Exercise 0.1: Computing determinants: ten direct calculations

For each line, compute the requested determinant, or the requested value, by the shortest route. Give fractions exactly, as $\frac{3}{4}$ or $3/4$ .

Look at the matrix before computing: a triangular matrix is read off its diagonal, a row or column full of zeros points to a cofactor expansion, with the sign $(-1)^{i+j}$ of the position, and a row with a common factor is simplified first. Adding a multiple of one row to another does not change the determinant, and that is how zeros are created. What is left is a $2\times 2$ , $ad-bc$ , or the rule of Sarrus for a $3\times 3$ .

a) $\begin{vmatrix} -6 & 4 \\ 5 & -3 \end{vmatrix}$ .
b) $\begin{vmatrix} \frac{2}{3} & -\frac{1}{2} \\ \frac{3}{4} & \frac{3}{2} \end{vmatrix}$ .
c) $\begin{vmatrix} 2 & 0 & 0 & 0 \\ 7 & -1 & 0 & 0 \\ -3 & 4 & 3 & 0 \\ 9 & 5 & -8 & -2 \end{vmatrix}$ .
d) $\begin{vmatrix} 4 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 5 & 1 \end{vmatrix}$ , by the rule of Sarrus.
e) $\begin{vmatrix} 2 & 1 & 4 \\ 0 & 0 & -3 \\ 5 & -1 & 2 \end{vmatrix}$ , by cofactor expansion along the second row.
f) $\begin{vmatrix} 1 & 3 & -2 \\ 4 & 10 & -5 \\ -2 & -4 & 7 \end{vmatrix}$ , after first creating zeros with $R_{2}-4R_{1}$ and $R_{3}+2R_{1}$ .
g) $\begin{vmatrix} 6 & -9 & 3 \\ 4 & 2 & -2 \\ 1 & 0 & 5 \end{vmatrix}$ , after first factoring the common factors out of the first two rows.
h) $\begin{vmatrix} 1 & 4 & 0 & -1 \\ 2 & 1 & 0 & 3 \\ 3 & 0 & 0 & 1 \\ -1 & 2 & 2 & 5 \end{vmatrix}$ .
i) Find the values of $x$ for which $\begin{vmatrix} x-1 & 3 \\ 4 & x+3 \end{vmatrix}=0$ .
j) Solve the system $\begin{cases} 4x-3y=5 \\ 2x+5y=-1 \end{cases}$ by Cramer's rule, then give $x$ and $y$ .

Type your answers, the page tells you right or wrong 0/12

a)

determinant

b)

determinant

c)

determinant

d)

determinant

e)

determinant

f)

determinant

g)

determinant

h)

determinant

i)

the smaller value

the larger value

j)

x

y

Show the solution

Answers

a) $18-20=-2$
b) $1+\frac{3}{8}=\frac{11}{8}$
c) $2(-1)(3)(-2)=12$
d) $32-(-43)=75$
e) $(-3)(-1)(-7)=-21$
f) $\begin{vmatrix} -2 & 3 \\ 2 & 3 \end{vmatrix}=-12$
g) $6\times 42=252$
h) $2\times(-1)\times 32=-64$
i) $(x+5)(x-3)=0$ ; $x=-5$ or $x=3$
j) $x=\frac{22}{26}=\frac{11}{13}$ ; $y=-\frac{14}{26}=-\frac{7}{13}$

The move that decides everything: choose the route before computing. A diagonal, a row of zeros, a common factor or an operation $R_{i}-kR_{j}$ replaces a long expansion, and every cofactor carries the sign of its position in the checkerboard $+,-,+$ .

a)

$ad-bc=(-6)(-3)-(4)(5)=18-20=-2$ .

b)

$\frac{2}{3}\cdot\frac{3}{2}-\left(-\frac{1}{2}\right)\cdot\frac{3}{4}=1+\frac{3}{8}=\frac{11}{8}$ . Pitfall: $-bc$ with $b$ negative gives a PLUS.

c)

The matrix is lower triangular: the determinant is the product of the diagonal, $2\times(-1)\times 3\times(-2)=12$ . The entries below the diagonal do not count.

d)

Downward diagonals: $4\cdot 0\cdot 1+(-1)(-2)(1)+2\cdot 3\cdot 5=0+2+30=32$ . Upward diagonals: $2\cdot 0\cdot 1+4(-2)(5)+(-1)(3)(1)=0-40-3=-43$ . So $32-(-43)=75$ . Pitfall: subtracting the upward diagonals without brackets, which gives $32-43=-11$ .

e)

The second row has a single nonzero entry, $-3$ , in position $(2,3)$ , with sign $(-1)^{2+3}=-1$ . Its minor is $\begin{vmatrix} 2 & 1 \\ 5 & -1 \end{vmatrix}=-2-5=-7$ . So $\det=(-3)\times(-1)\times(-7)=-21$ . Pitfall: forgetting the sign of the position, which gives $+21$ .

f)

$R_{2}-4R_{1}$ gives $(0, -2, 3)$ and $R_{3}+2R_{1}$ gives $(0, 2, 3)$ , without changing the determinant. Expand along the first column: $1\times\begin{vmatrix} -2 & 3 \\ 2 & 3 \end{vmatrix}=-6-6=-12$ .

g)

Factor $3$ out of the first row and $2$ out of the second: $\det=3\times 2\times\begin{vmatrix} 2 & -3 & 1 \\ 2 & 1 & -1 \\ 1 & 0 & 5 \end{vmatrix}$ . This last one equals $2(5-0)-(-3)(10+1)+1(0-1)=10+33-1=42$ , so $\det=6\times 42=252$ . Pitfall: factoring a number out of a row without putting it back in front.

h)

The third column has a single nonzero entry, $2$ , in position $(4,3)$ , with sign $(-1)^{4+3}=-1$ . Its minor, without row 4 and column 3, is $\begin{vmatrix} 1 & 4 & -1 \\ 2 & 1 & 3 \\ 3 & 0 & 1 \end{vmatrix}=1(1-0)-4(2-9)+(-1)(0-3)=1+28+3=32$ . So $\det=2\times(-1)\times 32=-64$ . Pitfall: the sign of position $(4,3)$ .

i)

$(x-1)(x+3)-12=x^{2}+2x-15=(x+5)(x-3)$ , zero for $x=-5$ or $x=3$ . Check: for $x=3$ , $\begin{vmatrix} 2 & 3 \\ 4 & 6 \end{vmatrix}=12-12=0$ .

j)

$\det A=4(5)-(-3)(2)=26$ . Replace the column of $x$ by the right-hand side: $\det A_{x}=\begin{vmatrix} 5 & -3 \\ -1 & 5 \end{vmatrix}=25-3=22$ , then the column of $y$ : $\det A_{y}=\begin{vmatrix} 4 & 5 \\ 2 & -1 \end{vmatrix}=-4-10=-14$ . So $x=\frac{22}{26}=\frac{11}{13}$ and $y=-\frac{14}{26}=-\frac{7}{13}$ . Pitfall: $(-3)(-1)=+3$ , which is subtracted in $\det A_{x}$ .

Check: a $3\times 3$ determinant is redone along another row or by Sarrus, and the two results must agree. For Cramer, substitute back into the system: $4\cdot\frac{11}{13}-3\cdot\left(-\frac{7}{13}\right)=\frac{44+21}{13}=5$ .

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 1: Determinants of order 2 and 3, two methods

Consider the matrices $A=\begin{pmatrix} 3 & -2 \\ 5 & 4 \end{pmatrix}$ and $B=\begin{pmatrix} 2 & -1 & 3 \\ 0 & 4 & 5 \\ 1 & 2 & -2 \end{pmatrix}$ .

a) Compute $\det(A)$ .
b) Compute $\det(B)$ by the rule of Sarrus.
c) Compute $\det(B)$ again by cofactor expansion along the first column, and check that you get the same number.
d) Are the matrices $A$ and $B$ invertible? Justify.

Show the solution

Answers

a) $\det(A)=22$
b) $\det(B)=-53$
c) $C_{11}=-18$ , $C_{31}=-17$ : $-53$
d) Both are invertible

a)

$\det(A)=(3)(4)-(-2)(5)=12+10=22$ . Watch the double negative sign: it is the most frequent error on a $2\times 2$ determinant.

b)

Rule of Sarrus: copy the first two columns to the right, then add the three downward diagonals and subtract the three upward diagonals. Downward diagonals: $(2)(4)(-2)=-16$ , $(-1)(5)(1)=-5$ , $(3)(0)(2)=0$ , with sum $-21$ . Upward diagonals: $(1)(4)(3)=12$ , $(2)(5)(2)=20$ , $(-2)(0)(-1)=0$ , with sum $32$ . So $\det(B)=-21-32=-53$ .

c)

The first column contains a zero, so it is a good choice. $\det(B)=2\,C_{11}+0\,C_{21}+1\,C_{31}$ , and the middle term vanishes. We have $C_{11}=(+1)\begin{vmatrix} 4 & 5 \\ 2 & -2 \end{vmatrix}=(4)(-2)-(5)(2)=-8-10=-18$ and $C_{31}=(+1)\begin{vmatrix} -1 & 3 \\ 4 & 5 \end{vmatrix}=(-1)(5)-(3)(4)=-5-12=-17$ . The sign of $C_{31}$ is positive because $(-1)^{3+1}=+1$ . So $\det(B)=2(-18)+1(-17)=-36-17=-53$ . The two methods agree.

An important reminder: the rule of Sarrus works ONLY for $3\times 3$ matrices. It does not extend to order $4$ , contrary to what many students attempt on exams. Beyond order $3$ , only cofactor expansion is valid.

d)

$\det(A)=22\neq 0$ , so $A$ is invertible. $\det(B)=-53\neq 0$ , so $B$ is invertible as well. A negative determinant is no obstacle: a zero determinant, and only a zero determinant, rules out invertibility.

Exercise 2: A determinant of order 4 by cofactors

Consider the matrix of order $4$ : $C=\begin{pmatrix} 2 & 0 & 1 & 3 \\ 0 & 1 & 0 & 0 \\ 4 & 2 & 3 & 1 \\ 1 & 0 & 2 & 0 \end{pmatrix}$ .

a) Which row or column would you choose for the cofactor expansion? Justify your choice.
b) Compute $\det(C)$ .
c) Without any new calculation, give $\det(C^{T})$ and justify.
d) Without computing another determinant, give $\det(2C)$ and justify.

Show the solution

Answers

a) Second row: three zeros
b) $\det(C)=12$
c) $\det(C^T)=12$
d) $\det(2C)=2^4\times 12=192$

a)

The second row, $(0, 1, 0, 0)$ , contains three zeros: it is the best possible choice, since the expansion reduces to a single term instead of four. The second column, $(0, 1, 2, 0)$ , contains two, so it would be the second choice. Expanding along the first row would mean computing three $3\times 3$ determinants for nothing.

b)

Expand along the second row. Only the entry $a_{22}=1$ is nonzero, and its sign is $(-1)^{2+2}=+1$ . So $\det(C)=1\times\begin{vmatrix} 2 & 1 & 3 \\ 4 & 3 & 1 \\ 1 & 2 & 0 \end{vmatrix}$ , the minor obtained by deleting the second row and the second column.

Compute this $3\times 3$ determinant by expanding along its first row: $2\big((3)(0)-(1)(2)\big)-1\big((4)(0)-(1)(1)\big)+3\big((4)(2)-(3)(1)\big)=2(-2)-1(-1)+3(8-3)=-4+1+15=12$ . So $\det(C)=12$ .

c)

$\det(C^{T})=\det(C)=12$ . A matrix and its transpose always have the same determinant. That is precisely why one may expand along a row or along a column alike: transposing swaps the two roles without changing the result.

d)

$\det(2C)=2^{4}\det(C)=16\times 12=192$ . The error to avoid here is writing $\det(2C)=2\det(C)$ . Multiplying the whole matrix by $2$ multiplies EACH of its four rows by $2$ , and each multiplied row brings a factor $2$ out of the determinant, hence $2^{4}$ . The general rule is $\det(kA)=k^{n}\det(A)$ for an $n\times n$ matrix, and the exponent $n$ is the part regularly forgotten on exams.

Exercise 3: Properties of the determinant and elementary row operations

Let $A$ be a square matrix of order $3$ about which we only know that $\det(A)=6$ . None of its entries is known.

In each case, give the requested value and justify it by the property used, without ever trying to rebuild $A$ .

a) $A_{1}$ is obtained by swapping the first and third rows of $A$ . What is $\det(A_{1})$ ?
b) $A_{2}$ is obtained by multiplying the second row of $A$ by $5$ . What is $\det(A_{2})$ ? What is $\det(5A)$ ?
c) $A_{3}$ is obtained by adding four times the first row of $A$ to its second row. What is $\det(A_{3})$ ?
d) Give $\det(A^{2})$ , $\det(A^{-1})$ and $\det(A^{T})$ . What would $\det(A+A)$ be?

Show the solution

Answers

a) $-6$
b) 30; $\det(5A)=750$
c) 6
d) 36; $\frac16$ ; 6; $\det(A+A)=48$

a)

Swapping two rows changes the sign of the determinant: $\det(A_{1})=-6$ .

b)

Multiplying ONE row by $5$ multiplies the determinant by $5$ : $\det(A_{2})=5\times 6=30$ . On the other hand, $5A$ multiplies ALL THREE rows by $5$ , so $\det(5A)=5^{3}\times 6=125\times 6=750$ . This is exactly the distinction the exam is testing when it asks the two questions side by side.

c)

Adding a multiple of one row to another row does not change the determinant: $\det(A_{3})=6$ . It is the most useful property of all, because it lets you create zeros freely before a cofactor expansion, with nothing to correct afterwards. Geometrically, adding a multiple of one column to another SLIDES the parallelogram without changing its base or its height: its area, hence the determinant, stays the same.

d)

$\det(A^{2})=\det(A)\det(A)=6\times 6=36$ , by the property $\det(AB)=\det(A)\det(B)$ . Since $\det(A)=6\neq 0$ , the matrix $A$ is invertible and $\det(A^{-1})=\dfrac{1}{\det(A)}=\dfrac{1}{6}$ . Finally $\det(A^{T})=\det(A)=6$ .

For $\det(A+A)$ : the trap is to write $\det(A)+\det(A)=12$ . That is wrong, because the determinant is NOT additive. You have to see that $A+A=2A$ , so $\det(A+A)=\det(2A)=2^{3}\times 6=48$ . Remember the formula that does not exist: $\det(A+B)\neq\det(A)+\det(B)$ in general. The determinant is multiplicative, never additive.

Exercise 4: Adjugate matrix and inverse by cofactors

Consider the matrix $A=\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$ .

a) Compute $\det(A)$ and conclude about the invertibility of $A$ .
b) Compute the nine cofactors of $A$ .
c) Deduce the adjugate matrix $\text{adj}(A)$ , then $A^{-1}$ .
d) Check your result by computing the product $A\,A^{-1}$ .

Show the solution

Answers

a) $\det(A)=1$ : invertible
b) Cofactors $\begin{pmatrix}-24&20&-5\\18&-15&4\\5&-4&1\end{pmatrix}$
c) $A^{-1}=\begin{pmatrix}-24&18&5\\20&-15&-4\\-5&4&1\end{pmatrix}$
d) $AA^{-1}=I_3$

a)

Expansion along the first row: $\det(A)=1\big((1)(0)-(4)(6)\big)-2\big((0)(0)-(4)(5)\big)+3\big((0)(6)-(1)(5)\big)=1(-24)-2(-20)+3(-5)=-24+40-15=1$ . The determinant is $1\neq 0$ , so $A$ is invertible. A determinant equal to $1$ will make what follows especially easy to read, since dividing by $1$ changes nothing.

b)

Apply $C_{ij}=(-1)^{i+j}M_{ij}$ , following the checkerboard of signs $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$ .

$C_{11}=+\begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix}=0-24=-24$ , $\ C_{12}=-\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix}=-(0-20)=20$ , $\ C_{13}=+\begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix}=0-5=-5$ .

$C_{21}=-\begin{vmatrix} 2 & 3 \\ 6 & 0 \end{vmatrix}=-(0-18)=18$ , $\ C_{22}=+\begin{vmatrix} 1 & 3 \\ 5 & 0 \end{vmatrix}=0-15=-15$ , $\ C_{23}=-\begin{vmatrix} 1 & 2 \\ 5 & 6 \end{vmatrix}=-(6-10)=4$ .

$C_{31}=+\begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix}=8-3=5$ , $\ C_{32}=-\begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix}=-(4-0)=-4$ , $\ C_{33}=+\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix}=1-0=1$ .

c)

The matrix of cofactors is $\begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix}$ . The adjugate is its TRANSPOSE: $\text{adj}(A)=\begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$ . Forgetting to transpose is the number one error in this exercise, and it stays invisible until the final product is checked.

Finally $A^{-1}=\dfrac{1}{\det(A)}\text{adj}(A)=\dfrac{1}{1}\text{adj}(A)=\begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$ .

d)

First row of the product $A\,A^{-1}$ : $(1)(-24)+(2)(20)+(3)(-5)=-24+40-15=1$ , then $(1)(18)+(2)(-15)+(3)(4)=18-30+12=0$ , then $(1)(5)+(2)(-4)+(3)(1)=5-8+3=0$ . We do get $(1, 0, 0)$ . Carrying on in the same way with the other two rows gives the identity matrix $I_{3}$ , which confirms the inverse. This check takes one minute and catches every sign error: never skip it.

Exercise 5: Computing a determinant cleverly: triangular, proportional rows, factors

This exercise never asks for a brute-force expansion along the first row. Each determinant is quick to compute once its structure has been spotted.

a) Compute $\det(U)$ with $U=\begin{pmatrix} 3 & 7 & -2 \\ 0 & -2 & 5 \\ 0 & 0 & 4 \end{pmatrix}$ , and state the rule used.
b) Without expanding, explain why $\det(P)=0$ with $P=\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 5 & 0 & 1 \end{pmatrix}$ .
c) Compute $\det(M)$ with $M=\begin{pmatrix} 1 & 2 & -1 \\ 2 & 5 & 1 \\ 3 & 7 & 2 \end{pmatrix}$ by first creating zeros in the first column with elementary row operations.
d) Compute $\det(N)$ with $N=\begin{pmatrix} 2 & 4 & 6 \\ 3 & 3 & 3 \\ 1 & 5 & 2 \end{pmatrix}$ by first factoring the common factors out of the rows.

Show the solution

Answers

a) $-24$
b) Proportional rows: 0
c) $\det(M)=2$
d) $\det(N)=42$

a)

The matrix $U$ is upper triangular: every entry below the diagonal is zero. The determinant of a triangular matrix is the product of its diagonal entries: $\det(U)=(3)(-2)(4)=-24$ . No expansion is needed, and the rule also holds for lower triangular and diagonal matrices. This is exactly the point of Gaussian elimination here: triangularize, then multiply the diagonal.

b)

The second row of $P$ is twice the first: $(2, 4, 6)=2(1, 2, 3)$ . Two proportional rows form a linearly dependent set, so the determinant is zero: $\det(P)=0$ . It can also be seen as a consequence of the effect of row operations: replacing $R_{2}$ by $R_{2}-2R_{1}$ gives a row of zeros, and a determinant with a row of zeros equals $0$ . No Sarrus calculation is required.

c)

Adding a multiple of one row to another does not change the value of a determinant. Create zeros below the first pivot: $R_{2}\leftarrow R_{2}-2R_{1}$ gives $(0, 1, 3)$ , and $R_{3}\leftarrow R_{3}-3R_{1}$ gives $(0, 1, 5)$ . The determinant becomes $\begin{vmatrix} 1 & 2 & -1 \\ 0 & 1 & 3 \\ 0 & 1 & 5 \end{vmatrix}$ . Expand along the first column, which now has a single term: $\det(M)=1\times\begin{vmatrix} 1 & 3 \\ 1 & 5 \end{vmatrix}=1(5-3)=2$ . Direct check by Sarrus: $1(10-7)-2(4-3)+(-1)(14-15)=3-2+1=2$ , everything agrees.

d)

Each row carries a common factor. The first is $2(1, 2, 3)$ , the second is $3(1, 1, 1)$ . Multiplying a row by a scalar multiplies the determinant by that scalar, so both factors come out: $\det(N)=2\times 3\times\begin{vmatrix} 1 & 2 & 3 \\ 1 & 1 & 1 \\ 1 & 5 & 2 \end{vmatrix}$ . The remaining determinant equals $1(1\cdot 2-1\cdot 5)-2(1\cdot 2-1\cdot 1)+3(1\cdot 5-1\cdot 1)=1(-3)-2(1)+3(4)=-3-2+12=7$ . So $\det(N)=6\times 7=42$ . The trap would be to factor a number out and then forget to carry it into the final product: every scalar taken out must reappear as a factor.

Exercise 6: Proofs and a singular matrix with a parameter

This exercise mixes proof and the search for a parameter. Questions a) and b) are done without a single numerical entry.

a) Prove that if $A$ is a square matrix with two identical rows, then $\det(A)=0$ . Use the effect of a row interchange.
b) Let $A$ be an invertible matrix. Prove that $\det(A^{-1})=\dfrac{1}{\det(A)}$ .
c) Find all values of $k$ for which the matrix $M=\begin{pmatrix} k & 1 & 2 \\ 1 & k & 1 \\ 2 & 1 & k \end{pmatrix}$ is singular.
d) For the integer value of $k$ found in c), exhibit an explicit linear dependence relation between the rows of $M$ , and explain why such a relation had to exist.

Show the solution

Answers

a) $\det(A)=-\det(A)$ : zero
b) $\det(A)\det(A^{-1})=1$
c) $k=2$ or $k=-1\pm\sqrt3$
d) $R_1-R_3=0$

a)

Let $A$ be a matrix whose rows $R_{i}$ and $R_{j}$ are identical, with $i\neq j$ . Swap these two rows: the resulting matrix, call it $A'$ , is exactly the same as $A$ , since the two rows swapped had the same entries. So $\det(A')=\det(A)$ . But by the row-interchange property, $\det(A')=-\det(A)$ . Combining: $\det(A)=-\det(A)$ , so $2\det(A)=0$ , so $\det(A)=0$ . The proof takes three lines and uses no entry at all.

A consistent geometric consequence: two identical rows mean two equal vectors among the rows, hence a linearly dependent set, hence a zero volume. For two vectors in the plane the picture is even clearer: if they are COLLINEAR, the parallelogram they span is flattened, its area is zero, so the determinant is zero and the matrix is not invertible.

b)

By definition of the inverse, $A\,A^{-1}=I$ . Take the determinant of both sides: $\det(A\,A^{-1})=\det(I)$ . The left side factors by the multiplicative property: $\det(A)\det(A^{-1})$ . The right side equals $1$ , since the determinant of the identity matrix is $1$ . So $\det(A)\det(A^{-1})=1$ . Since $A$ is invertible, $\det(A)\neq 0$ , and we may divide: $\det(A^{-1})=\dfrac{1}{\det(A)}$ .

Notice in passing what this equality proves between the lines: the inverse of a matrix with integer entries has integer entries only if $\det(A)=\pm 1$ .

c)

Expand along the first row: $\det(M)=k\begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix}-1\begin{vmatrix} 1 & 1 \\ 2 & k \end{vmatrix}+2\begin{vmatrix} 1 & k \\ 2 & 1 \end{vmatrix}=k(k^{2}-1)-1(k-2)+2(1-2k)$ .

Expanding: $k^{3}-k-k+2+2-4k=k^{3}-6k+4$ . The matrix is singular when this polynomial vanishes. Look for an obvious root among the divisors of $4$ : for $k=2$ , $8-12+4=0$ . So $k=2$ is a root, and we factor out $k-2$ : $k^{3}-6k+4=(k-2)(k^{2}+2k-2)$ .

The quadratic factor gives $k=\dfrac{-2\pm\sqrt{4+8}}{2}=\dfrac{-2\pm 2\sqrt{3}}{2}=-1\pm\sqrt{3}$ . The matrix $M$ is therefore singular for $k=2$ , $k=-1+\sqrt{3}\approx 0.73$ and $k=-1-\sqrt{3}\approx -2.73$ . Three values, not one: a determinant of order $3$ with a parameter on the diagonal gives a polynomial of degree $3$ , and all of its roots must be found.

d)

For $k=2$ , the matrix becomes $M=\begin{pmatrix} 2 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 2 \end{pmatrix}$ . Rows $R_{1}$ and $R_{3}$ are identical, so $R_{1}-R_{3}=(0, 0, 0)$ is an explicit dependence relation, with coefficients $(1, 0, -1)$ not all zero.

Such a relation had to exist: $\det(M)=0$ is equivalent to saying that the rows of $M$ are linearly dependent, and a linearly dependent set is precisely one that admits a zero linear combination with coefficients not all zero. A zero determinant and the dependence relation are two statements of the same fact, one numerical and the other structural. Here, question a) even gives the result with no calculation: two identical rows are enough to conclude that $\det(M)=0$ .

Exercise 7: Problem: three tutoring formats solved by Cramer's rule

A homework help centre bills three formats: a one-on-one session, a session for two and a group workshop. The unit rates are unknown, but three weekly statements are available.

Week 1: $2$ one-on-one sessions, $1$ session for two and $1$ workshop cancelled and refunded (so counted negatively) give a balance of $2$ billing units. Week 2: $1$ one-on-one, $3$ sessions for two and $1$ refunded workshop give $2$ units. Week 3: $3$ one-on-one sessions, $1$ refunded session for two and $2$ workshops give $11$ units.

Writing $x$ , $y$ and $z$ for the rates of the three formats, we get the system $\begin{cases} 2x+y-z=2 \\ x+3y-z=2 \\ 3x-y+2z=11 \end{cases}$ .

a) Write the system in the matrix form $AX=B$ and compute $\det(A)$ .
b) Does Cramer's rule apply? What can be said about the number of solutions before even computing them?
c) Compute $\det(A_{x})$ and deduce $x$ .
d) Compute $\det(A_{y})$ and $\det(A_{z})$ , then deduce $y$ and $z$ .
e) Check the solution in the three original equations.
f) The third statement is corrected: its right-hand side goes from $11$ to $14$ , the coefficients being unchanged. Without redoing all the work, explain what must be recomputed and what remains valid, then give the new value of $x$ .

Show the solution

Answers

a) $\det(A)=15$
b) Cramer applies: unique solution
c) $\det(A_x)=30$ : $x=2$
d) $y=1$ , $z=3$
e) All three equations check
f) $\det(A)$ unchanged; $x=\frac{36}{15}=2.4$

a)

$A=\begin{pmatrix} 2 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & -1 & 2 \end{pmatrix}$ , $X=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ and $B=\begin{pmatrix} 2 \\ 2 \\ 11 \end{pmatrix}$ .

Expansion along the first row: $\det(A)=2\big((3)(2)-(-1)(-1)\big)-1\big((1)(2)-(-1)(3)\big)+(-1)\big((1)(-1)-(3)(3)\big)=2(6-1)-1(2+3)-1(-1-9)=10-5+10=15$ .

b)

The determinant is $15\neq 0$ , so $A$ is invertible and Cramer's rule applies. We can state BEFORE any calculation that the system has a unique solution. That is exactly the role of the determinant of the system: it settles existence and uniqueness without giving the solution. Had $\det(A)$ been zero, Cramer would have been unusable and Gauss-Jordan elimination would have been needed to tell no solution from infinitely many.

c)

Replace the first column of $A$ by $B$ : $\det(A_{x})=\begin{vmatrix} 2 & 1 & -1 \\ 2 & 3 & -1 \\ 11 & -1 & 2 \end{vmatrix}=2\big(6-1\big)-1\big(4+11\big)+(-1)\big(-2-33\big)=10-15+35=30$ . So $x=\dfrac{\det(A_{x})}{\det(A)}=\dfrac{30}{15}=2$ .

d)

Replace the second column by $B$ : $\det(A_{y})=\begin{vmatrix} 2 & 2 & -1 \\ 1 & 2 & -1 \\ 3 & 11 & 2 \end{vmatrix}=2\big((2)(2)-(-1)(11)\big)-2\big((1)(2)-(-1)(3)\big)+(-1)\big((1)(11)-(2)(3)\big)=2(15)-2(5)-1(5)=30-10-5=15$ . So $y=\dfrac{15}{15}=1$ .

Then the third column: $\det(A_{z})=\begin{vmatrix} 2 & 1 & 2 \\ 1 & 3 & 2 \\ 3 & -1 & 11 \end{vmatrix}=2\big((3)(11)-(2)(-1)\big)-1\big((1)(11)-(2)(3)\big)+2\big((1)(-1)-(3)(3)\big)=2(35)-1(5)+2(-10)=70-5-20=45$ . So $z=\dfrac{45}{15}=3$ .

e)

Equation 1: $2(2)+1-3=4+1-3=2$ , as required. Equation 2: $2+3(1)-3=2+3-3=2$ , as required. Equation 3: $3(2)-1+2(3)=6-1+6=11$ , as required. The solution is $(x, y, z)=(2, 1, 3)$ , that is rates of $2$ , $1$ and $3$ billing units. Checking in ALL THREE equations, and not only in the one used for the last calculation, is what separates a sure answer from a hopeful one.

f)

The coefficients of the system do not change, so the matrix $A$ is unchanged and $\det(A)=15$ remains valid: nothing to recompute on that side, and the conclusion about existence and uniqueness of the solution still holds. On the other hand, each of the three determinants $\det(A_{x})$ , $\det(A_{y})$ and $\det(A_{z})$ contains the column $B$ , so all three must be recomputed.

For $x$ : $\det(A_{x})=\begin{vmatrix} 2 & 1 & -1 \\ 2 & 3 & -1 \\ 14 & -1 & 2 \end{vmatrix}=2(6-1)-1(4+14)+(-1)(-2-42)=10-18+44=36$ , hence $x=\dfrac{36}{15}=\dfrac{12}{5}=2.4$ .

This is the structural advantage of Cramer over Gauss-Jordan when only the right-hand side moves: the denominator, which is the costly part of the work, is computed once and for all. And if only one unknown is asked for, Cramer gives a single determinant to compute where Gauss-Jordan requires solving the whole system.

Exercise 8: Proof: the Vandermonde determinant of order 3

The Vandermonde determinant is $V(a, b, c)=\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{vmatrix}$ , which appears as soon as one looks for a polynomial through three points.

a) Compute $V(a, b, c)$ in terms of $a$ , $b$ , $c$ , and factor the result completely.
b) Deduce a simple condition on $a$ , $b$ and $c$ for the Vandermonde matrix to be invertible.
c) Numerical application: compute $V(1, 2, 4)$ in two ways, by the factored formula and by a direct expansion.
d) A second-degree polynomial $p(x)=\alpha+\beta x+\gamma x^{2}$ must pass through three points with $x$ -coordinates $a$ , $b$ , $c$ . Explain how the result of b) guarantees that such a polynomial exists and is unique.

Show the solution

Answers

a) $V=(b-a)(c-a)(c-b)$
b) $a$ , $b$ , $c$ distinct
c) $V(1,2,4)=6$
d) System with determinant $V\neq 0$ : unique parabola

a)

Create zeros in the first row by subtracting columns, which does not change the determinant: $C_{2}\leftarrow C_{2}-C_{1}$ and $C_{3}\leftarrow C_{3}-C_{1}$ give $\begin{vmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ a^{2} & b^{2}-a^{2} & c^{2}-a^{2} \end{vmatrix}$ . Expand along the first row: $V=\begin{vmatrix} b-a & c-a \\ b^{2}-a^{2} & c^{2}-a^{2} \end{vmatrix}$ . Factor each column: $b^{2}-a^{2}=(b-a)(b+a)$ and $c^{2}-a^{2}=(c-a)(c+a)$ , so $(b-a)$ comes out of the first column and $(c-a)$ out of the second: $V=(b-a)(c-a)\begin{vmatrix} 1 & 1 \\ b+a & c+a \end{vmatrix}=(b-a)(c-a)\big((c+a)-(b+a)\big)=(b-a)(c-a)(c-b)$ .

b)

The Vandermonde matrix is invertible if and only if $V(a, b, c)\neq 0$ , that is, if and only if none of the three factors $(b-a)$ , $(c-a)$ , $(c-b)$ is zero. In other words, it is invertible exactly when $a$ , $b$ and $c$ are pairwise distinct. A single coincidence between two of the three values is enough to make the determinant vanish.

c)

By the formula: $V(1, 2, 4)=(2-1)(4-1)(4-2)=(1)(3)(2)=6$ . By direct expansion: $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 16 \end{vmatrix}=1(2\cdot 16-4\cdot 4)-1(1\cdot 16-4\cdot 1)+1(1\cdot 4-2\cdot 1)=1(16)-1(12)+1(2)=6$ . The two methods agree, which confirms the factorization.

d)

Prescribing $p(a)$ , $p(b)$ , $p(c)$ amounts to solving a system of three linear equations whose unknowns are $\alpha$ , $\beta$ , $\gamma$ , and whose coefficient matrix is exactly the transpose of the Vandermonde matrix. Since a matrix and its transpose have the same determinant, the determinant of the system is $V(a, b, c)$ . When $a$ , $b$ , $c$ are distinct, this determinant is nonzero: the system has a unique solution, so there is one and only one polynomial of degree at most $2$ through the three points. This is the algebraic foundation of interpolation: three points with distinct $x$ -coordinates determine a unique parabola.

Exercise 9: Problem: the determinant as an area and as a volume

The determinant is not only a test for invertibility: its absolute value measures an area in the plane and a volume in space. This geometric reading is used here in both dimensions.

a) Compute the area of the triangle with vertices $A(1, 1)$ , $B(4, 2)$ and $C(2, 5)$ using a $2\times 2$ determinant.
b) Are the points $P(0, 0)$ , $Q(2, 1)$ and $R(6, 3)$ collinear? Answer with a determinant, and give the geometric meaning of its value.
c) Find the value of $t$ for which the points $(1, 2)$ , $(3, 5)$ and $(5, t)$ are collinear.
d) Compute the volume of the tetrahedron built on the origin and the three vectors $\vec{u}=(1, 0, 1)$ , $\vec{v}=(2, 1, 0)$ , $\vec{w}=(0, 3, 1)$ , then find the value of $m$ for which the vectors $(1, 1, 1)$ , $(2, 0, 1)$ and $(1, -1, m)$ are coplanar.

Show the solution

Answers

a) $\frac{11}{2}=5.5$
b) Zero determinant: collinear
c) $t=8$
d) Volume $\frac76$ ; $m=0$

a)

The area of a triangle $ABC$ is half the absolute value of the determinant formed by $\vec{AB}$ and $\vec{AC}$ . Here $\vec{AB}=(3, 1)$ and $\vec{AC}=(1, 4)$ , so the area is $\dfrac{1}{2}\left|\begin{vmatrix} 3 & 1 \\ 1 & 4 \end{vmatrix}\right|=\dfrac{1}{2}|12-1|=\dfrac{11}{2}=5.5$ square units. The absolute value is essential: the determinant alone can be negative depending on the orientation, but an area never is.

b)

$\vec{PQ}=(2, 1)$ and $\vec{PR}=(6, 3)$ . The determinant is $\begin{vmatrix} 2 & 6 \\ 1 & 3 \end{vmatrix}=6-6=0$ . Three points are collinear if and only if this determinant is zero, because a zero determinant means a zero area, hence a flattened triangle: the three points lie on one line. Here $\vec{PR}=3\vec{PQ}$ , so the collinearity is explicit.

c)

The points $(1, 2)$ , $(3, 5)$ , $(5, t)$ are collinear when $\vec{AB}=(2, 3)$ and $\vec{AC}=(4, t-2)$ are collinear, that is, when $\begin{vmatrix} 2 & 4 \\ 3 & t-2 \end{vmatrix}=0$ . This gives $2(t-2)-3(4)=2t-4-12=2t-16=0$ , so $t=8$ . For any other value of $t$ , the three points form a genuine triangle with nonzero area.

d)

The volume of the tetrahedron built on three vectors from a common vertex is one sixth of the absolute value of their determinant: $\begin{vmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 3 & 1 \end{vmatrix}=1(1-0)-0(2-0)+1(6-0)=1+6=7$ , so the volume is $\dfrac{1}{6}|7|=\dfrac{7}{6}$ cubic units. For coplanarity, require the determinant of the three vectors to be zero: $\begin{vmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 1 & -1 & m \end{vmatrix}=1(0\cdot m-1\cdot(-1))-1(2m-1)+1(-2-0)=1-2m+1-2=-2m$ . This determinant vanishes for $m=0$ : it is the only value that makes the three vectors coplanar, that is, that flattens to zero the volume of the parallelepiped they span.

Exercise 10: Proof: adjugate matrix, the fundamental identity and the determinant of the adjugate

Recall that the adjugate matrix $\text{adj}(A)$ is the transpose of the matrix of cofactors of $A$ , and that for $A$ invertible, $A^{-1}=\dfrac{1}{\det(A)}\text{adj}(A)$ .

a) Prove that for every invertible square matrix $A$ of order $n$ , $A\,\text{adj}(A)=\det(A)\,I_{n}$ .
b) Deduce that $\det(\text{adj}(A))=(\det(A))^{n-1}$ .
c) Application with $A=\begin{pmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 2 \end{pmatrix}$ : compute $\det(A)$ and $\text{adj}(A)$ , check the identity of a), and confirm the formula of b).
d) A matrix $B$ of order $3$ satisfies $\det(B)=1$ . What are $\det(\text{adj}(B))$ and $\det(\text{adj}(\text{adj}(B)))$ ? Justify using b).

Show the solution

Answers

a) $A\,\text{adj}(A)=\det(A)\,I_n$
b) $\det(\text{adj}(A))=\det(A)^{n-1}$
c) $\det(A)=5$ ; $\det(\text{adj}(A))=25$
d) 1 and 1

a)

Start from the equality that defines the inverse by cofactors: $A^{-1}=\dfrac{1}{\det(A)}\text{adj}(A)$ , valid because $A$ is invertible, so $\det(A)\neq 0$ . Multiply both sides on the left by $A$ : $A\,A^{-1}=\dfrac{1}{\det(A)}A\,\text{adj}(A)$ . The left side equals $I_{n}$ . Multiplying both sides by the scalar $\det(A)$ gives $\det(A)\,I_{n}=A\,\text{adj}(A)$ , which is the stated identity.

b)

Take the determinant of the identity $A\,\text{adj}(A)=\det(A)\,I_{n}$ . The left side factors: $\det(A)\det(\text{adj}(A))$ . The right side is the determinant of the scalar $\det(A)$ times $I_{n}$ , that is $(\det(A))^{n}\det(I_{n})=(\det(A))^{n}$ , since multiplying an $n\times n$ matrix by a scalar $k$ multiplies its determinant by $k^{n}$ and $\det(I_{n})=1$ . So $\det(A)\det(\text{adj}(A))=(\det(A))^{n}$ . Since $\det(A)\neq 0$ , divide: $\det(\text{adj}(A))=(\det(A))^{n-1}$ .

c)

Expansion along the first column: $\det(A)=2\begin{vmatrix} 1 & 0 \\ 1 & 2 \end{vmatrix}-1\begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix}+0=2(2)-1(-1)=5$ . The nine cofactors give the matrix of cofactors $\begin{pmatrix} 2 & -2 & 1 \\ 1 & 4 & -2 \\ -1 & 1 & 2 \end{pmatrix}$ , and the adjugate is its transpose: $\text{adj}(A)=\begin{pmatrix} 2 & 1 & -1 \\ -2 & 4 & 1 \\ 1 & -2 & 2 \end{pmatrix}$ . Check the identity: $A\,\text{adj}(A)=\begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}=5I_{3}=\det(A)I_{3}$ . Finally $\det(\text{adj}(A))=25=5^{2}=(\det(A))^{3-1}$ , which confirms the formula of b) with $n=3$ .

d)

With $n=3$ and $\det(B)=1$ : $\det(\text{adj}(B))=(\det(B))^{n-1}=1^{2}=1$ . Applying the formula again to the matrix $\text{adj}(B)$ , itself of order $3$ : $\det(\text{adj}(\text{adj}(B)))=(\det(\text{adj}(B)))^{2}=1^{2}=1$ . One can go further: the known identity $\text{adj}(\text{adj}(B))=(\det(B))^{n-2}B$ gives here $\text{adj}(\text{adj}(B))=(\det(B))^{1}B=B$ , so the adjugate of the adjugate gives back $B$ itself when $\det(B)=1$ and $n=3$ .

Exercise 11: What the multiplicativity of the determinant rules out

We use the property $\det(AB)=\det(A)\det(B)$ for square matrices of the same order, together with $\det(kA)=k^{n}\det(A)$ for a matrix of order $n$ .

a) Check the property with $A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B=\begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$ .
b) Is there a square matrix $A$ of order 3, with real entries, such that $A^{2}=-I_{3}$ ?
c) Compute $A^{2}$ for $A=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ . Why does the argument of b) not apply to matrices of order 2?
d) A matrix $N$ satisfies $N^{3}=0$ . What is $\det(N)$ ? Give a nonzero example of order 2 with $N^{2}=0$ .
e) A matrix $Q$ satisfies $Q^{T}Q=I$ . What values can $\det(Q)$ take? Compute it for $Q=\begin{pmatrix} 0.6 & -0.8 \\ 0.8 & 0.6 \end{pmatrix}$ .

Show the solution

Answers

a) $-12=(-2)\times 6$
b) No: $\det(A)^2=-1$
c) $A^2=-I_2$ ; $\det(-I_2)=1$
d) $\det(N)=0$
e) $\pm 1$ ; here 1

a)

$\det A=4-6=-2$ and $\det B=6-0=6$ . $AB=\begin{pmatrix} 2+2 & 0+6 \\ 6+4 & 0+12 \end{pmatrix}=\begin{pmatrix} 4 & 6 \\ 10 & 12 \end{pmatrix}$ , with determinant $48-60=-12=(-2)\times 6$ .

b)

No. If $A^{2}=-I_{3}$ , then $\det(A)^{2}=\det(-I_{3})=(-1)^{3}=-1$ . The square of a real number cannot equal $-1$ : no real matrix of order 3 works, and the same argument holds for every odd order.

c)

$A^{2}=\begin{pmatrix} 0-1 & 0+0 \\ 0+0 & -1+0 \end{pmatrix}=-I_{2}$ . For order 2, $\det(-I_{2})=(-1)^{2}=1$ and $\det(A)^{2}=1^{2}=1$ : no contradiction. This matrix is the rotation by a quarter turn, which applied twice gives the half turn $-I$ .

d)

$\det(N)^{3}=\det(N^{3})=\det(0)=0$ , so $\det(N)=0$ : a nilpotent matrix is never invertible. Example: $N=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ , nonzero, with $N^{2}=0$ and $\det N=0$ .

e)

$\det(Q^{T})\det(Q)=\det(Q)^{2}=\det(I)=1$ , so $\det(Q)=1$ or $\det(Q)=-1$ . For the given matrix: $0.36+0.64=1$ . It is a rotation; a reflection such as $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ would give $-1$ .

The common thread: taking the determinant of a matrix equation turns it into an equation between numbers, which is often impossible.

Exercise 12: A tridiagonal determinant of order n

For $n\geq 1$ , let $D_{n}$ be the determinant of the square matrix of order $n$ with $2$ on the diagonal, $-1$ just above and just below it, and $0$ everywhere else. For example $D_{3}=\begin{vmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{vmatrix}$ .

a) Compute $D_{1}$ and $D_{2}$ .
b) Compute $D_{3}$ directly.
c) By expanding along the first row, show that $D_{n}=2D_{n-1}-D_{n-2}$ for $n\geq 3$ .
d) Conjecture an expression for $D_{n}$ , prove it by induction and give $D_{10}$ .
e) Are all these matrices invertible?

Show the solution

Answers

a) $D_1=2$ , $D_2=3$
b) $D_3=4$
c) $D_n=2D_{n-1}-D_{n-2}$
d) $D_n=n+1$ ; $D_{10}=11$
e) Yes

a)

$D_{1}=2$ and $D_{2}=\begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix}=4-1=3$ .

b)

$D_{3}=2(4-1)-(-1)\big((-1)(2)-(-1)(0)\big)+0=6-2=4$ .

c)

The first entry, $2$ , has as its minor the same matrix of order $n-1$ : contribution $2D_{n-1}$ . The second, $-1$ , carries the sign $-$ ; its minor has a first column reduced to $-1$ at the top followed by zeros, and expands to $(-1)D_{n-2}$ . Contribution: $(-1)\times(-1)\times(-1)D_{n-2}=-D_{n-2}$ . The other entries of the row are zero. So $D_{n}=2D_{n-1}-D_{n-2}$ .

d)

$2$ , $3$ , $4$ : we conjecture $D_{n}=n+1$ . True for $n=1$ and $n=2$ . If $D_{n-1}=n$ and $D_{n-2}=n-1$ , then $D_{n}=2n-(n-1)=n+1$ . By induction, $D_{n}=n+1$ for every $n$ , and $D_{10}=11$ .

e)

Yes: $D_{n}=n+1\neq 0$ for every $n$ . These matrices, which model a string or a bar discretized at $n$ points, are always invertible.

The common thread: a determinant with a regular pattern reduces to a recurrence relation by expanding along the first row.

Exercise 13: Lines and planes by determinants

Three points of the plane are collinear if and only if $\begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}=0$ . Likewise, four points in space are coplanar if and only if the determinant of order 4 whose rows are $(x_{i}, y_{i}, z_{i}, 1)$ is zero.

a) Expand $\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 8 & 1 \end{vmatrix}=0$ and give the slope-intercept equation of the line obtained.
b) Why do the points $(1, 2)$ and $(3, 8)$ satisfy this equation without any calculation?
c) Is the point $(5, 14)$ on this line?
d) Expand $\begin{vmatrix} x & y & z & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 2 & 0 & 1 \\ 0 & 0 & 3 & 1 \end{vmatrix}=0$ along the first row and give the equation of the plane.
e) Are the points $(1, 0, 0)$ , $(0, 2, 0)$ , $(0, 0, 3)$ and $(2, -2, 3)$ coplanar?

Show the solution

Answers

a) $y=3x-1$
b) Two equal rows
c) Yes
d) $6x+3y+2z=6$
e) No: 12 instead of 6

a)

$x(2-8)-y(1-3)+1(8-6)=-6x+2y+2=0$ , that is $y=3x-1$ .

b)

Replacing $(x, y)$ by $(1, 2)$ makes the first row equal to the second: a determinant with two equal rows is zero. Same thing with $(3, 8)$ and the third row. So the line passes through both points.

c)

$3\times 5-1=14$ : yes, the point is on the line.

d)

The minors of the first row are $6$ , $-3$ , $2$ and $6$ , deleting each column in turn; with alternating signs: $6x-(-3)y+2z-6=0$ , that is $6x+3y+2z=6$ , or $\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=1$ , the form that shows the three points where the plane meets the axes.

e)

$6(2)+3(-2)+2(3)=12-6+6=12\neq 6$ : the fourth point is not in the plane, so the four points are not coplanar.

The common thread: an equation “the determinant is zero” describes the set of points that make the set of rows linearly dependent, a line or a plane depending on the dimension.

Exercise 14: Problem: the area of a lot by the shoelace formula

A lot has the shape of the pentagon with vertices $A(0, 0)$ , $B(4, 0)$ , $C(5, 3)$ , $D(2, 5)$ and $E(-1, 3)$ , listed counterclockwise. One unit is $10$ metres. The pentagon is convex.

The area of the triangle $OPQ$ , where $O$ is a vertex and $\vec{OP}=(a, b)$ , $\vec{OQ}=(c, d)$ , is $\frac{1}{2}|ad-bc|$ .

a) Why can the pentagon be cut into three triangles from $A$ ?
b) Compute the areas of the triangles $ABC$ , $ACD$ and $ADE$ .
c) Find the total area again by the shoelace formula: $\mathcal{A}=\frac{1}{2}\left|\sum_{i}(x_{i}y_{i+1}-x_{i+1}y_{i})\right|$ , the last vertex being followed by the first.
d) What does the sum in c), without the absolute value, become if the vertices are listed clockwise?
e) Give the actual area of the lot in square metres.

Show the solution

Answers

a) Convex pentagon
b) 6; $9.5$ ; $5.5$
c) $\frac{42}{2}=21$
d) $-42$
e) 2,100 m²

a)

Since the pentagon is convex, the segments $AC$ and $AD$ stay inside it: the triangles $ABC$ , $ACD$ and $ADE$ cover the lot without overlapping.

b)

$ABC$ : $\frac{1}{2}|4\times 3-5\times 0|=6$ . $ACD$ : $\frac{1}{2}|5\times 5-2\times 3|=\frac{19}{2}=9.5$ . $ADE$ : $\frac{1}{2}|2\times 3-(-1)\times 5|=\frac{11}{2}=5.5$ . Total: $6+9.5+5.5=21$ .

c)

The five terms: $0\times 0-4\times 0=0$ ; $4\times 3-5\times 0=12$ ; $5\times 5-2\times 3=19$ ; $2\times 3-(-1)\times 5=11$ ; $(-1)\times 0-0\times 3=0$ . The sum is $42$ , so $\mathcal{A}=21$ . Each term is twice the signed area of a triangle from the origin: the formula adds up $2\times 2$ determinants.

d)

Listing the vertices in the opposite order changes the sign of each determinant, since it swaps its two rows: the sum is $-42$ . The absolute value makes the formula independent of the direction, and the sign tells which direction was used.

e)

One square unit is $10\times 10=100\ \mathrm{m^{2}}$ , so the lot measures $21\times 100=2100\ \mathrm{m^{2}}$ .

The common thread: the area of a polygon is a sum of $2\times 2$ determinants, each giving the signed area of a triangle.

Exercise 15: Problem: the circle through three points

We look for the circle through $P_{1}(2, 0)$ , $P_{2}(0, 4)$ and $P_{3}(-2, 0)$ , in the form $x^{2}+y^{2}+Dx+Ey+F=0$ .

a) Write the system satisfied by $D$ , $E$ and $F$ , compute its determinant and solve it.
b) Find the centre and the radius of the circle.
c) Assume that the circle also has the equation $\begin{vmatrix} x^{2}+y^{2} & x & y & 1 \\ 4 & 2 & 0 & 1 \\ 16 & 0 & 4 & 1 \\ 4 & -2 & 0 & 1 \end{vmatrix}=0$ . The minors of the first row are $16$ , $0$ , $-48$ and $64$ . Expand and compare with b).
d) Are the points $(2, 3)$ and $(1.5, 4)$ on the circle?
e) Why does no circle pass through $(0, 0)$ , $(1, 1)$ and $(2, 2)$ ? Answer with the determinant of the system of a).

Show the solution

Answers

a) $D=0$ , $E=-3$ , $F=-4$
b) Centre $(0, 1.5)$ , radius $2.5$
c) $16(x^2+y^2-3y-4)=0$
d) $(2, 3)$ yes; $(1.5, 4)$ no
e) Zero determinant: collinear points

a)

$P_{1}$ : $4+2D+F=0$ . $P_{2}$ : $16+4E+F=0$ . $P_{3}$ : $4-2D+F=0$ . The determinant of the coefficients of $(D, E, F)$ is $\begin{vmatrix} 2 & 0 & 1 \\ 0 & 4 & 1 \\ -2 & 0 & 1 \end{vmatrix}=2(4-0)-0+1(0+8)=16\neq 0$ : unique solution. Subtracting the third equation from the first gives $4D=0$ , so $D=0$ , then $F=-4$ and $E=\frac{-16+4}{4}=-3$ . The circle is $x^{2}+y^{2}-3y-4=0$ .

b)

$x^{2}+\left(y-\frac{3}{2}\right)^{2}=4+\frac{9}{4}=\frac{25}{4}$ : centre $(0, 1.5)$ , radius $2.5$ .

c)

With alternating signs: $16(x^{2}+y^{2})-0\cdot x+(-48)y-64=0$ , that is $16(x^{2}+y^{2}-3y-4)=0$ : the same equation, multiplied by $16$ . Replacing $(x, y)$ by any of the three points makes the first row equal to another one, which makes the determinant zero: the circle does pass through the three points.

d)

$(2, 3)$ : $4+9-9-4=0$ , yes. $(1.5, 4)$ : $2.25+16-12-4=2.25\neq 0$ , no, the point is outside.

e)

The system becomes $F=0$ , $2+D+E+F=0$ , $8+2D+2E+F=0$ , with determinant $\begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{vmatrix}=0$ , the first two columns being equal. It would give $D+E=-2$ and $D+E=-4$ : inconsistent. Three collinear points lie on no circle.

The common thread: a curve defined by three linear conditions on its coefficients exists and is unique when the determinant of the system is not zero.

Corrected exercises: determinants, cofactors and Cramer's rule (201-SN4-RE)

Course recap

Calculations: the determinants to compute

Exercise 0.1: Computing determinants: ten direct calculations

Part A: the basics (/50)

Exercise 1: Determinants of order 2 and 3, two methods

Exercise 2: A determinant of order 4 by cofactors

Exercise 3: Properties of the determinant and elementary row operations

Exercise 4: Adjugate matrix and inverse by cofactors

Exercise 5: Computing a determinant cleverly: triangular, proportional rows, factors

Part B: exam level (/50)

Exercise 6: Proofs and a singular matrix with a parameter

Exercise 7: Problem: three tutoring formats solved by Cramer's rule

Exercise 8: Proof: the Vandermonde determinant of order 3

Exercise 9: Problem: the determinant as an area and as a volume

Exercise 10: Proof: adjugate matrix, the fundamental identity and the determinant of the adjugate

Part C: the classics (/50)

Exercise 11: What the multiplicativity of the determinant rules out

Exercise 12: A tridiagonal determinant of order n

Exercise 13: Lines and planes by determinants

Exercise 14: Problem: the area of a lot by the shoelace formula

Exercise 15: Problem: the circle through three points

See also

Looking for a linear algebra tutor in Montreal?