inverse of order 2, and its rules →
Solving the system with the inverse
Read the theory: 6. Solving the system with the inverse →Let $\mathbf{A}=\begin{pmatrix}5&3\\3&2\end{pmatrix}.$
(a) Find $\mathbf{A}^{-1}.$
(b) Solve $\mathbf{A}\mathbf{x}=\mathbf{b}$ for the three right-hand sides $\mathbf{b}=(11,7),$ $(2,1)$ and $(0,1).$
(c) Find the matrix $\mathbf{Y}$ with $\mathbf{Y}\mathbf{A}=\begin{pmatrix}1&0\\2&1\end{pmatrix},$ and show that $\mathbf{A}^{-1}\begin{pmatrix}1&0\\2&1\end{pmatrix}$ is not it.
Hint
$|\mathbf{A}|=1.$ In (b) each solution is one product $\mathbf{A}^{-1}\mathbf{b}.$ In (c), $\mathbf{A}$ stands to the right of $\mathbf{Y}.$
(a) $|\mathbf{A}|=10-9=1,$ so $\mathbf{A}^{-1}=\begin{pmatrix}2&-3\\-3&5\end{pmatrix}.$
(b) Three products. $$ \begin{aligned} \mathbf{A}^{-1}\begin{pmatrix}11\\7\end{pmatrix}&=\begin{pmatrix}1\\2\end{pmatrix},\\[4pt] \mathbf{A}^{-1}\begin{pmatrix}2\\1\end{pmatrix}&=\begin{pmatrix}1\\-1\end{pmatrix},\\[4pt] \mathbf{A}^{-1}\begin{pmatrix}0\\1\end{pmatrix}&=\begin{pmatrix}-3\\5\end{pmatrix}. \end{aligned} $$ Check the first: $5+6=11,$ $3+4=7.$ The inverse was computed once; each new right-hand side cost one product.
(c) From the right. By Fact 14.24 $$ \begin{aligned} \mathbf{Y}&=\begin{pmatrix}1&0\\2&1\end{pmatrix}\mathbf{A}^{-1}\\ &=\begin{pmatrix}2&-3\\1&-1\end{pmatrix}. \end{aligned} $$ Check: $\mathbf{Y}\mathbf{A}$ has rows $(10-9,\ 6-6)=(1,0)$ and $(5-3,\ 3-2)=(2,1).$
(c) The wrong side. $$ \mathbf{A}^{-1}\begin{pmatrix}1&0\\2&1\end{pmatrix}=\begin{pmatrix}-4&-3\\7&5\end{pmatrix}, $$ whose product with $\mathbf{A}$ on the right has first row $(-20-9,\ -12-6)=(-29,-18),$ not $(1,0).$
Answer
(a) $\begin{pmatrix}2&-3\\-3&5\end{pmatrix}.$ (b) $(1,2),$ $(1,-1),$ $(-3,5).$ (c) $\mathbf{Y}=\begin{pmatrix}2&-3\\1&-1\end{pmatrix}.$
How did it go?
Stuck? Read 6. Solving the system with the inverse →
Mixed, as on the exam
Consider the system $$ \begin{aligned} x+y+z&=1\\ x+2y+az&=2\\ x+4y+a^{2}z&=c\,, \end{aligned} $$ where $a$ and $c$ are numbers.
(a) For which $a$ does the system have exactly one solution for every $c?$
(b) For each remaining value of $a,$ find the values of $c$ for which there are solutions, and give all solutions for those $c.$
Hint
The determinant depends on $a$ only. For (b), subtract the first equation from the other two.
(a) The determinant. Along the first row, $$ \begin{aligned} |\mathbf{A}|&=(2a^{2}-4a)-(a^{2}-a)\\ &\quad+(4-2)\\ &=a^{2}-3a+2\\ &=(a-1)(a-2)\,. \end{aligned} $$ Exactly one solution for every $c$ if and only if $|\mathbf{A}|\ne0$ (Fact 14.26): for $a\ne1$ and $a\ne2.$
(b) $a=1.$ Subtracting row $1$ from rows $2$ and $3,$ that is $R_2-R_1$ and $R_3-R_1,$ gives $y=1$ and $3y=c-1.$ These agree only if $c-1=3,$ that is $c=4.$ For $c\ne4$ there is no solution. For $c=4$ we are left with $x+z=0$ and $y=1{:}$ $$ \begin{aligned} (x,y,z)&=(-t,\,1,\,t)\,,\\ t&\in\mathbb{R}\,. \end{aligned} $$
(b) $a=2.$ The same operations give $y+z=1$ and $3y+3z=c-1.$ Again they agree only if $c=4.$ For $c=4,$ put $z=t{:}$ $y=1-t$ and $x=1-y-z=0,$ so $$ \begin{aligned} (x,y,z)&=(0,\,1-t,\,t)\,,\\ t&\in\mathbb{R}\,. \end{aligned} $$
What it shows. For $a=1$ and $a=2$ the determinant is $0$ whatever $c$ is. It settles uniqueness — never exactly one — but existence depends on the right-hand side: none for $c\ne4,$ infinitely many for $c=4$ (Watch out 14.27).
Answer
(a) $a\ne1$ and $a\ne2.$ (b) For $a=1$ and for $a=2{:}$ solutions only when $c=4;$ then $(-t,1,t)$ for $a=1$ and $(0,1-t,t)$ for $a=2,$ $t\in\mathbb{R}.$
How did it go?
Stuck? Read 6. Solving the system with the inverse →
As on the exam
3 problems · 28 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 cubic determinant, and the values it leaves
For a real number $a,$ consider the linear system $$ \left\{ \begin{aligned} ax+3y+z&=2\\ x+2y+az&=1\\ 2x+ay-2z&=2 \end{aligned} \right. $$ and let $\mathbf{A}$ be its coefficient matrix.
- (a)4 p
Compute $\det\mathbf{A}$ and write it as a product of factors of degree one in $a.$
- (b)2 p
For which values of $a$ does the system have exactly one solution? Justify your answer.
- (c)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.
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(a) Expansion along the first row (Method 14.7): $$ \begin{aligned} \det\mathbf{A}&=\begin{vmatrix} a&3&1\\ 1&2&a\\ 2&a&-2\end{vmatrix}\\ &=a\begin{vmatrix}2&a\\a&-2\end{vmatrix} -3\begin{vmatrix}1&a\\2&-2\end{vmatrix}\\ &\quad+1\begin{vmatrix}1&2\\2&a\end{vmatrix}. \end{aligned} $$
(a) Simplify. $$ \begin{aligned} \det\mathbf{A}&=a(-4-a^{2})\\ &\quad-3(-2-2a)\\ &\quad+(a-4)\\ &=-a^{3}-4a+6\\ &\quad+6a+a-4\\ &=-a^{3}+3a+2\\ &=-(a^{3}-3a-2)\,. \end{aligned} $$
(a) One root. $p(a)=a^{3}-3a-2$ has integer coefficients, so by the integer root test (Fact 1.18) an integer root divides $-2{:}$ the candidates are $\pm1$ and $\pm2.$ Trying them, $p(-1)=-1+3-2=0.$
(a) Divide it out. By the factor theorem (Fact 1.17) $a+1$ is a factor, and division gives $$ \begin{aligned} &a^{3}-3a-2\\ &=(a+1)(a^{2}-a-2)\\ &=(a+1)(a+1)(a-2)\,. \end{aligned} $$ So $$\det\mathbf{A}=-(a+1)^{2}(a-2)\,.$$
(b) Exactly one solution. A square system has exactly one solution precisely when $\det\mathbf{A}\ne0$ (Fact 14.18, Fact 14.26). That holds for every $a$ except $a=-1$ and $a=2.$ So the system has exactly one solution if and only if $a\ne-1$ and $a\ne2.$
(c) $a=2.$ Put the second equation first and eliminate, writing $R_i$ for row $i{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c} 1&2&2&1\\ 2&3&1&2\\ 2&2&-2&2\end{array}\right)\\ &\quad\big\downarrow\ {\scriptstyle R_2-2R_1,\ R_3-2R_1}\\ &\left(\begin{array}{ccc|c} 1&2&2&1\\ 0&-1&-3&0\\ 0&-2&-6&0\end{array}\right)\\ &\quad\big\downarrow\ {\scriptstyle R_3-2R_2}\\ &\left(\begin{array}{ccc|c} 1&2&2&1\\ 0&-1&-3&0\\ 0&0&0&0\end{array}\right). \end{aligned} $$
(c) $a=2{:}$ all solutions. The last row is $0=0,$ and column $3$ has no leading entry, so $z$ is free: put $z=t.$ The second row gives $-y-3t=0,$ so $y=-3t,$ and the first $$ \begin{aligned} x&=1-2y-2z\\ &=1+6t-2t=1+4t\,. \end{aligned} $$ There are infinitely many solutions, $$ \begin{aligned} (x,y,z)&=(1+4t,\,-3t,\,t)\,,\\ t&\in\mathbb{R}\,. \end{aligned} $$ Check, equation by equation: $$ \begin{aligned} 2(1+4t)-9t+t&=2\,,\\ (1+4t)-6t+2t&=1\,,\\ 2(1+4t)-6t-2t&=2\,. \end{aligned} $$
(c) $a=-1.$ Again with the second equation first: $$ \begin{aligned} &\left(\begin{array}{ccc|c} 1&2&-1&1\\ -1&3&1&2\\ 2&-1&-2&2\end{array}\right)\\ &\quad\big\downarrow\ {\scriptstyle R_2+R_1,\ R_3-2R_1}\\ &\left(\begin{array}{ccc|c} 1&2&-1&1\\ 0&5&0&3\\ 0&-5&0&0\end{array}\right)\\ &\quad\big\downarrow\ {\scriptstyle R_3+R_2}\\ &\left(\begin{array}{ccc|c} 1&2&-1&1\\ 0&5&0&3\\ 0&0&0&3\end{array}\right). \end{aligned} $$ The last row reads $0=3,$ which no $(x,y,z)$ satisfies: for $a=-1$ the system has no solution. In words: the second row says $5y=3$ and the third $-5y=0.$
Note. Both remaining values make $\det\mathbf{A}=0,$ yet they end differently: whether solutions exist is decided by $\mathbf{b},$ and only elimination tells (Watch out 14.27).
Answer
(a) $$ \begin{aligned} \det\mathbf{A}&=-a^{3}+3a+2\\ &=-(a+1)^{2}(a-2)\,. \end{aligned} $$
(b) Exactly one solution for all $a$ except $a=-1$ and $a=2.$
(c) $a=2{:}$ infinitely many, $(x,y,z)=(1+4t,\,-3t,\,t),$ $t\in\mathbb{R}.$ $a=-1{:}$ no solution.
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
- Keeping one root of the determinant. $a=-1$ is the visible one; stopping there loses $a=2,$ and with it half of (c). In the live vote on the same step for a $2\times2$ system about one answer in five kept one root.
- Reading $\det\mathbf{A}=0$ as "no solution" for both values. The determinant says not exactly one; at $a=2$ there are infinitely many.
- A sign lost in the expansion: the middle term of the first row is $-3\begin{vmatrix}1&a\\2&-2\end{vmatrix}.$ With $+3$ the cubic becomes $-a^{3}-9a-10,$ which is $-(a+1)(a^{2}-a+10),$ whose quadratic factor has no real roots: the value $a=2$ disappears from the answer.
- Writing "infinitely many" at $a=2$ without the solutions, or giving one solution, such as $(1,0,0),$ instead of all of them.
Matrix form, a determinant, elimination — and a singular matrix
Consider the linear system $$ \left\{ \begin{aligned} x-y+2z&=-3\\ 2x+y+z&=6\\ 3x-2y+4z&=-4 \end{aligned} \right. $$
- (a)3 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, and compute $\det\mathbf{A}.$ What does its value tell you about the number of solutions?
- (b)4 p
Solve the system using Gaussian elimination.
- (c)3 p
Find all values of $k$ for which the matrix $$\begin{pmatrix} k&1&2\\ 3&k&0\\ 1&0&1\end{pmatrix}$$ is not invertible.
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(a) Matrix form. With $$ \begin{aligned} \mathbf{A}&=\begin{pmatrix}1&-1&2\\2&1&1\\3&-2&4\end{pmatrix},\\[4pt] \mathbf{v}&=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}-3\\6\\-4\end{pmatrix}, \end{aligned} $$ the system is $\mathbf{A}\mathbf{v}=\mathbf{b}.$
(a) The determinant. Expanding along the first row, $$ \begin{aligned} \det\mathbf{A}&=1\cdot\begin{vmatrix}1&1\\-2&4\end{vmatrix}\\ &\quad-(-1)\begin{vmatrix}2&1\\3&4\end{vmatrix}\\ &\quad+2\begin{vmatrix}2&1\\3&-2\end{vmatrix}\\ &=6+5+2\cdot(-7)=-3\,. \end{aligned} $$
(a) What it says. $\det\mathbf{A}=-3\ne0,$ so $\mathbf{A}$ is invertible and the system has exactly one solution (Fact 14.18, Fact 14.26) — for this $\mathbf{b}$ and for every other right-hand side.
(b) Elimination, writing $R_i$ for row $i{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c}1&-1&2&-3\\2&1&1&6\\3&-2&4&-4\end{array}\right)\\ &\quad\big\downarrow\ {\scriptstyle R_2-2R_1,\ R_3-3R_1}\\ &\left(\begin{array}{ccc|c}1&-1&2&-3\\0&3&-3&12\\0&1&-2&5\end{array}\right)\\ &\quad\big\downarrow\ {\scriptstyle \frac13R_2}\\ &\left(\begin{array}{ccc|c}1&-1&2&-3\\0&1&-1&4\\0&1&-2&5\end{array}\right)\\ &\quad\big\downarrow\ {\scriptstyle R_3-R_2}\\ &\left(\begin{array}{ccc|c}1&-1&2&-3\\0&1&-1&4\\0&0&-1&1\end{array}\right). \end{aligned} $$
(b) Back substitution. The last row gives $-z=1,$ so $z=-1;$ the second $y=4+z=3;$ the first $$ \begin{aligned} x&=-3+y-2z\\ &=-3+3+2=2\,. \end{aligned} $$ So $(x,y,z)=(2,3,-1).$ Check: $2-3-2=-3,$ $4+3-1=6,$ $6-6-4=-4.$
(c) The criterion. A square matrix is not invertible exactly when its determinant is $0$ (Fact 14.18). Expanding along the first row, $$ \begin{aligned} &\begin{vmatrix} k&1&2\\ 3&k&0\\ 1&0&1\end{vmatrix}\\ &=k(k-0)-1\cdot(3-0)\\ &\quad+2(0-k)\\ &=k^{2}-2k-3\,. \end{aligned} $$
(c) Both roots. $k^{2}-2k-3=(k-3)(k+1),$ which is $0$ exactly for $k=3$ and $k=-1.$ So the matrix is not invertible for $k=3$ and for $k=-1,$ and invertible for every other $k.$
Answer
(a) $\mathbf{A}=\begin{pmatrix}1&-1&2\\2&1&1\\3&-2&4\end{pmatrix},$ $\mathbf{v}=(x,y,z)^{\top},$ $\mathbf{b}=(-3,6,-4)^{\top};$ $\det\mathbf{A}=-3\ne0,$ so exactly one solution.
(b) $(x,y,z)=(2,3,-1).$
(c) $k=3$ and $k=-1.$
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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
- In (c), solving $k^{2}-2k-3=0$ and giving one root. A quadratic has two, and both make the matrix singular.
- The middle term of an expansion with the wrong sign: in (a) the entry $-1$ at $(1,2)$ gets the sign $-,$ so the term is $-(-1)\cdot5=+5.$
- In (a), "$\det\mathbf{A}\ne0,$ so the system has a solution": the point of the determinant is that the solution is unique.
- In (b), row operations done without being named, or a row replaced by a multiple of another row instead of by a combination with itself.
A matrix equation, solved from the right
Let $$ \begin{aligned} \mathbf{M}&=\begin{pmatrix}3&1\\4&2\end{pmatrix},\\[4pt] \mathbf{P}&=\begin{pmatrix}13&5\\-6&0\\14&6\end{pmatrix},\quad \mathbf{Q}=\begin{pmatrix}1&0\\-1&1\\2&1\end{pmatrix}. \end{aligned} $$ We look for every matrix $\mathbf{N}$ with $$\mathbf{N}\mathbf{M}=\mathbf{P}-2\mathbf{Q}\,.$$
- (a)2 p
Compute $\mathbf{C}=\mathbf{P}-2\mathbf{Q}.$ What order must $\mathbf{N}$ have? Explain.
- (b)3 p
Show that $\mathbf{M}$ is invertible, and find $\mathbf{M}^{-1}.$
- (c)3 p
Find $\mathbf{N}.$ Explain on which side you multiply, and check your answer.
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(a) The right-hand side. $$ \begin{aligned} \mathbf{C}&=\begin{pmatrix}13-2&5-0\\-6+2&0-2\\14-4&6-2\end{pmatrix}\\ &=\begin{pmatrix}11&5\\-4&-2\\10&4\end{pmatrix}. \end{aligned} $$
(a) The order of N. If $\mathbf{N}$ is $m\times n,$ the product $\mathbf{N}\mathbf{M}$ exists only if $n=2,$ the number of rows of $\mathbf{M},$ and it is then $m\times2.$ It must equal the $3\times2$ matrix $\mathbf{C},$ so $m=3{:}$ $\mathbf{N}$ is $3\times2.$
(b) Invertible. $|\mathbf{M}|=3\cdot2-1\cdot4=2,$ and $2\ne0,$ so $\mathbf{M}$ is invertible (Fact 14.18).
(b) The inverse. By Fact 14.20 — swap the diagonal, change the sign of the other two, divide by $2{:}$ $$ \begin{aligned} \mathbf{M}^{-1}&=\frac12\begin{pmatrix}2&-1\\-4&3\end{pmatrix}\\ &=\begin{pmatrix}1&-\tfrac12\\-2&\tfrac32\end{pmatrix}. \end{aligned} $$ Check: the rows of $\mathbf{M}\mathbf{M}^{-1}$ are $$ \begin{aligned} (3-2,\ -\tfrac32+\tfrac32)&=(1,0)\,,\\ (4-4,\ -2+3)&=(0,1)\,, \end{aligned} $$ so it is $\mathbf{I}_{2},$ and by Fact 14.21 one product is enough.
(c) The side. $\mathbf{M}$ stands to the right of $\mathbf{N},$ so multiply $\mathbf{N}\mathbf{M}=\mathbf{C}$ on the right by $\mathbf{M}^{-1}{:}$ $$ \begin{aligned} \mathbf{N}&=\mathbf{N}\mathbf{M}\mathbf{M}^{-1}\\ &=\mathbf{C}\mathbf{M}^{-1} \end{aligned} $$ (Fact 14.24). The product $\mathbf{M}^{-1}\mathbf{C}$ does not even exist — a $2\times2$ times a $3\times2$ matrix — and there is no $\mathbf{C}/\mathbf{M}$ (Watch out 14.17).
(c) The product. $$ \begin{aligned} \mathbf{N}&=\begin{pmatrix}11&5\\-4&-2\\10&4\end{pmatrix} \begin{pmatrix}1&-\tfrac12\\-2&\tfrac32\end{pmatrix}\\ &=\begin{pmatrix}11-10&-\tfrac{11}2+\tfrac{15}2\\-4+4&2-3\\10-8&-5+6\end{pmatrix}\\ &=\begin{pmatrix}1&2\\0&-1\\2&1\end{pmatrix}. \end{aligned} $$
(c) Check. $$ \begin{aligned} \mathbf{N}\mathbf{M}&=\begin{pmatrix}3+8&1+4\\-4&-2\\6+4&2+2\end{pmatrix}\\ &=\begin{pmatrix}11&5\\-4&-2\\10&4\end{pmatrix}=\mathbf{C}\,. \end{aligned} $$ Since $\mathbf{M}$ is invertible, this $\mathbf{N}$ is the only one.
Answer
(a) $\mathbf{C}=\begin{pmatrix}11&5\\-4&-2\\10&4\end{pmatrix};$ $\mathbf{N}$ is $3\times2.$
(b) $|\mathbf{M}|=2\ne0;$ $\mathbf{M}^{-1}=\begin{pmatrix}1&-\tfrac12\\-2&\tfrac32\end{pmatrix}.$
(c) $\mathbf{N}=\mathbf{C}\mathbf{M}^{-1}=\begin{pmatrix}1&2\\0&-1\\2&1\end{pmatrix}.$
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
- $\mathbf{N}=\mathbf{M}^{-1}\mathbf{C}{:}$ the inverse on the wrong side. Here the product does not even exist, which is the quickest way to notice.
- $|\mathbf{M}|$ forgotten in the inverse: $\begin{pmatrix}2&-1\\-4&3\end{pmatrix}$ instead of half of it. The check $\mathbf{M}\mathbf{M}^{-1}=\mathbf{I}_{2}$ gives $2\mathbf{I}_{2}$ and catches it.
- The signs in the inverse: the off-diagonal entries are negated and the diagonal entries swapped, not the other way round. The lecture's worst hub template asks exactly for one entry of an inverse ($67\%$ right).