Exercise 0.1: Computing matrix products: ten direct calculations

For each line, carry out the requested calculation, then give the indicated entries of the result. We write $M_{ij}$ for the entry in row $i$ and column $j$ of the matrix $M$ .

Before any product, write down the sizes: $(m\times n)(n\times p)$ gives an $m\times p$ matrix, and the product exists only if the two $n$ match. The entry $(AB)_{ij}$ is row $i$ of $A$ times column $j$ of $B$ : multiply term by term and add. A linear combination is computed entry by entry; a power is computed by successive products, never by raising each entry to the power.

a) $A=\begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix}$ and $B=\begin{pmatrix} -1 & 2 \\ 5 & 0 \end{pmatrix}$ : compute $2A+3B$ , then give $(2A+3B)_{12}$ and $(2A+3B)_{21}$ .
b) $C=\begin{pmatrix} 1 & -2 \\ 0 & 3 \\ 4 & 1 \end{pmatrix}$ : write $C^{T}$ , then compute $CC^{T}$ ; give $(C^{T})_{23}$ and $(CC^{T})_{13}$ .
c) $\begin{pmatrix} 2 & 5 \\ -1 & 3 \end{pmatrix}\begin{pmatrix} 4 & -1 \\ 2 & 6 \end{pmatrix}$ : give entries $(1,2)$ and $(2,1)$ of the product.
d) $\begin{pmatrix} 2 & -1 & 4 \end{pmatrix}\begin{pmatrix} 3 \\ 5 \\ -2 \end{pmatrix}$ : give the result.
e) $\begin{pmatrix} 3 \\ 5 \\ -2 \end{pmatrix}\begin{pmatrix} 2 & -1 & 4 \end{pmatrix}$ : give entries $(2,3)$ and $(3,2)$ of the product.
f) $\begin{pmatrix} 1 & 0 & -2 \\ 3 & 1 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ -1 & 4 \\ 3 & 0 \end{pmatrix}$ : give entries $(1,1)$ and $(2,2)$ of the product.
g) $\begin{pmatrix} 1 & 2 & 0 \\ -1 & 1 & 3 \\ 2 & 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & -1 & 1 \\ 0 & 3 & 2 \\ 1 & 1 & -2 \end{pmatrix}$ : give entries $(2,3)$ and $(3,2)$ of the product.
h) $M=\begin{pmatrix} 1 & -3 \\ 2 & 4 \end{pmatrix}$ : compute $M^{2}$ , then give $(M^{2})_{11}$ and $(M^{2})_{22}$ .
i) $N=\begin{pmatrix} 2 & 1 \\ 0 & -1 \end{pmatrix}$ : compute $N^{3}$ , then give $(N^{3})_{12}$ and $(N^{3})_{22}$ .
j) $P=\begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}$ : compute $P^{2}-3P+2I$ , then give its entries $(1,1)$ and $(1,2)$ .

Type your answers, the page tells you right or wrong 0/19

a)

(2A+3B)_{12}

(2A+3B)_{21}

b)

(C^{T})_{23}

(CC^{T})_{13}

c)

Entry

(1,2)

Entry

(2,1)

d)

The result

e)

Entry

(2,3)

Entry

(3,2)

f)

Entry

(1,1)

Entry

(2,2)

g)

Entry

(2,3)

Entry

(3,2)

h)

(M^{2})_{11}

(M^{2})_{22}

i)

(N^{3})_{12}

(N^{3})_{22}

j)

Entry

(1,1)

Entry

(1,2)

Show the solution

Answers

a) $2A+3B=\begin{pmatrix} 3 & 2 \\ 17 & 8 \end{pmatrix}$ ; $2$ and $17$
b) $(C^{T})_{23}=1$ ; $(CC^{T})_{13}=2$
c) $\begin{pmatrix} 18 & 28 \\ 2 & 19 \end{pmatrix}$ ; $28$ and $2$
d) $6-5-8=-7$
e) a $3\times 3$ matrix; $20$ and $2$
f) $\begin{pmatrix} -4 & 1 \\ 11 & 7 \end{pmatrix}$ ; $-4$ and $7$
g) $\begin{pmatrix} 2 & 5 & 5 \\ 1 & 7 & -5 \\ 5 & -1 & 0 \end{pmatrix}$ ; $-5$ and $-1$
h) $M^{2}=\begin{pmatrix} -5 & -15 \\ 10 & 10 \end{pmatrix}$ ; $-5$ and $10$
i) $N^{3}=\begin{pmatrix} 8 & 3 \\ 0 & -1 \end{pmatrix}$ ; $3$ and $-1$
j) $P^{2}-3P+2I=\begin{pmatrix} -1 & 2 \\ -2 & 1 \end{pmatrix}$ ; $-1$ and $2$

The move that decides everything: write the sizes down BEFORE computing. They tell you whether the product exists, what shape the result has, and which row meets which column for each requested entry.

a)

$2A=\begin{pmatrix} 6 & -4 \\ 2 & 8 \end{pmatrix}$ and $3B=\begin{pmatrix} -3 & 6 \\ 15 & 0 \end{pmatrix}$ , so $2A+3B=\begin{pmatrix} 3 & 2 \\ 17 & 8 \end{pmatrix}$ : $(2A+3B)_{12}=2$ and $(2A+3B)_{21}=17$ . Pitfall: reading $(\cdot)_{12}$ as row 2, column 1; the ROW index always comes first.

b)

The transpose swaps rows and columns: $C^{T}=\begin{pmatrix} 1 & 0 & 4 \\ -2 & 3 & 1 \end{pmatrix}$ , so $(C^{T})_{23}=C_{32}=1$ . $CC^{T}$ is a $(3\times 2)(2\times 3)$ product, hence a $3\times 3$ matrix; its entry $(1,3)$ is row 1 of $C$ times column 3 of $C^{T}$ , that is, times row 3 of $C$ : $1\times 4+(-2)\times 1=2$ . Pitfall: computing $C^{T}C$ , which is a $2\times 2$ matrix and has no column 3.

c)

A $(2\times 2)(2\times 2)$ product. Row 1 times column 2: $2\times(-1)+5\times 6=28$ . Row 2 times column 1: $(-1)\times 4+3\times 2=2$ . The full product is $\begin{pmatrix} 18 & 28 \\ 2 & 19 \end{pmatrix}$ . Pitfall: multiplying entry by entry, which gives $\begin{pmatrix} 8 & -5 \\ -2 & 18 \end{pmatrix}$ .

d)

A $(1\times 3)(3\times 1)$ product gives a $1\times 1$ matrix, that is, a number: $2\times 3+(-1)\times 5+4\times(-2)=6-5-8=-7$ .

e)

The same factors as in d), in the other order: $(3\times 1)(1\times 3)$ gives a $3\times 3$ matrix, whose entry $(i,j)$ is the $i$ -th term of the column times the $j$ -th term of the row. Entry $(2,3)$ : $5\times 4=20$ . Entry $(3,2)$ : $(-2)\times(-1)=2$ . Pitfall: expecting to get $-7$ again; here, changing the order changes even the size.

f)

$(2\times 3)(3\times 2)$ gives a $2\times 2$ matrix. Row 1 times column 1: $1\times 2+0\times(-1)+(-2)\times 3=-4$ . Row 2 times column 2: $3\times 1+1\times 4+2\times 0=7$ . The full product is $\begin{pmatrix} -4 & 1 \\ 11 & 7 \end{pmatrix}$ .

g)

Row 2 times column 3: $(-1)\times 1+1\times 2+3\times(-2)=-5$ . Row 3 times column 2: $2\times(-1)+0\times 3+1\times 1=-1$ . The full product is $\begin{pmatrix} 2 & 5 & 5 \\ 1 & 7 & -5 \\ 5 & -1 & 0 \end{pmatrix}$ . Pitfall: taking column 2 of the FIRST matrix; the first factor always supplies the row.

h)

$M^{2}=MM=\begin{pmatrix} 1-6 & -3-12 \\ 2+8 & -6+16 \end{pmatrix}=\begin{pmatrix} -5 & -15 \\ 10 & 10 \end{pmatrix}$ , so $(M^{2})_{11}=-5$ and $(M^{2})_{22}=10$ . Pitfall: squaring each entry, which gives $\begin{pmatrix} 1 & 9 \\ 4 & 16 \end{pmatrix}$ .

i)

$N^{2}=\begin{pmatrix} 4 & 2-1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 4 & 1 \\ 0 & 1 \end{pmatrix}$ , then $N^{3}=N^{2}N=\begin{pmatrix} 8 & 4-1 \\ 0 & -1 \end{pmatrix}=\begin{pmatrix} 8 & 3 \\ 0 & -1 \end{pmatrix}$ . The diagonal of a triangular matrix can be computed on its own, $2^{3}=8$ and $(-1)^{3}=-1$ , but not the off-diagonal entry. Pitfall: writing $1^{3}=1$ in position $(1,2)$ .

j)

$P^{2}=\begin{pmatrix} 4-1 & 2+3 \\ -2-3 & -1+9 \end{pmatrix}=\begin{pmatrix} 3 & 5 \\ -5 & 8 \end{pmatrix}$ , $-3P=\begin{pmatrix} -6 & -3 \\ 3 & -9 \end{pmatrix}$ and $2I=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ . The sum is $\begin{pmatrix} -1 & 2 \\ -2 & 1 \end{pmatrix}$ : entries $-1$ and $2$ . Pitfall: adding $2$ to every entry instead of $2I$ , which gives $4$ in position $(1,2)$ .

Check: the rule $(m\times n)(n\times p)=m\times p$ tells you the shape of each result in advance, and an entry is cross-checked by recomputing ONE row times ONE column. In j), the trace checks the whole calculation: $\operatorname{tr}(P^{2})-3\operatorname{tr}(P)+2\operatorname{tr}(I)=11-15+4=0$ , and the trace found is indeed $-1+1=0$ .

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Exercise 0.2: Inverting a matrix: eight direct calculations

For each line, compute the requested inverse, or the solution of the system, then give the indicated entries. A matrix that is not invertible is reported as such.

For a $2\times 2$ matrix, first compute $\det A=ad-bc$ : if it is zero, the matrix is not invertible; otherwise swap $a$ and $d$ , change the sign of $b$ and $c$ , and divide by the determinant. For a $3\times 3$ matrix, form $[A\,|\,I]$ and apply elementary row operations until you reach $[I\,|\,A^{-1}]$ . For a system $AX=B$ , write $X=A^{-1}B$ , with the inverse on the LEFT.

a) $P=\begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix}$ : compute $P^{-1}$ , then give $(P^{-1})_{11}$ and $(P^{-1})_{12}$ .
b) $Q=\begin{pmatrix} 4 & -2 \\ 3 & 1 \end{pmatrix}$ : compute $Q^{-1}$ , then give $(Q^{-1})_{12}$ and $(Q^{-1})_{21}$ .
c) $R=\begin{pmatrix} 6 & -4 \\ -9 & 6 \end{pmatrix}$ : compute $\det R$ and say whether $R$ is invertible.
d) $D=\begin{pmatrix} -5 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & \frac{1}{3} \end{pmatrix}$ : compute $D^{-1}$ , then give $(D^{-1})_{11}$ and $(D^{-1})_{33}$ .
e) $U=\begin{pmatrix} 1 & 2 & -1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}$ : compute $U^{-1}$ by the $[U\,|\,I]$ method, then give $(U^{-1})_{13}$ and $(U^{-1})_{23}$ .
f) $M=\begin{pmatrix} 1 & 2 & 1 \\ 2 & 5 & 2 \\ 1 & 3 & 2 \end{pmatrix}$ : compute $M^{-1}$ by the $[M\,|\,I]$ method, then give $(M^{-1})_{11}$ and $(M^{-1})_{32}$ .
g) Write the system $\begin{cases} 3x+5y=1 \\ 2x+4y=-2 \end{cases}$ in the form $AX=B$ , then solve it with $X=A^{-1}B$ .
h) $K=\begin{pmatrix} 3 & 1 & 0 \\ 2 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}$ has inverse $K^{-1}=\begin{pmatrix} 1 & -1 & 0 \\ -2 & 3 & 0 \\ 1 & -2 & 1 \end{pmatrix}$ : solve $KX=\begin{pmatrix} 5 \\ 3 \\ 4 \end{pmatrix}$ , then give $y$ and $z$ .

Type your answers, the page tells you right or wrong 0/16

a)

(P^{-1})_{11}

(P^{-1})_{12}

b)

(Q^{-1})_{12}

(Q^{-1})_{21}

c)

\det R

Is

R

invertible?

d)

(D^{-1})_{11}

(D^{-1})_{33}

e)

(U^{-1})_{13}

(U^{-1})_{23}

f)

(M^{-1})_{11}

(M^{-1})_{32}

g)

x

y

h)

y

z

Show the solution

Answers

a) $P^{-1}=\begin{pmatrix} 2 & -3 \\ -3 & 5 \end{pmatrix}$ ; $2$ and $-3$
b) $Q^{-1}=\frac{1}{10}\begin{pmatrix} 1 & 2 \\ -3 & 4 \end{pmatrix}$ ; $0.2$ and $-0.3$
c) $\det R=0$ ; not invertible
d) $D^{-1}=\operatorname{diag}\left(-\frac{1}{5},\frac{1}{4},3\right)$ ; $-0.2$ and $3$
e) $U^{-1}=\begin{pmatrix} 1 & -2 & 7 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix}$ ; $7$ and $-3$
f) $M^{-1}=\begin{pmatrix} 4 & -1 & -1 \\ -2 & 1 & 0 \\ 1 & -1 & 1 \end{pmatrix}$ ; $4$ and $-1$
g) $X=A^{-1}B=\begin{pmatrix} 7 \\ -4 \end{pmatrix}$ ; $x=7$ , $y=-4$
h) $X=\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ ; $y=-1$ , $z=3$

The move that decides everything: the determinant first. If it is zero, the matrix has no inverse and you stop; if not, it is the common denominator of every entry of the inverse. Each result is then checked with one row of the product $AA^{-1}=I$ .

a)

$\det P=5\times 2-3\times 3=1$ . Swap $5$ and $2$ , change the sign of both $3$ s: $P^{-1}=\begin{pmatrix} 2 & -3 \\ -3 & 5 \end{pmatrix}$ . Check on row 1 of $PP^{-1}$ : $5\times 2+3\times(-3)=1$ and $5\times(-3)+3\times 5=0$ .

b)

$\det Q=4\times 1-(-2)\times 3=4+6=10$ , so $Q^{-1}=\frac{1}{10}\begin{pmatrix} 1 & 2 \\ -3 & 4 \end{pmatrix}$ : $(Q^{-1})_{12}=\frac{2}{10}=0.2$ and $(Q^{-1})_{21}=-\frac{3}{10}=-0.3$ . Pitfall: writing $\det Q=4-6=-2$ , forgetting that $b=-2$ is already negative.

c)

$\det R=6\times 6-(-4)\times(-9)=36-36=0$ : $R$ is not invertible. The second row is $-\frac{3}{2}$ times the first, which could be seen before any calculation. Pitfall: applying the formula anyway and dividing by zero.

d)

A diagonal matrix is inverted entry by entry: $D^{-1}=\begin{pmatrix} -\frac{1}{5} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & 3 \end{pmatrix}$ , so $(D^{-1})_{11}=-\frac{1}{5}=-0.2$ and $(D^{-1})_{33}=3$ . Pitfall: losing the sign, when $\frac{1}{-5}$ stays negative.

e)

Start from $[U\,|\,I]$ and work upward. $R_{2}\to R_{2}-3R_{3}$ gives $(0\ \ 1\ \ 0\,|\,0\ \ 1\ \ -3)$ ; $R_{1}\to R_{1}+R_{3}$ gives $(1\ \ 2\ \ 0\,|\,1\ \ 0\ \ 1)$ ; $R_{1}\to R_{1}-2R_{2}$ gives $(1\ \ 0\ \ 0\,|\,1\ \ -2\ \ 7)$ . So $U^{-1}=\begin{pmatrix} 1 & -2 & 7 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix}$ : $(U^{-1})_{13}=7$ and $(U^{-1})_{23}=-3$ . Pitfall: only changing the off-diagonal signs, which puts $1$ in position $(1,3)$ ; the $7=2\times 3+1$ mixes the two other entries.

f)

$[M\,|\,I]$ , then $R_{2}\to R_{2}-2R_{1}$ : $(0\ \ 1\ \ 0\,|\,-2\ \ 1\ \ 0)$ ; $R_{3}\to R_{3}-R_{1}$ : $(0\ \ 1\ \ 1\,|\,-1\ \ 0\ \ 1)$ ; $R_{3}\to R_{3}-R_{2}$ : $(0\ \ 0\ \ 1\,|\,1\ \ -1\ \ 1)$ . Now work upward: $R_{1}\to R_{1}-2R_{2}$ gives $(1\ \ 0\ \ 1\,|\,5\ \ -2\ \ 0)$ , then $R_{1}\to R_{1}-R_{3}$ gives $(1\ \ 0\ \ 0\,|\,4\ \ -1\ \ -1)$ . So $M^{-1}=\begin{pmatrix} 4 & -1 & -1 \\ -2 & 1 & 0 \\ 1 & -1 & 1 \end{pmatrix}$ : $(M^{-1})_{11}=4$ and $(M^{-1})_{32}=-1$ . Check, row 1 of $M$ times column 1 of $M^{-1}$ : $4-4+1=1$ .

g)

$A=\begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix}$ , $X=\begin{pmatrix} x \\ y \end{pmatrix}$ , $B=\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ . $\det A=12-10=2$ , so $A^{-1}=\frac{1}{2}\begin{pmatrix} 4 & -5 \\ -2 & 3 \end{pmatrix}$ and $X=\frac{1}{2}\begin{pmatrix} 4\times 1+(-5)\times(-2) \\ -2\times 1+3\times(-2) \end{pmatrix}=\frac{1}{2}\begin{pmatrix} 14 \\ -8 \end{pmatrix}=\begin{pmatrix} 7 \\ -4 \end{pmatrix}$ . Check: $3\times 7+5\times(-4)=1$ and $2\times 7+4\times(-4)=-2$ .

h)

$X=K^{-1}\begin{pmatrix} 5 \\ 3 \\ 4 \end{pmatrix}$ , row by row: $x=5-3+0=2$ , $y=-10+9+0=-1$ , $z=5-6+4=3$ . Check in $K$ : $3\times 2-1=5$ , $2\times 2-1=3$ and $2-1+3=4$ . Pitfall: writing $X=BK^{-1}$ , a $(3\times 1)(3\times 3)$ product that does not even exist.

Check: each inverse is verified with one row of $AA^{-1}$ , which must give back a row of $I$ , and each solution is substituted into the original system. A $2\times 2$ inverse whose entries are not all divided by the determinant is wrong from the start.

Exercise 1: Matrix operations and non-commutativity

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

a) Compute $A+B$ and $3A-2B$ .
b) Find the transpose $A^{T}$ .
c) Compute both products $AB$ and $BA$ .
d) Compare $AB$ and $BA$ . What can you conclude about the commutativity of the matrix product?

Show the solution

Answers

a) $A+B=\begin{pmatrix}3&3\\1&5\end{pmatrix}$ , $3A-2B=\begin{pmatrix}4&-11\\13&-10\end{pmatrix}$
b) $A^T=\begin{pmatrix}2&3\\-1&0\end{pmatrix}$
c) $AB=\begin{pmatrix}4&3\\3&12\end{pmatrix}$ , $BA=\begin{pmatrix}14&-1\\11&2\end{pmatrix}$
d) $AB\neq BA$ : not commutative

a)

$A+B=\begin{pmatrix} 2+1 & -1+4 \\ 3-2 & 0+5 \end{pmatrix}=\begin{pmatrix} 3 & 3 \\ 1 & 5 \end{pmatrix}$ . For $3A-2B$ : $3A=\begin{pmatrix} 6 & -3 \\ 9 & 0 \end{pmatrix}$ and $2B=\begin{pmatrix} 2 & 8 \\ -4 & 10 \end{pmatrix}$ , so $3A-2B=\begin{pmatrix} 4 & -11 \\ 13 & -10 \end{pmatrix}$ .

b)

$A^{T}=\begin{pmatrix} 2 & 3 \\ -1 & 0 \end{pmatrix}$ : the first row of $A$ becomes the first column.

c)

$AB=\begin{pmatrix} (2)(1)+(-1)(-2) & (2)(4)+(-1)(5) \\ (3)(1)+(0)(-2) & (3)(4)+(0)(5) \end{pmatrix}=\begin{pmatrix} 4 & 3 \\ 3 & 12 \end{pmatrix}$ . And $BA=\begin{pmatrix} (1)(2)+(4)(3) & (1)(-1)+(4)(0) \\ (-2)(2)+(5)(3) & (-2)(-1)+(5)(0) \end{pmatrix}=\begin{pmatrix} 14 & -1 \\ 11 & 2 \end{pmatrix}$ .

d)

$AB=\begin{pmatrix} 4 & 3 \\ 3 & 12 \end{pmatrix}\neq\begin{pmatrix} 14 & -1 \\ 11 & 2 \end{pmatrix}=BA$ . The matrix product is therefore NOT commutative: the order of the factors changes the result. This is the most important difference from the product of real numbers, and the number one source of errors in this chapter.

Exercise 2: The matrix product and writing a system

Consider the matrix $A=\begin{pmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 3 & 2 & 1 \end{pmatrix}$ and the matrix $C=\begin{pmatrix} 1 & 0 \\ 2 & 1 \\ 0 & 3 \end{pmatrix}$ .

a) Is the product $AC$ defined? If so, give its size, then compute it.
b) Is the product $CA$ defined? Justify.
c) Write the system $\begin{cases} 2x+y-z=1 \\ x-3y+2z=1 \\ 3x+2y+z=10 \end{cases}$ in the matrix form $AX=B$ , specifying $X$ and $B$ .
d) By computing the product $AX$ , check that $X=\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ is a solution of the system.

Show the solution

Answers

a) $3\times 2$ : $\begin{pmatrix}4&-2\\-5&3\\7&5\end{pmatrix}$
b) No: 2 columns, 3 rows
c) $X=(x,y,z)$ , $B=(1,1,10)$
d) $AX=B$ : a solution

a)

$A$ is $3\times 3$ and $C$ is $3\times 2$ : the number of columns of $A$ (3) equals the number of rows of $C$ (3), so $AC$ is defined and has size $3\times 2$ . $AC=\begin{pmatrix} (2)(1)+(1)(2)+(-1)(0) & (2)(0)+(1)(1)+(-1)(3) \\ (1)(1)+(-3)(2)+(2)(0) & (1)(0)+(-3)(1)+(2)(3) \\ (3)(1)+(2)(2)+(1)(0) & (3)(0)+(2)(1)+(1)(3) \end{pmatrix}=\begin{pmatrix} 4 & -2 \\ -5 & 3 \\ 7 & 5 \end{pmatrix}$ .

b)

$C$ is $3\times 2$ and $A$ is $3\times 3$ : the number of columns of $C$ (2) differs from the number of rows of $A$ (3), so $CA$ is NOT defined. One more illustration that order matters: $AC$ exists but $CA$ does not.

c)

$A=\begin{pmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 3 & 2 & 1 \end{pmatrix}$ , $X=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ and $B=\begin{pmatrix} 1 \\ 1 \\ 10 \end{pmatrix}$ . The system reads $AX=B$ .

d)

$AX=\begin{pmatrix} (2)(1)+(1)(2)+(-1)(3) \\ (1)(1)+(-3)(2)+(2)(3) \\ (3)(1)+(2)(2)+(1)(3) \end{pmatrix}=\begin{pmatrix} 2+2-3 \\ 1-6+6 \\ 3+4+3 \end{pmatrix}=\begin{pmatrix} 1 \\ 1 \\ 10 \end{pmatrix}=B$ . The product gives back exactly $B$ , so $X=(1,\ 2,\ 3)$ is indeed a solution.

Exercise 3: Solving by Gauss-Jordan elimination

We return to the system of the previous exercise, but this time we solve it without knowing the answer in advance:

$\begin{cases} 2x+y-z=1 \\ x-3y+2z=1 \\ 3x+2y+z=10 \end{cases}$

a) Write the augmented matrix $[A\,|\,B]$ of the system.
b) Using Gauss-Jordan elimination, reduce this matrix to reduced row echelon form. State the elementary row operations used at each step.
c) Deduce the solution of the system.

Show the solution

Answers

a) A $3\times 4$ matrix
b) Reduced form $[I\,|\,X]$
c) $(1,\ 2,\ 3)$

a)

$[A\,|\,B]=\left(\begin{array}{ccc|c} 2 & 1 & -1 & 1 \\ 1 & -3 & 2 & 1 \\ 3 & 2 & 1 & 10 \end{array}\right)$ .

b)

First swap $R_{1}$ and $R_{2}$ to put a 1 in the top left corner: $\left(\begin{array}{ccc|c} 1 & -3 & 2 & 1 \\ 2 & 1 & -1 & 1 \\ 3 & 2 & 1 & 10 \end{array}\right)$ . Then $R_{2}\to R_{2}-2R_{1}$ and $R_{3}\to R_{3}-3R_{1}$ : $\left(\begin{array}{ccc|c} 1 & -3 & 2 & 1 \\ 0 & 7 & -5 & -1 \\ 0 & 11 & -5 & 7 \end{array}\right)$ .

b, continued) $R_{2}\to \tfrac{1}{7}R_{2}$ gives the row $(0,\ 1,\ -\tfrac{5}{7}\,|\,-\tfrac{1}{7})$ . Clear column 2 everywhere else: $R_{1}\to R_{1}+3R_{2}$ and $R_{3}\to R_{3}-11R_{2}$ . The third row becomes $(0,\ 0,\ \tfrac{20}{7}\,|\,\tfrac{60}{7})$ , which after $R_{3}\to \tfrac{7}{20}R_{3}$ is the row $(0,\ 0,\ 1\,|\,3)$ : so $z=3$ .

c)

Working back up (or finishing the reduction): row 2 gives $y-\tfrac{5}{7}z=-\tfrac{1}{7}$ , hence $y=\tfrac{-1+5(3)}{7}=\tfrac{14}{7}=2$ . Row 1 gives $x-3y+2z=1$ , hence $x=1+3(2)-2(3)=1$ . The reduced row echelon form is therefore $[I\,|\,X]$ with $X=(1,\ 2,\ 3)$ . The system has the unique solution $x=1,\ y=2,\ z=3$ , which confirms the check made in exercise 2.

Exercise 4: Determinants and the inverse matrix

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

a) Compute $\det M$ .
b) Is $M$ invertible? If so, find $M^{-1}$ , then check that $M\,M^{-1}=I$ .
c) Compute $\det A$ by cofactor expansion along the first row.
d) Using $\det A$ , what can you conclude about the number of solutions of the system $AX=B$ of exercise 3, without solving it?

Show the solution

Answers

a) $\det M=3$
b) $M^{-1}=\begin{pmatrix}0&\frac13\\-1&\frac23\end{pmatrix}$
c) $\det A=-20$
d) Unique solution

a)

$\det M=(2)(0)-(-1)(3)=0+3=3$ .

b)

$\det M=3\neq 0$ , so $M$ is invertible. $M^{-1}=\dfrac{1}{3}\begin{pmatrix} 0 & 1 \\ -3 & 2 \end{pmatrix}=\begin{pmatrix} 0 & \frac{1}{3} \\ -1 & \frac{2}{3} \end{pmatrix}$ . Check: $M\,M^{-1}=\begin{pmatrix} 2 & -1 \\ 3 & 0 \end{pmatrix}\begin{pmatrix} 0 & \frac{1}{3} \\ -1 & \frac{2}{3} \end{pmatrix}=\begin{pmatrix} 0+1 & \frac{2}{3}-\frac{2}{3} \\ 0+0 & 1+0 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=I$ .

c)

$\det A=2\,\det\begin{pmatrix} -3 & 2 \\ 2 & 1 \end{pmatrix}-1\,\det\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}+(-1)\,\det\begin{pmatrix} 1 & -3 \\ 3 & 2 \end{pmatrix}$ . Now $\det\begin{pmatrix} -3 & 2 \\ 2 & 1 \end{pmatrix}=-3-4=-7$ , $\det\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}=1-6=-5$ and $\det\begin{pmatrix} 1 & -3 \\ 3 & 2 \end{pmatrix}=2+9=11$ . Therefore $\det A=2(-7)-1(-5)-1(11)=-14+5-11=-20$ .

d)

$\det A=-20\neq 0$ , so $A$ is invertible and the system $AX=B$ has a unique solution, whatever the column $B$ . We know this without solving: it is consistent with the unique solution $(1,\ 2,\ 3)$ found in exercise 3.

Exercise 5: Inverse of a 3x3 matrix by Gauss-Jordan elimination

Consider the matrix $A=\begin{pmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8 \end{pmatrix}$ . We want its inverse by the method $[A\,|\,I]\to[I\,|\,A^{-1}]$ , which is different from the cofactor formula.

a) Write the augmented matrix $[A\,|\,I_{3}]$ , then apply Gauss-Jordan elimination, stating the elementary row operations. Deduce $A^{-1}$ .
b) Check your result by computing the product $A\,A^{-1}$ .
c) Use $A^{-1}$ to solve the system $AX=B$ with $B=\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ , by writing $X=A^{-1}B$ .
d) Explain in one sentence why this method becomes especially worthwhile when several systems $AX=B$ must be solved with the same matrix $A$ but different columns $B$ .

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Answers

a) $A^{-1}=\begin{pmatrix}-11&2&2\\-4&0&1\\6&-1&-1\end{pmatrix}$
b) $AA^{-1}=I_3$
c) $(-1,\ -1,\ 1)$
d) Inversion done once

a)

Start from $[A\,|\,I]=\left(\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0 \\ 2 & -1 & 3 & 0 & 1 & 0 \\ 4 & 1 & 8 & 0 & 0 & 1 \end{array}\right)$ . Clear the first column below the pivot: $R_{2}\to R_{2}-2R_{1}$ and $R_{3}\to R_{3}-4R_{1}$ give $\left(\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0 \\ 0 & -1 & -1 & -2 & 1 & 0 \\ 0 & 1 & 0 & -4 & 0 & 1 \end{array}\right)$ .

Make the pivot of the second row equal to $1$ with $R_{2}\to -R_{2}$ : row $(0,\ 1,\ 1\,|\,2,\ -1,\ 0)$ . Then $R_{3}\to R_{3}-R_{2}$ gives $(0,\ 0,\ -1\,|\,-6,\ 1,\ 1)$ , and $R_{3}\to -R_{3}$ gives $(0,\ 0,\ 1\,|\,6,\ -1,\ -1)$ : the pivot of the third row is $1$ .

It remains to clear the third column above the last pivot: $R_{2}\to R_{2}-R_{3}$ gives $(0,\ 1,\ 0\,|\,-4,\ 0,\ 1)$ , and $R_{1}\to R_{1}-2R_{3}$ gives $(1,\ 0,\ 0\,|\,-11,\ 2,\ 2)$ . The left block is now $I_{3}$ , so $A^{-1}=\begin{pmatrix} -11 & 2 & 2 \\ -4 & 0 & 1 \\ 6 & -1 & -1 \end{pmatrix}$ .

b)

$A\,A^{-1}=\begin{pmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8 \end{pmatrix}\begin{pmatrix} -11 & 2 & 2 \\ -4 & 0 & 1 \\ 6 & -1 & -1 \end{pmatrix}$ . First row: $(1)(-11)+(0)(-4)+(2)(6)=1$ , $(1)(2)+(0)(0)+(2)(-1)=0$ , $(1)(2)+(0)(1)+(2)(-1)=0$ . Continuing with the other two rows gives $I_{3}$ , which confirms the inverse.

c)

$X=A^{-1}B=\begin{pmatrix} -11 & 2 & 2 \\ -4 & 0 & 1 \\ 6 & -1 & -1 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}=\begin{pmatrix} -11+4+6 \\ -4+0+3 \\ 6-2-3 \end{pmatrix}=\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}$ . The solution is $x=-1$ , $y=-1$ , $z=1$ . Check in the first equation $x+2z=1$ : $-1+2(1)=1$ , as required.

d)

Once $A^{-1}$ is known, each new system with the same matrix is solved by a simple product $A^{-1}B$ , without redoing the reduction: the expensive work, the inversion, is done only once.

Exercise 6: A proof, and the singular case

This exercise combines a general proof and the study of a system that depends on a parameter.

a) Let $A=\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ and $B=\begin{pmatrix} 4 & 1 \\ 2 & 5 \end{pmatrix}$ . Compute $(AB)^{T}$ and $B^{T}A^{T}$ , and check that they are equal. The general property is $(AB)^{T}=B^{T}A^{T}$ : transposition reverses the order of the factors.
b) Consider the matrix $A_{k}=\begin{pmatrix} 1 & 2 \\ 3 & k \end{pmatrix}$ . For which value of $k$ is $A_{k}$ not invertible?
c) Depending on the value of $c$ , determine the number of solutions of the system $\begin{cases} x+2y=3 \\ 3x+6y=c \end{cases}$ . Relate your answer to the fact that the determinant of the coefficient matrix is zero.

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Answers

a) $(AB)^T=B^TA^T=\begin{pmatrix}8&6\\11&15\end{pmatrix}$
b) $k=6$
c) $c=9$ : infinitely many; otherwise none

a)

$AB=\begin{pmatrix} (1)(4)+(2)(2) & (1)(1)+(2)(5) \\ (0)(4)+(3)(2) & (0)(1)+(3)(5) \end{pmatrix}=\begin{pmatrix} 8 & 11 \\ 6 & 15 \end{pmatrix}$ , so $(AB)^{T}=\begin{pmatrix} 8 & 6 \\ 11 & 15 \end{pmatrix}$ . Also $B^{T}=\begin{pmatrix} 4 & 2 \\ 1 & 5 \end{pmatrix}$ and $A^{T}=\begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}$ , so $B^{T}A^{T}=\begin{pmatrix} (4)(1)+(2)(2) & (4)(0)+(2)(3) \\ (1)(1)+(5)(2) & (1)(0)+(5)(3) \end{pmatrix}=\begin{pmatrix} 8 & 6 \\ 11 & 15 \end{pmatrix}$ . The two results agree: $(AB)^{T}=B^{T}A^{T}$ . Note that $A^{T}B^{T}$ , in the wrong order, would give something else: the reversed order is indeed the correct one.

b)

$\det A_{k}=(1)(k)-(2)(3)=k-6$ . The matrix is not invertible when its determinant is zero, that is, for $k=6$ .

c)

The coefficient matrix is $\begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix}$ , with determinant $(1)(6)-(2)(3)=0$ : so the system does not have a unique solution. In detail, the second equation is $3x+6y=c$ , that is $3(x+2y)=c$ . Since the first equation forces $x+2y=3$ , we would need $3(3)=c$ , that is $c=9$ . So: if $c=9$ , the two equations are proportional and the system has infinitely many solutions (the whole line $x+2y=3$ ); if $c\neq 9$ , the two lines are distinct parallel lines and the system has no solution. The zero determinant signals exactly this switch between infinitely many solutions and none.

Exercise 7: Problem: ticket sales, by Cramer's rule

A concert hall sells three types of tickets, student, regular and VIP, at 15 dollars, 25 dollars and 40 dollars respectively. For one show, let $e$ , $r$ and $v$ be the numbers of student, regular and VIP tickets sold. We know that:

380 tickets were sold in all; the total revenue was 8400 dollars; twice as many regular tickets as VIP tickets were sold.

a) Translate the three conditions into a system of three equations in the unknowns $e$ , $r$ and $v$ , then write the coefficient matrix $A$ and compute $\det A$ .
b) Solve the system by Cramer's rule, that is, by computing $e=\dfrac{\det A_{e}}{\det A}$ , $r=\dfrac{\det A_{r}}{\det A}$ and $v=\dfrac{\det A_{v}}{\det A}$ .
c) Synthesis question: keep the same 380 tickets sold and the same condition $r=2v$ , but replace the revenue by a target value $S$ . Treating $e$ , $r$ and $v$ as real numbers for now, for which values of $S$ is there a valid distribution, that is, one with $e\geq 0$ , $r\geq 0$ and $v\geq 0$ ? Interpret the two bounds.
d) Tickets are sold one at a time: $e$ , $r$ and $v$ are whole numbers. What is the largest revenue actually possible, and for which distribution?

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Answers

a) $\det A=-45$
b) 200, 120, 60
c) $5\,700\leq S\leq 11\,400$
d) $11\,370$ dollars, with $e=2$ , $r=252$ and $v=126$

a)

The conditions give $\begin{cases} e+r+v=380 \\ 15e+25r+40v=8400 \\ r-2v=0 \end{cases}$ . The coefficient matrix is $A=\begin{pmatrix} 1 & 1 & 1 \\ 15 & 25 & 40 \\ 0 & 1 & -2 \end{pmatrix}$ . By cofactor expansion along the first column: $\det A=1\,\det\begin{pmatrix} 25 & 40 \\ 1 & -2 \end{pmatrix}-15\,\det\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}+0=1(-50-40)-15(-2-1)=-90+45=-45$ . Since $\det A=-45\neq 0$ , the solution is unique.

b)

Replace each column in turn by $B=\begin{pmatrix} 380 \\ 8400 \\ 0 \end{pmatrix}$ . This gives $\det A_{e}=-9000$ , $\det A_{r}=-5400$ and $\det A_{v}=-2700$ . Hence $e=\dfrac{-9000}{-45}=200$ , $r=\dfrac{-5400}{-45}=120$ and $v=\dfrac{-2700}{-45}=60$ . Check: $200+120+60=380$ , $15(200)+25(120)+40(60)=3000+3000+2400=8400$ and $120=2(60)$ . The hall sold 200 student tickets, 120 regular tickets and 60 VIP tickets.

c)

With $r=2v$ , the first equation gives $e+2v+v=380$ , so $e=380-3v$ . The revenue becomes $S=15(380-3v)+25(2v)+40v=5700+45v$ , hence $v=\dfrac{S-5700}{45}$ . The constraint $v\geq 0$ forces $S\geq 5700$ : this is the minimum revenue, reached when $v=0$ (no regular or VIP tickets, 380 student tickets, revenue $15\times 380=5700$ dollars). The constraint $e=380-3v\geq 0$ forces $v\leq \dfrac{380}{3}$ , so $S\leq 5700+45\cdot\dfrac{380}{3}=5700+5700=11400$ : this is the upper bound, the revenue the hall would make if the student tickets disappeared entirely in favour of regular and VIP tickets. A valid distribution therefore exists if and only if $5700\leq S\leq 11400$ dollars. The actual revenue of 8400 dollars is indeed in this interval, which is consistent with the solution found in b).

d)

With whole tickets, $v$ is a whole number and $e=380-3v\geq 0$ forces $v\leq 126$ , since $3\times 127=381>380$ . The revenue $S=5700+45v$ increases with $v$ , so it is largest for $v=126$ : $r=252$ , $e=380-378=2$ and $S=5700+45\times 126=11370$ dollars. The bound $11400$ from question c) is not reached, because it would take $v=\dfrac{380}{3}$ VIP tickets. More generally, a revenue $S$ matches an exact ticket count only if $\dfrac{S-5700}{45}$ is a whole number between $0$ and $126$ .

Exercise 8: Homogeneous system, non-trivial solutions and the solution space

Consider the homogeneous system $\begin{cases} x+2y-z=0 \\ 2x-y+3z=0 \\ 3x+y+2z=0 \end{cases}$ , whose coefficient matrix is $A=\begin{pmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 3 & 1 & 2 \end{pmatrix}$ .

a) Compute $\det A$ . What can you deduce about the existence of solutions other than the zero solution?
b) Solve the system by Gaussian elimination and give its general solution in parametric form.
c) Interpret the solution set geometrically in space.
d) Show that every homogeneous linear system has at least one solution, and state the exact condition for it to have infinitely many.

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Answers

a) $\det A=0$ : non-trivial solutions
b) $t(-1,\ 1,\ 1)$
c) Line through the origin
d) Infinitely many if $\det A=0$

a)

By cofactor expansion along the first row: $\det A=1\big((-1)(2)-(3)(1)\big)-2\big((2)(2)-(3)(3)\big)+(-1)\big((2)(1)-(-1)(3)\big)=1(-5)-2(-5)-1(5)=-5+10-5=0$ . The determinant is zero, so the matrix is singular: the homogeneous system has non-trivial solutions, that is, solutions other than $(0,\ 0,\ 0)$ . One remark speeds everything up: the third row is the sum of the first two, $R_{3}=R_{1}+R_{2}$ , which made the dependence visible without any calculation.

b)

Keep the first two equations, the third being redundant. From $x+2y-z=0$ and $2x-y+3z=0$ , eliminate $x$ : $R_{2}\to R_{2}-2R_{1}$ gives $-5y+5z=0$ , so $y=z$ . Substituting into the first: $x=z-2y=z-2z=-z$ . Setting $z=t$ as a free parameter, the general solution is $(x,\ y,\ z)=(-t,\ t,\ t)=t(-1,\ 1,\ 1)$ . The direction $(1,\ -1,\ -1)$ works just as well, by changing the sign of the parameter.

c)

The solution set is a line through the origin, with direction vector $(-1,\ 1,\ 1)$ . This is consistent with the geometry: each homogeneous equation represents a plane through the origin, and since the system has rank $2$ (the third equation adds nothing), the three planes do not meet at the origin alone but along a whole line.

d)

A homogeneous system reads $AX=\vec{0}$ . The zero vector $X=\vec{0}$ always satisfies it, since $A\vec{0}=\vec{0}$ : so there is always at least the trivial solution. The system also has non-trivial solutions, hence infinitely many, if and only if $\det A=0$ (the matrix is singular, its columns are linearly dependent). If $\det A\neq 0$ , the only solution is the zero solution. Our system is in the first case, with $\det A=0$ and a whole line of solutions. This set is closed under addition and scalar multiplication: it is a subspace.

Exercise 9: A system with two parameters: case analysis by rank

Consider the system $\begin{cases} x+y+z=2 \\ x+2y+3z=5 \\ 2x+3y+az=b \end{cases}$ , where $a$ and $b$ are two real parameters.

a) Compute the determinant of the coefficient matrix as a function of $a$ . For which value of $a$ can the system fail to have a unique solution?
b) For $a\neq 4$ , what can you say about the number of solutions, without computing them?
c) Take the case $a=4$ . Using Gaussian elimination, determine the number of solutions depending on the value of $b$ .
d) For $a=4$ and the value of $b$ that gives infinitely many solutions, write the general solution in parametric form.

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Answers

a) $\det=a-4$ : $a=4$
b) Unique solution
c) $b=7$ : infinitely many; otherwise none
d) $(-1,\ 3,\ 0)+t(1,\ -2,\ 1)$

a)

The determinant of $\begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & a \end{pmatrix}$ , expanded along the first row, is $1(2a-9)-1(a-6)+1(3-4)=2a-9-a+6-1=a-4$ . It vanishes for $a=4$ : this is the only value for which the system can fail to have a unique solution.

b)

For $a\neq 4$ , $\det\neq 0$ , the matrix is invertible and the system has a unique solution, whatever the value of $b$ . Cramer's rule or the inverse apply without restriction.

c)

For $a=4$ , the coefficient matrix becomes $\begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{pmatrix}$ , and we notice that $R_{3}=R_{1}+R_{2}$ . Consistency therefore depends on the same test applied to the right-hand side: we need $b=2+5=7$ . Row reduction shows it: $R_{2}\to R_{2}-R_{1}$ and $R_{3}\to R_{3}-2R_{1}$ give the rows $(0,\ 1,\ 2\,|\,3)$ and $(0,\ 1,\ 2\,|\,b-4)$ ; then $R_{3}\to R_{3}-R_{2}$ gives $(0,\ 0,\ 0\,|\,b-7)$ . If $b\neq 7$ , this row says $0=b-7\neq 0$ , a contradiction: the system has no solution. If $b=7$ , the row becomes $0=0$ : the system has infinitely many solutions.

d)

For $a=4$ and $b=7$ , two independent equations remain: $x+y+z=2$ and $y+2z=3$ . Take $z=t$ as the free parameter: $y=3-2t$ and $x=2-y-z=2-(3-2t)-t=-1+t$ . The general solution is $(x,\ y,\ z)=(-1+t,\ 3-2t,\ t)=(-1,\ 3,\ 0)+t(1,\ -2,\ 1)$ . It is a line of solutions, consistent with the rank $2$ of the matrix when $a=4$ .

Exercise 10: Proofs: properties of the inverse and special matrices

Questions a) to c) are general proofs about invertible square matrices of order $n$ . Question d) illustrates the result of a) on an example.

a) Let $A$ and $B$ be two invertible matrices of the same order. Prove that $AB$ is invertible and that $(AB)^{-1}=B^{-1}A^{-1}$ , paying close attention to the order of the factors.
b) Prove that if $A$ is invertible, then so is $A^{T}$ , and $(A^{T})^{-1}=(A^{-1})^{T}$ .
c) Prove that if a matrix $A$ satisfies $A^{2}=A$ and is invertible, then $A=I$ .
d) Illustrate a) with $A=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ : compute $(AB)^{-1}$ and $B^{-1}A^{-1}$ , and check that they are equal but that $A^{-1}B^{-1}$ is different.

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Answers

a) $(AB)^{-1}=B^{-1}A^{-1}$
b) $(A^T)^{-1}=(A^{-1})^T$
c) $A^{-1}A^2=A^{-1}A$ : $A=I$
d) $\begin{pmatrix}\frac12&-1\\0&1\end{pmatrix}$ ; $A^{-1}B^{-1}$ is different

a)

It is enough to exhibit a matrix which, multiplied by $AB$ , gives the identity on both sides. Let $C=B^{-1}A^{-1}$ . Then $(AB)C=(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=A\,I\,A^{-1}=AA^{-1}=I$ , using the associativity of the product. Likewise $C(AB)=B^{-1}(A^{-1}A)B=B^{-1}IB=B^{-1}B=I$ . So $AB$ is invertible and its inverse is $B^{-1}A^{-1}$ . The order is reversed: writing $A^{-1}B^{-1}$ would be wrong, because the inner factors would no longer cancel. It is the same reversal as for the transpose, $(AB)^{T}=B^{T}A^{T}$ .

b)

Transpose the identity $AA^{-1}=I$ . Since $(XY)^{T}=Y^{T}X^{T}$ and $I^{T}=I$ , we get $(A^{-1})^{T}A^{T}=I$ . Likewise, transposing $A^{-1}A=I$ gives $A^{T}(A^{-1})^{T}=I$ . The matrix $(A^{-1})^{T}$ is therefore the inverse of $A^{T}$ on both sides: $A^{T}$ is invertible and $(A^{T})^{-1}=(A^{-1})^{T}$ .

c)

Start from $A^{2}=A$ . Since $A$ is invertible, we can multiply both sides on the left by $A^{-1}$ : $A^{-1}A^{2}=A^{-1}A$ . The left side is $A^{-1}(AA)=(A^{-1}A)A=IA=A$ , and the right side is $I$ . So $A=I$ . In other words, the only invertible idempotent matrix is the identity: the other matrices satisfying $A^{2}=A$ (the non-trivial projections) are all singular.

d)

$AB=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 2 & 2 \\ 0 & 1 \end{pmatrix}$ , with determinant $2$ , so $(AB)^{-1}=\dfrac{1}{2}\begin{pmatrix} 1 & -2 \\ 0 & 2 \end{pmatrix}=\begin{pmatrix} \frac{1}{2} & -1 \\ 0 & 1 \end{pmatrix}$ . Also $A^{-1}=\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{pmatrix}$ and $B^{-1}=\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ , so $B^{-1}A^{-1}=\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} \frac{1}{2} & -1 \\ 0 & 1 \end{pmatrix}=(AB)^{-1}$ . On the other hand $A^{-1}B^{-1}=\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ 0 & 1 \end{pmatrix}\neq(AB)^{-1}$ : the wrong order does give a different result.

Exercise 11: Matrix equations: where to put the inverse

Consider $A=\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$ . We are looking for matrices $X$ of order 2.

a) Show that $A$ is invertible and compute $A^{-1}$ .
b) Solve $AX=B$ .
c) Solve $XA=B$ . Is the solution the same as in b)?
d) Solve $AX+X=B$ .
e) For b), a student writes “ $X=\frac{B}{A}$ ”. What should you tell them?

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Answers

a) $A^{-1}=\begin{pmatrix}3&-1\\-5&2\end{pmatrix}$
b) $X=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$
c) $X=\begin{pmatrix}3&-1\\1&0\end{pmatrix}$ : different
d) $X=\frac17\begin{pmatrix}2&-1\\1&3\end{pmatrix}$
e) No matrix division

a)

$\det A=6-5=1\neq 0$ , so $A$ is invertible and $A^{-1}=\frac{1}{1}\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}=\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$ .

b)

Multiply ON THE LEFT by $A^{-1}$ : $X=A^{-1}B=\begin{pmatrix} 3-2 & -1 \\ -5+4 & 2 \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$ . Check: $AX=\begin{pmatrix} 2-1 & -2+2 \\ 5-3 & -5+6 \end{pmatrix}=B$ .

c)

Multiply ON THE RIGHT by $A^{-1}$ : $X=BA^{-1}=\begin{pmatrix} 3 & -1 \\ 6-5 & -2+2 \end{pmatrix}=\begin{pmatrix} 3 & -1 \\ 1 & 0 \end{pmatrix}$ . Check: $XA=\begin{pmatrix} 6-5 & 3-3 \\ 2 & 1 \end{pmatrix}=B$ . The solution differs from that of b): since the product is not commutative, the side on which $X$ sits decides the side on which $A^{-1}$ goes.

d)

Factor $X$ out on the right: $AX+X=AX+IX=(A+I)X=B$ . Writing $(A+1)X$ would make no sense: you do not add a number to a matrix. $A+I=\begin{pmatrix} 3 & 1 \\ 5 & 4 \end{pmatrix}$ , with determinant $7$ and inverse $\frac{1}{7}\begin{pmatrix} 4 & -1 \\ -5 & 3 \end{pmatrix}$ . So $X=\frac{1}{7}\begin{pmatrix} 4-2 & -1 \\ -5+6 & 3 \end{pmatrix}=\frac{1}{7}\begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}$ .

e)

Division by a matrix does not exist: $\frac{B}{A}$ does not say whether to multiply by $A^{-1}$ on the left or on the right, and that is exactly what changes the answer between b) and c). Always write $A^{-1}B$ or $BA^{-1}$ , never a fraction.

The thread: to isolate $X$ , multiply by the inverse on the same side as the factor to remove; $X$ is factored out with $I$ , not with $1$ .

Exercise 12: Triangular matrices: products, powers and inverse

Consider the upper triangular matrices $T=\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ and $U=\begin{pmatrix} 2 & -1 \\ 0 & 1 \end{pmatrix}$ , and the diagonal matrix $D=\begin{pmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ .

a) Compute $TU$ . What shape does the product have, and what are its diagonal entries?
b) Compute $T^{2}$ and $T^{3}$ , conjecture a formula for $T^{n}$ , then prove it by induction.
c) Compute $T^{-1}$ . Is it still triangular?
d) Compute $D^{5}$ and $D^{-1}$ .
e) Under what condition is a triangular matrix invertible? Is the matrix $\begin{pmatrix} 1 & 4 & 5 \\ 0 & 0 & 6 \\ 0 & 0 & 2 \end{pmatrix}$ invertible?

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Answers

a) $TU=\begin{pmatrix}2&1\\0&3\end{pmatrix}$
b) $T^n=\begin{pmatrix}1&3^n-1\\0&3^n\end{pmatrix}$
c) $T^{-1}=\begin{pmatrix}1&-\frac23\\0&\frac13\end{pmatrix}$
d) $\text{diag}(32,-1,243)$ ; $\text{diag}\left(\frac12,-1,\frac13\right)$
e) No zero on the diagonal; here, not invertible

a)

$TU=\begin{pmatrix} 2 & -1+2 \\ 0 & 3 \end{pmatrix}=\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ : still upper triangular, and its diagonal is the term-by-term product of the diagonals, $1\times 2$ and $3\times 1$ .

b)

$T^{2}=\begin{pmatrix} 1 & 2+6 \\ 0 & 9 \end{pmatrix}=\begin{pmatrix} 1 & 8 \\ 0 & 9 \end{pmatrix}$ and $T^{3}=T^{2}T=\begin{pmatrix} 1 & 2+24 \\ 0 & 27 \end{pmatrix}=\begin{pmatrix} 1 & 26 \\ 0 & 27 \end{pmatrix}$ . Conjecture: $T^{n}=\begin{pmatrix} 1 & 3^{n}-1 \\ 0 & 3^{n} \end{pmatrix}$ , true for $n=1$ . If it holds for $n$ , then $T^{n+1}=T^{n}T=\begin{pmatrix} 1 & 2+3(3^{n}-1) \\ 0 & 3^{n+1} \end{pmatrix}=\begin{pmatrix} 1 & 3^{n+1}-1 \\ 0 & 3^{n+1} \end{pmatrix}$ : the inductive step holds.

c)

$\det T=3$ and $T^{-1}=\frac{1}{3}\begin{pmatrix} 3 & -2 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{pmatrix}$ : upper triangular, with a diagonal made of the reciprocals $1$ and $\frac{1}{3}$ .

d)

For a diagonal matrix, raise each diagonal entry to the power: $D^{5}=\text{diag}(32,\ -1,\ 243)$ , and invert each one: $D^{-1}=\text{diag}\left(\frac{1}{2},\ -1,\ \frac{1}{3}\right)$ .

e)

The determinant of a triangular matrix is the product of its diagonal entries: the matrix is invertible if and only if none of them is zero. The given matrix has a $0$ on the diagonal, so its determinant is $1\times 0\times 2=0$ : it is not invertible, despite its large off-diagonal entries.

The thread: products, powers and inverses of triangular matrices stay triangular, and their diagonal is computed term by term.

Exercise 13: The trace of a product

The trace of a square matrix, written $\operatorname{tr}$ , is the sum of its diagonal entries. Consider $A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B=\begin{pmatrix} 0 & 1 \\ 5 & -2 \end{pmatrix}$ .

a) Compute $AB$ and $BA$ , then their traces. Are the two products equal?
b) Prove that $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ for all square matrices $A$ and $B$ of order 2.
c) Are there two square matrices of order 2 such that $AB-BA=I$ ? Justify.
d) With $E=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ , $F=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $G=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ , compute $\operatorname{tr}(EFG)$ and $\operatorname{tr}(FEG)$ . What do you conclude?
e) Show that $\operatorname{tr}(P^{-1}AP)=\operatorname{tr}(A)$ for every invertible matrix $P$ , and check it with $P=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ .

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Answers

a) Both traces equal 5, the products differ
b) The same four products
c) No: $0\neq 2$
d) 1 and 0: the trace is only cyclic
e) Trace 5 preserved

a)

$AB=\begin{pmatrix} 10 & -3 \\ 20 & -5 \end{pmatrix}$ and $BA=\begin{pmatrix} 3 & 4 \\ -1 & 2 \end{pmatrix}$ . The products are different, but $\operatorname{tr}(AB)=10-5=5$ and $\operatorname{tr}(BA)=3+2=5$ .

b)

With $A=(a_{ij})$ and $B=(b_{ij})$ : $\operatorname{tr}(AB)=(a_{11}b_{11}+a_{12}b_{21})+(a_{21}b_{12}+a_{22}b_{22})$ and $\operatorname{tr}(BA)=(b_{11}a_{11}+b_{12}a_{21})+(b_{21}a_{12}+b_{22}a_{22})$ . These are the same four products of numbers, arranged differently: the two traces are equal. The same calculation, $\sum_{i,j}a_{ij}b_{ji}$ , works for any order $n$ .

c)

No. The trace is linear, so $\operatorname{tr}(AB-BA)=\operatorname{tr}(AB)-\operatorname{tr}(BA)=0$ , whereas $\operatorname{tr}(I)=2$ . No pair of matrices can satisfy this equality, in any finite order.

d)

$EF=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ , then $EFG=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ , with trace $1$ . $FE=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ , so $FEG$ is zero, with trace $0$ . The factors can be rotated, $\operatorname{tr}(EFG)=\operatorname{tr}(GEF)$ , but not permuted freely: the property is CYCLIC, not commutative.

e)

By b) applied to $P^{-1}A$ and $P$ : $\operatorname{tr}\big((P^{-1}A)P\big)=\operatorname{tr}\big(P(P^{-1}A)\big)=\operatorname{tr}(A)$ . With $P^{-1}=\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ : $P^{-1}A=\begin{pmatrix} -2 & -2 \\ 3 & 4 \end{pmatrix}$ and $P^{-1}AP=\begin{pmatrix} -2 & -4 \\ 3 & 7 \end{pmatrix}$ , with trace $5=\operatorname{tr}(A)$ . Two similar matrices have the same trace, which will be useful for eigenvalues.

The thread: the trace does not see the order of a product of two factors; that is what rules out $AB-BA=I$ and makes the trace invariant under a change of basis.

Exercise 14: Problem: traffic flows in a neighbourhood

Four intersections $A$ , $B$ , $C$ and $D$ form a square of one-way streets: $x_{1}$ cars per hour go from $A$ to $B$ , $x_{2}$ from $B$ to $C$ , $x_{3}$ from $C$ to $D$ and $x_{4}$ from $D$ to $A$ .

Outside streets bring traffic in and take it out, in cars per hour: at $A$ , $500$ enter and $200$ leave; at $B$ , $300$ enter and $400$ leave; at $C$ , $200$ enter and $600$ leave; at $D$ , $400$ enter and $200$ leave. At each intersection, as many cars enter as leave.

In the figure, the green arrows come into the neighbourhood and the orange arrows leave it.

a) Write the four conservation equations.
b) Find the rank of the system. Why is one equation redundant?
c) Express $x_{2}$ , $x_{3}$ and $x_{4}$ in terms of $x_{1}=t$ , then compute them for $t=600$ .
d) No flow can be negative, since the streets are one-way. What is the smallest possible value of $x_{1}$ ?
e) Roadwork closes the street from $C$ to $D$ . What are the other flows? Could the street from $B$ to $C$ also be limited to $300$ cars per hour?

Show the solution

Answers

a) $x_1-x_4=300$ , $x_1-x_2=100$ , $x_2-x_3=400$ , $x_3-x_4=-200$
b) Rank 3
c) $t-100$ , $t-500$ , $t-300$ ; 500, 100, 300
d) $x_1\geq 500$
e) $x_2=400$ , $x_4=200$ ; the limit is impossible

a)

At $A$ : $500+x_{4}=x_{1}+200$ , that is $x_{1}-x_{4}=300$ . At $B$ : $x_{1}+300=x_{2}+400$ , that is $x_{1}-x_{2}=100$ . At $C$ : $x_{2}+200=x_{3}+600$ , that is $x_{2}-x_{3}=400$ . At $D$ : $x_{3}+400=x_{4}+200$ , that is $x_{3}-x_{4}=-200$ .

b)

The equations at $B$ , $C$ and $D$ , added together, give $x_{1}-x_{4}=300$ , the equation at $A$ . The rank is $3$ : four unknowns and three independent equations, hence one free parameter. The redundancy comes from the overall balance: $1\,400$ cars per hour enter the neighbourhood and $1\,400$ leave it, so the last intersection is balanced as soon as the other three are.

c)

$x_{2}=t-100$ , $x_{3}=x_{2}-400=t-500$ and $x_{4}=x_{3}+200=t-300$ . For $t=600$ : $x_{2}=500$ , $x_{3}=100$ and $x_{4}=300$ . Check at $A$ : $600-300=300$ .

d)

We need $t-100\geq 0$ , $t-500\geq 0$ and $t-300\geq 0$ : the strongest constraint is $t\geq 500$ . At the minimum, $500$ cars per hour go from $A$ to $B$ , and the street from $C$ to $D$ is then empty.

e)

$x_{3}=0$ forces $t=500$ , so $x_{1}=500$ , $x_{2}=400$ and $x_{4}=200$ : only one distribution is possible. Limiting $x_{2}$ to $300$ would require $t\leq 400$ , which is incompatible with $t\geq 500$ : it is impossible without changing the inflows and outflows, the neighbourhood would be saturated.

The thread: a flow network translates into a system whose rank gives the number of degrees of freedom; the physical constraints then cut out the interval of acceptable solutions.

Exercise 15: Problem: a bakery's production in matrices

A bakery makes croissants, pains au chocolat and brioches. The matrix $M$ gives, in grams, the flour, butter and sugar (rows) used for one piece of each product (columns): $M=\begin{pmatrix} 50 & 55 & 60 \\ 25 & 25 & 15 \\ 5 & 10 & 15 \end{pmatrix}$ .

The purchase price of the ingredients, in dollars per gram, is given by the row matrix $p=\begin{pmatrix} 0.002 & 0.012 & 0.003 \end{pmatrix}$ .

a) An order consists of $100$ croissants, $80$ pains au chocolat and $50$ brioches, written $q=\begin{pmatrix} 100 \\ 80 \\ 50 \end{pmatrix}$ . Compute $Mq$ and interpret it.
b) Compute $pM$ and interpret each of its entries.
c) Compute the ingredient cost of the order in two ways, $p(Mq)$ and $(pM)q$ .
d) Today's stock is $11\,900$ grams of flour, $5\,100$ grams of butter and $2\,000$ grams of sugar. Compute $\det M$ , then find the production that uses up exactly all the stock.
e) Why does question d) require $\det M\neq 0$ , and what would a solution with a negative component mean?

Show the solution

Answers

a) $12\,400$ g, $5\,250$ g, $2\,050$ g
b) $0.415$ ; $0.44$ ; $0.345$ dollar
c) $93.95$ dollars
d) $\det M=2\,250$ ; 80, 100, 40
e) Unique solution; negative: impossible

a)

$Mq=\begin{pmatrix} 5\,000+4\,400+3\,000 \\ 2\,500+2\,000+750 \\ 500+800+750 \end{pmatrix}=\begin{pmatrix} 12\,400 \\ 5\,250 \\ 2\,050 \end{pmatrix}$ : the order needs $12\,400$ grams of flour, $5\,250$ of butter and $2\,050$ of sugar. The product is defined because $M$ has three columns, one per product, just as $q$ has three rows.

b)

$pM=\begin{pmatrix} 0.1+0.3+0.015 & 0.11+0.3+0.03 & 0.12+0.18+0.045 \end{pmatrix}=\begin{pmatrix} 0.415 & 0.44 & 0.345 \end{pmatrix}$ : the ingredient cost of one croissant, one pain au chocolat and one brioche, in dollars.

c)

$p(Mq)=0.002\times 12\,400+0.012\times 5\,250+0.003\times 2\,050=24.80+63.00+6.15=93.95$ dollars. $(pM)q=0.415\times 100+0.44\times 80+0.345\times 50=41.50+35.20+17.25=93.95$ dollars. Same result: this is the associativity of the matrix product, read first by ingredients, then by products.

d)

$\det M=50(375-150)-55(375-75)+60(250-125)=11\,250-16\,500+7\,500=2\,250\neq 0$ . The system $Mq=s$ has a unique solution; check that $q=\begin{pmatrix} 80 \\ 100 \\ 40 \end{pmatrix}$ works: $4\,000+5\,500+2\,400=11\,900$ , $2\,000+2\,500+600=5\,100$ and $400+1\,000+600=2\,000$ . The bakery can make $80$ croissants, $100$ pains au chocolat and $40$ brioches.

e)

If $\det M=0$ , the recipes would be linked, for example one recipe a combination of the other two: the system would have infinitely many solutions or none, never a unique production. A negative component would be a mathematical solution with no physical meaning: to use up the stock exactly, you would have to “unmake” pastries, and you would then have to give up using everything.

The thread: the rows and columns of a matrix carry a meaning; a matrix product is defined only if those meanings link up, and a solution is acceptable only if it respects reality.

Corrected exercises: matrices, linear systems and determinants (201-SN4-RE)

Course recap

Calculations: matrix products to compute

Exercise 0.1: Computing matrix products: ten direct calculations

Calculations: inverses to compute

Exercise 0.2: Inverting a matrix: eight direct calculations

Part A: Operations and systems (/50)

Exercise 1: Matrix operations and non-commutativity

Exercise 2: The matrix product and writing a system

Exercise 3: Solving by Gauss-Jordan elimination

Exercise 4: Determinants and the inverse matrix

Exercise 5: Inverse of a 3x3 matrix by Gauss-Jordan elimination

Part B: Exam level (/50)

Exercise 6: A proof, and the singular case

Exercise 7: Problem: ticket sales, by Cramer's rule

Exercise 8: Homogeneous system, non-trivial solutions and the solution space

Exercise 9: A system with two parameters: case analysis by rank

Exercise 10: Proofs: properties of the inverse and special matrices

Part C: the classics (/50)

Exercise 11: Matrix equations: where to put the inverse

Exercise 12: Triangular matrices: products, powers and inverse

Exercise 13: The trace of a product

Exercise 14: Problem: traffic flows in a neighbourhood

Exercise 15: Problem: a bakery's production in matrices

See also

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