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1. 1.Computation · 7 min

   Let $A=\left(\begin{array}{cc}2& -1\\ 4& 3\end{array}\right)$ and $B=\left(\begin{array}{cc}1& 5\\ 0& -2\end{array}\right)$. State the size of every intermediate matrix you form.

   1. (a)Compute $3A-2B^T$.
   2. (b)Write $A=S+K$ with $S$ symmetric and $K$ skew-symmetric, and check both types by inspecting entries.
   3. (c)Compute $AB$ and $BA$. Confirm that $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ even though $AB\ne BA$, and check whether either equals $\mathrm{tr}(A)\mathrm{tr}(B)$.
   Show the solution to question 1

   (a) Transpose, scale, then subtract. Transposing swaps the off-diagonal entries: $B^T=\left(\begin{array}{cc}1& 0\\ 5& -2\end{array}\right)$, still $2\times 2$, so the combination is defined. Then

   $$3A=\left(\begin{array}{cc}6& -3\\ 12& 9\end{array}\right),2B^T=\left(\begin{array}{cc}2& 0\\ 10& -4\end{array}\right),3A-2B^T=\left(\begin{array}{cc}4& -3\\ 2& 13\end{array}\right).$$

   (b) Use the formulas $S=\frac{1}{2}(A+A^T)$ and $K=\frac{1}{2}(A-A^T)$. With $A^T=\left(\begin{array}{cc}2& 4\\ -1& 3\end{array}\right)$,

   $$A+A^T=\left(\begin{array}{cc}4& 3\\ 3& 6\end{array}\right),S=\left(\begin{array}{cc}2& \frac{3}{2}\\ \frac{3}{2}& 3\end{array}\right),A-A^T=\left(\begin{array}{cc}0& -5\\ 5& 0\end{array}\right),K=\left(\begin{array}{cc}0& -\frac{5}{2}\\ \frac{5}{2}& 0\end{array}\right).$$

   $S$ is symmetric because $S_{12}=S_{21}=\frac{3}{2}$. $K$ is skew-symmetric because $K_{12}=-\frac{5}{2}=-K_{21}$ and its diagonal is zero, as skew-symmetry forces. Sanity check: $S+K=\left(\begin{array}{cc}2& \frac{3}{2}-\frac{5}{2}\\ \frac{3}{2}+\frac{5}{2}& 3\end{array}\right)=\left(\begin{array}{cc}2& -1\\ 4& 3\end{array}\right)=A$.

   (c) Row–column rule, four entries each. Both products are $2\times 2$:

   $$AB=\left(\begin{array}{cc}2\cdot 1+(-1)\cdot 0& 2\cdot 5+(-1)(-2)\\ 4\cdot 1+3\cdot 0& 4\cdot 5+3(-2)\end{array}\right)=\left(\begin{array}{cc}2& 12\\ 4& 14\end{array}\right),BA=\left(\begin{array}{cc}1\cdot 2+5\cdot 4& 1(-1)+5\cdot 3\\ 0\cdot 2+(-2)4& 0(-1)+(-2)3\end{array}\right)=\left(\begin{array}{cc}22& 14\\ -8& -6\end{array}\right).$$

   The two products differ in three of their four entries, so $AB\ne BA$. Their traces are $\mathrm{tr}(AB)=2+14=16$ and $\mathrm{tr}(BA)=22-6=16$: equal, exactly as $\mathrm{tr}(XY)=\mathrm{tr}(YX)$ guarantees. By contrast $\mathrm{tr}(A)\mathrm{tr}(B)=5\cdot (-1)=-5$, which matches neither — the trace is linear, not multiplicative.

2. 2.Proof · 7 min

   Throughout, $A$ is an $n\times n$ matrix and $A=S+K$ is its symmetric–skew decomposition, $S=\frac{1}{2}(A+A^T)$ and $K=\frac{1}{2}(A-A^T)$. Argue from the definitions and the transpose and trace laws; testing on examples earns nothing.

   1. (a)Prove that a matrix which is both symmetric and skew-symmetric must be the zero matrix.
   2. (b)Prove that $\mathrm{tr}(K)=0$ and hence $\mathrm{tr}(S)=\mathrm{tr}(A)$. Give two different proofs that $\mathrm{tr}(K)=0$: one entrywise, one using only the trace laws.
   3. (c)Prove that $\mathrm{tr}(SK)=0$. (Write $\mathrm{tr}(SK)$ as a double sum using the row–column rule and swap the names of the two summation indices.) Deduce that $\mathrm{tr}(A^2)=\mathrm{tr}(S^2)+\mathrm{tr}(K^2)$.
   Show the solution to question 2

   (a) Compare the two conditions entry by entry. Symmetric means $A_{ij}=A_{ji}$ for all $i,j$; skew-symmetric means $A_{ij}=-A_{ji}$. If both hold then $A_{ji}=A_{ij}=-A_{ji}$, so $2A_{ji}=0$ and $A_{ji}=0$. Every entry is zero, so $A=0$. $■$

   (This is the reason the decomposition is unique: two decompositions $S+K=S^{\prime }+K^{\prime }$ give $S-S^{\prime }=K^{\prime }-K$, a matrix that is symmetric by the left side and skew-symmetric by the right side, hence zero.)

   (b) Entrywise proof. Setting $i=j$ in $K_{ij}=-K_{ji}$ gives $K_{ii}=-K_{ii}$, so every diagonal entry of $K$ is $0$ and $\mathrm{tr}(K)={\sum }_i{K}_{ii}=0$.

   Proof from the trace laws. The trace is linear and $\mathrm{tr}(A^T)=\mathrm{tr}(A)$, so

   $$\mathrm{tr}(K)=\frac{1}{2}(\mathrm{tr}(A)-\mathrm{tr}(A^T))=\frac{1}{2}(\mathrm{tr}(A)-\mathrm{tr}(A))=0.$$

   Either way, linearity then gives $\mathrm{tr}(A)=\mathrm{tr}(S+K)=\mathrm{tr}(S)+\mathrm{tr}(K)=\mathrm{tr}(S)$. $■$

   (c) Write the trace as a double sum. By the row–column rule, $(SK{)}_{ii}={\sum }_j{S}_{ij}{K}_{ji}$, so

   $$\mathrm{tr}(SK)=\sum _{i=1}^n\sum _{j=1}^n{S}_{ij}{K}_{ji}.$$

   The names of the summation indices are arbitrary, so rename $i\leftrightarrow j$: the same number is ${\sum }_j{\sum }_i{S}_{ji}{K}_{ij}$. Now use $S_{ji}=S_{ij}$ and $K_{ij}=-K_{ji}$:

   $$\mathrm{tr}(SK)=\sum _{i,j}{S}_{ji}{K}_{ij}=\sum _{i,j}{S}_{ij}(-K_{ji})=-\mathrm{tr}(SK).$$

   Hence $2\mathrm{tr}(SK)=0$, so $\mathrm{tr}(SK)=0$. $■$ (A second route: $\mathrm{tr}(SK)=\mathrm{tr}((SK{)}^T)=\mathrm{tr}(K^T{S}^T)=\mathrm{tr}(-KS)=-\mathrm{tr}(KS)=-\mathrm{tr}(SK)$, using $\mathrm{tr}(M^T)=\mathrm{tr}(M)$, the reversal rule, and $\mathrm{tr}(XY)=\mathrm{tr}(YX)$.)

   Deduction. Expand without assuming anything commutes: $A^2=(S+K)(S+K)=S^2+SK+KS+K^2$. Taking traces and using linearity, $\mathrm{tr}(A^2)=\mathrm{tr}(S^2)+\mathrm{tr}(SK)+\mathrm{tr}(KS)+\mathrm{tr}(K^2)$. The middle terms are $\mathrm{tr}(SK)=0$ and $\mathrm{tr}(KS)=\mathrm{tr}(SK)=0$, leaving $\mathrm{tr}(A^2)=\mathrm{tr}(S^2)+\mathrm{tr}(K^2)$. $■$

3. 3.Computation · 7 min

   Let $A=\left(\begin{array}{ccc}1& 0& 2\\ -1& 3& 1\end{array}\right)$ and $B=\left(\begin{array}{cc}2& 1\\ 0& -1\\ 1& 4\end{array}\right)$.

   1. (a)For each of $AB$, $BA$, $A^2$, $A{A}^T$, $A^T{B}^T$ and $B^T{A}^T$, decide from the shapes alone whether it is defined, and give the size of each one that is.
   2. (b)Compute $AB$ and $BA$, and verify that their traces agree.
   3. (c)Compute $B^T{A}^T$ directly and compare it with $(AB{)}^T$. Then, without multiplying anything out, say which transpose $A^T{B}^T$ is equal to, and why.
   Show the solution to question 3

   (a) Match inner dimensions. $A$ is $2\times 3$ and $B$ is $3\times 2$, so $A^T$ is $3\times 2$ and $B^T$ is $2\times 3$.

   $AB$: $(2\times \underline{3})(\underline{3}\times 2)$, defined, size $2\times 2$. $BA$: $(3\times \underline{2})(\underline{2}\times 3)$, defined, size $3\times 3$. $A^2=AA$: $(2\times 3)(2\times 3)$, inner dimensions $3\ne 2$, undefined — only square matrices have powers. $A{A}^T$: $(2\times 3)(3\times 2)$, defined, $2\times 2$. $A^T{B}^T$: $(3\times 2)(2\times 3)$, defined, $3\times 3$. $B^T{A}^T$: $(2\times 3)(3\times 2)$, defined, $2\times 2$.

   Note that $A^T{B}^T$ is defined here only because $A$ and $B$ have reversed shapes, and it cannot be $(AB{)}^T$, which is $2\times 2$.

   (b) Row–column rule. For $AB$, row $1$ of $A$ is $(1,0,2)$ and row $2$ is $(-1,3,1)$:

   $$AB=\left(\begin{array}{cc}1\cdot 2+0\cdot 0+2\cdot 1& 1\cdot 1+0(-1)+2\cdot 4\\ -1\cdot 2+3\cdot 0+1\cdot 1& -1\cdot 1+3(-1)+1\cdot 4\end{array}\right)=\left(\begin{array}{cc}4& 9\\ -1& 0\end{array}\right).$$

   For $BA$ use the row rule: row $i$ of $BA$ is row $i$ of $B$ times $A$, a combination of the two rows of $A$. Row $1$ of $B$ is $(2,1)$, giving $2(1,0,2)+1(-1,3,1)=(1,3,5)$; row $2$ is $(0,-1)$, giving $-(-1,3,1)=(1,-3,-1)$; row $3$ is $(1,4)$, giving $(1,0,2)+4(-1,3,1)=(-3,12,6)$. So

   $$BA=\left(\begin{array}{ccc}1& 3& 5\\ 1& -3& -1\\ -3& 12& 6\end{array}\right).$$

   Traces: $\mathrm{tr}(AB)=4+0=4$ and $\mathrm{tr}(BA)=1-3+6=4$. A $2\times 2$ and a $3\times 3$ matrix with the same trace, exactly as $\mathrm{tr}(XY)=\mathrm{tr}(YX)$ promises for any compatible pair.

   (c) Compute, then compare. $B^T=\left(\begin{array}{ccc}2& 0& 1\\ 1& -1& 4\end{array}\right)$ and $A^T=\left(\begin{array}{cc}1& -1\\ 0& 3\\ 2& 1\end{array}\right)$, so

   $$B^T{A}^T=\left(\begin{array}{cc}2\cdot 1+0\cdot 0+1\cdot 2& 2(-1)+0\cdot 3+1\cdot 1\\ 1\cdot 1+(-1)0+4\cdot 2& 1(-1)+(-1)3+4\cdot 1\end{array}\right)=\left(\begin{array}{cc}4& -1\\ 9& 0\end{array}\right).$$

   Transposing the answer to (b) gives $(AB{)}^T=\left(\begin{array}{cc}4& -1\\ 9& 0\end{array}\right)$ — the same, confirming the reversal rule $(AB{)}^T=B^T{A}^T$.

   For $A^T{B}^T$, apply the reversal rule to the other product: $(BA{)}^T=A^T{B}^T$. So $A^T{B}^T$ is the transpose of $BA$, namely the $3\times 3$ matrix $\left(\begin{array}{ccc}1& 1& -3\\ 3& -3& 12\\ 5& -1& 6\end{array}\right)$. Nothing new needs multiplying; the rule tells you which product to transpose.

4. 4.Proof · 7 min

   Let $A=\left(\begin{array}{cc}3& 1\\ 0& 3\end{array}\right)$.

   1. (a)Compute $A^2$ and $A^3$.
   2. (b)Conjecture a formula for $A^n$ valid for every integer $n\ge 1$ and prove it by induction on $n$.
   3. (c)Write $A=3I+N$ and compute $N^2$. Explain exactly why the binomial theorem may be applied to $(3I+N{)}^n$ even though matrix multiplication is not commutative, and use it to rederive your formula.
   Show the solution to question 4

   (a) Multiply directly.

   $$A^2=\left(\begin{array}{cc}3\cdot 3+1\cdot 0& 3\cdot 1+1\cdot 3\\ 0& 3\cdot 3\end{array}\right)=\left(\begin{array}{cc}9& 6\\ 0& 9\end{array}\right),A^3=A^{2}A=\left(\begin{array}{cc}27& 9\cdot 1+6\cdot 3\\ 0& 27\end{array}\right)=\left(\begin{array}{cc}27& 27\\ 0& 27\end{array}\right).$$

   (b) Conjecture. The diagonal is $3^n$; the corner runs $1,6,27$, which is $1\cdot 3^0,2\cdot 3^1,3\cdot 3^2$. So conjecture

   $$A^n=\left(\begin{array}{cc}{3}^n& n{3}^{n-1}\\ 0& 3^n\end{array}\right).$$

   Proof by induction. Base case $n=1$: the formula gives $\left(\begin{array}{cc}3& 1\\ 0& 3\end{array}\right)=A$. Inductive step: suppose the formula holds for some $n\ge 1$. Since $A^{n+1}=A^{n}A$ by the definition of powers (associativity makes the bracketing irrelevant),

   $$A^{n+1}=\left(\begin{array}{cc}{3}^n& n{3}^{n-1}\\ 0& 3^n\end{array}\right)\left(\begin{array}{cc}3& 1\\ 0& 3\end{array}\right)=\left(\begin{array}{cc}{3}^{n+1}& 3^n+3n{3}^{n-1}\\ 0& 3^{n+1}\end{array}\right)=\left(\begin{array}{cc}{3}^{n+1}& (n+1)3^n\\ 0& 3^{n+1}\end{array}\right),$$

   because $3^n+n{3}^n=(n+1)3^n$. This is the formula with $n+1$ in place of $n$, so it holds for every $n\ge 1$. $■$

   (c) Split off the identity. $N=A-3I=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, and $N^2=\left(\begin{array}{cc}0\cdot 0+1\cdot 0& 0\cdot 1+1\cdot 0\\ 0& 0\end{array}\right)=0$. So $N$ is nilpotent and every power $N^k$ with $k\ge 2$ is zero.

   The binomial expansion of $(X+Y{)}^n$ is obtained by multiplying out $n$ brackets and collecting terms, and collecting $XYX$ together with $XXY$ requires $XY=YX$. So the theorem is valid for matrices exactly when the two summands commute. Here $X=3I$ and $Y=N$, and $3I$ commutes with everything: $(3I)N=3N=N(3I)$ by the product laws $r(AB)=(rA)B=A(rB)$ and $IN=N=NI$. Hence

   $$(3I+N{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(3I{)}^{n-k}{N}^k=3^{n}I+n{3}^{n-1}N,$$

   since every term with $k\ge 2$ contains $N^k=0$. That is $\left(\begin{array}{cc}{3}^n& 0\\ 0& 3^n\end{array}\right)+\left(\begin{array}{cc}0& n{3}^{n-1}\\ 0& 0\end{array}\right)$, the formula from (b). Sanity check at $n=3$: $27I+3\cdot 9N=\left(\begin{array}{cc}27& 27\\ 0& 27\end{array}\right)$, matching (a).

5. 5.Computation · 9 min

   Let $A=\left(\begin{array}{cc}4& 7\\ 1& 2\end{array}\right)$ and $B=\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)$. Every time you multiply by an inverse, say whether it is on the left or on the right.

   1. (a)Show that $A$ is invertible, find $A^{-1}$, and verify $A^{-1}A=I$ before using it.
   2. (b)Solve $AX=B$ and $XA=B$ for the unknown $2\times 2$ matrix $X$. Check each answer by substitution, and explain why the two solutions differ.
   3. (c)Now let $P$ and $Q$ be invertible $n\times n$ matrices. Solve $(PX{Q}^{-1}{)}^T=Q$ for $X$, and verify your answer by substituting back.
   Show the solution to question 5

   (a) Determinant, then swap-and-negate. $ad-bc=4\cdot 2-7\cdot 1=1\ne 0$, so $A$ is invertible and

   $$A^{-1}=\frac{1}{1}\left(\begin{array}{cc}2& -7\\ -1& 4\end{array}\right)=\left(\begin{array}{cc}2& -7\\ -1& 4\end{array}\right).$$

   Verify: $A^{-1}A=\left(\begin{array}{cc}2\cdot 4-7\cdot 1& 2\cdot 7-7\cdot 2\\ -1\cdot 4+4\cdot 1& -1\cdot 7+4\cdot 2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$.

   (b) Strip $A$ from the correct side. For $AX=B$, multiply on the left by $A^{-1}$: $A^{-1}AX=A^{-1}B$, so $X=A^{-1}B$:

   $$X_1=\left(\begin{array}{cc}2& -7\\ -1& 4\end{array}\right)\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)=\left(\begin{array}{cc}2-14& -7\\ -1+8& 4\end{array}\right)=\left(\begin{array}{cc}-12& -7\\ 7& 4\end{array}\right).$$

   Check: $A{X}_1=\left(\begin{array}{cc}-48+49& -28+28\\ -12+14& -7+8\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)=B$.

   For $XA=B$, multiply on the right by $A^{-1}$: $XA{A}^{-1}=B{A}^{-1}$, so $X=B{A}^{-1}$:

   $$X_2=\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)\left(\begin{array}{cc}2& -7\\ -1& 4\end{array}\right)=\left(\begin{array}{cc}2& -7\\ 4-1& -14+4\end{array}\right)=\left(\begin{array}{cc}2& -7\\ 3& -10\end{array}\right).$$

   Check: $X_{2}A=\left(\begin{array}{cc}8-7& 14-14\\ 12-10& 21-20\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)=B$.

   The solutions differ because $A^{-1}B\ne B{A}^{-1}$: multiplication is not commutative, so “divide by $A$” is meaningless until you say which side. Each solution is unique, by the same left- or right-multiplication that produced it.

   (c) Undo the operations in reverse order. The outermost operation is the transpose, so remove it first: transposing both sides and using $(M^T{)}^T=M$ gives $PX{Q}^{-1}=Q^T$. Next multiply on the left by $P^{-1}$: $X{Q}^{-1}=P^{-1}{Q}^T$. Finally multiply on the right by $Q$: $X=P^{-1}{Q}^{T}Q$.

   Verify: $PX{Q}^{-1}=P\left(P^{-1}{Q}^{T}Q\right)Q^{-1}=(P{P}^{-1})Q^T(Q{Q}^{-1})=Q^T$, regrouping by associativity, and transposing gives $(Q^T{)}^T=Q$. The order $Q^{T}Q$ cannot be swapped to $Q{Q}^T$: those are different matrices in general.

6. 6.Computation · 11 min

   Let $A=\left(\begin{array}{ccc}1& 0& 2\\ 2& 1& 3\\ 1& 1& 2\end{array}\right)$.

   1. (a)Compute $A^{-1}$ by row-reducing $[A\mid I]$. Name every row operation, and say at which point you can already be certain that $A$ is invertible.
   2. (b)Write each row operation you used as an elementary matrix $E_1,E_2,\dots$ in the order performed. State the identity relating $E_p\cdots E_1$, $A$ and $I$, explain from it why $A^{-1}=E_p\cdots E_1$, and hence express $A$ itself as a product of elementary matrices (you need not multiply that product out). Write $E_1^{-1}$ explicitly.
   3. (c)Use $A^{-1}$ to solve $A\mathbf{x}=(1,0,1{)}^T$, and check the answer in the original system.
   Show the solution to question 6

   (a) Reduce, carrying $I$ along.

   $$\left[\begin{array}{cccccc}1& 0& 2& 1& 0& 0\\ 2& 1& 3& 0& 1& 0\\ 1& 1& 2& 0& 0& 1\end{array}\right]\stackrel{R_2-2R_1,R_3-R_1}{\to }\left[\begin{array}{cccccc}1& 0& 2& 1& 0& 0\\ 0& 1& -1& -2& 1& 0\\ 0& 1& 0& -1& 0& 1\end{array}\right]\stackrel{R_3-R_2}{\to }\left[\begin{array}{cccccc}1& 0& 2& 1& 0& 0\\ 0& 1& -1& -2& 1& 0\\ 0& 0& 1& 1& -1& 1\end{array}\right].$$

   The left block is now upper triangular with three nonzero pivots, so it has $3$ pivot positions and the reduction can never produce a zero row: $A$ is invertible (Invertible Matrix Theorem, statement (c)), before the upward sweep is even started. Now clear upward: $R_2+R_3$ and $R_1-2R_3$ give

   $$\left[\begin{array}{cccccc}1& 0& 0& -1& 2& -2\\ 0& 1& 0& -1& 0& 1\\ 0& 0& 1& 1& -1& 1\end{array}\right],\text{so}{A}^{-1}=\left(\begin{array}{ccc}-1& 2& -2\\ -1& 0& 1\\ 1& -1& 1\end{array}\right).$$

   Sanity check two entries of $A{A}^{-1}$: row $2$ of $A$ times column $1$ of $A^{-1}$ is $2(-1)+1(-1)+3(1)=0$; row $3$ times column $3$ is $1(-2)+1(1)+2(1)=1$.

   (b) One elementary matrix per operation, first operation nearest $A$. In the order performed,

   $$E_1=L_{21}(-2)=\left(\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 0& 0& 1\end{array}\right),E_2=L_{31}(-1),E_3=L_{32}(-1),E_4=L_{23}(1),E_5=L_{13}(-2),$$

   where $L_{ij}(k)$ is $I$ with $k$ placed in position $(i,j)$. Because a row operation is left multiplication by its elementary matrix, the reduction says

   $$E_5{E}_4{E}_3{E}_2{E}_{1}A=I.$$

   So $M=E_5{E}_4{E}_3{E}_2{E}_1$ satisfies $MA=I$. Multiplying this on the right by $A^{-1}$, which exists by (a), gives $M=A^{-1}$; alternatively, the one-sided inverse corollary says a left inverse of a square matrix is the inverse. That is exactly what the right block computed: it is $MI=M$.

   Inverting the product with the reversal rule and $L_{ij}(k{)}^{-1}=L_{ij}(-k)$,

   $$A=E_1^{-1}{E}_2^{-1}{E}_3^{-1}{E}_4^{-1}{E}_5^{-1}=L_{21}(2)L_{31}(1)L_{32}(1)L_{23}(-1)L_{13}(2),E_1^{-1}=\left(\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 0& 0& 1\end{array}\right).$$

   (Multiplying the five factors out does return $A$; the first factor alone already restores the $2$ in position $(2,1)$.)

   (c) Multiply on the left by $A^{-1}$.

   $$\mathbf{x}=A^{-1}\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)=\left(\begin{array}{c}-1+0-2\\ -1+0+1\\ 1+0+1\end{array}\right)=\left(\begin{array}{c}-3\\ 0\\ 2\end{array}\right).$$

   Check: $A\mathbf{x}=(-3+0+4,-6+0+6,-3+0+4)=(1,0,1)$, and this is the only solution because $A$ is invertible.

7. 7.Critique · 9 min

   Three short arguments follow. Each contains an error. For each one, quote the exact sentence that is wrong, name the theorem or hypothesis being abused, and then give a correct treatment — which may or may not reach the same conclusion.

   1. (a)“Let $A$ be a square matrix with $A^2=A$. Then $AA=AI$. Cancelling the factor $A$ on the left gives $A=I$. So the identity is the only idempotent matrix.”
   2. (b)“Let $A=\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\end{array}\right)$ and $B=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right)$. A direct computation gives $AB=I_2$. By the corollary that a one-sided inverse suffices, $B=A^{-1}$, and therefore $BA=I_3$ as well.”
   3. (c)“Let $A$ be $3\times 3$ and suppose that the system $A\mathbf{x}=(1,2,3{)}^T$ has exactly one solution. Since $A\mathbf{x}=\mathbf{b}$ has a solution, statement (g) of the Invertible Matrix Theorem holds, so $A$ is invertible.”
   Show the solution to question 7

   (a) The wrong sentence is “Cancelling the factor $A$ on the left gives $A=I$.” There is no cancellation law for matrices: from $AA=AI$ one may conclude $A=I$ only by multiplying on the left by $A^{-1}$, which requires $A$ to be invertible — and nothing in the hypothesis says it is. The argument silently assumes the very thing that decides the question.

   Correct treatment: the deduction is valid if $A$ is invertible, so the true statement is “an invertible idempotent matrix is the identity”. A singular idempotent need not be: $A=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ satisfies $A^2=A$ and $A\ne I$. (It is singular, as it must be: $A\mathbf{x}=0$ has the nontrivial solution $(0,1{)}^T$, so statement (d) fails.) The conclusion of the argument is false.

   (b) The wrong sentence is “By the corollary that a one-sided inverse suffices, $B=A^{-1}$.” That corollary is a statement about $n\times n$ matrices — it is a consequence of the Invertible Matrix Theorem, which is a square-matrix theorem. Here $A$ is $2\times 3$ and $B$ is $3\times 2$; neither is square, so neither can be invertible in the first place (the definition $AC=CA=I$ needs both products to have the same size), and the symbol $A^{-1}$ has no meaning. The computation $AB=I_2$ is correct; the inference from it is not.

   Correct treatment: compute $BA$ directly. Row $i$ of $BA$ is row $i$ of $B$ times $A$, so rows $1$ and $2$ of $BA$ are rows $1$ and $2$ of $A$ and row $3$ is zero:

   $$BA=\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right)\ne I_3.$$

   Indeed $BA$ could never be $I_3$: $A\mathbf{x}=0$ has the nonzero solution $\mathbf{x}=(1,1,-1{)}^T$ (three unknowns, two equations), and then $BA\mathbf{x}=B0=0$, whereas $I_3\mathbf{x}=\mathbf{x}\ne 0$.

   (c) The wrong sentence is “Since $A\mathbf{x}=\mathbf{b}$ has a solution, statement (g) of the Invertible Matrix Theorem holds.” Statement (g) says $A\mathbf{x}=\mathbf{b}$ is solvable for every $\mathbf{b}\in {\mathbb{R}}^3$; solvability for one particular $\mathbf{b}$ establishes nothing. (The matrix $A=\mathrm{diag}(1,1,0)$ with $\mathbf{b}=(1,2,0{)}^T$ is a solvable system on a singular matrix.) The argument quotes the wrong statement — but its conclusion happens to be true, and the unused word “exactly” is what saves it.

   Correct treatment: use the uniqueness. Let ${\mathbf{x}}_0$ be the unique solution and suppose $A\mathbf{v}=0$. Then $A({\mathbf{x}}_0+\mathbf{v})=A{\mathbf{x}}_0+A\mathbf{v}=\mathbf{b}+0=\mathbf{b}$, so ${\mathbf{x}}_0+\mathbf{v}$ is also a solution, and uniqueness forces ${\mathbf{x}}_0+\mathbf{v}={\mathbf{x}}_0$, that is $\mathbf{v}=0$. Hence $A\mathbf{x}=0$ has only the trivial solution, which is statement (d), and by the Invertible Matrix Theorem $A$ is invertible. In words: a unique solution to any one consistent square system already certifies invertibility; a merely existing one does not.

8. 8.Computation · 10 min

   Let $M=\left(\begin{array}{cccc}1& 1& 2& 0\\ 0& 1& 1& 3\\ 0& 0& 1& 2\\ 0& 0& 1& 3\end{array}\right)$.

   1. (a)Partition $M$ into four $2\times 2$ blocks $\left(\begin{array}{cc}A& B\\ 0& D\end{array}\right)$, write down $A$, $B$ and $D$, and explain — without row-reducing $M$ — why $M$ is invertible.
   2. (b)Compute $M^{-1}$ using the block formula for a block upper triangular matrix. Keep the factors in the order the formula prescribes, and check one full row of $M{M}^{-1}$.
   3. (c)Replace the zero block by $C=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$ to obtain $N=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Compute the Schur complement $N/A=D-C{A}^{-1}B$ and decide whether $N$ is invertible, stating the criterion you use.
   Show the solution to question 8

   (a) Read off the blocks.

   $$A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right),B=\left(\begin{array}{cc}2& 0\\ 1& 3\end{array}\right),D=\left(\begin{array}{cc}1& 2\\ 1& 3\end{array}\right),$$

   and the lower-left block is $0$. A block upper triangular matrix with square diagonal blocks is invertible if and only if its diagonal blocks are. Here $\det A=1\cdot 1-1\cdot 0=1\ne 0$ and $\det D=1\cdot 3-2\cdot 1=1\ne 0$, so both are invertible by the $2\times 2$ criterion, and therefore $M$ is invertible.

   (b) Assemble the formula $M^{-1}=\left(\begin{array}{cc}{A}^{-1}& -A^{-1}B{D}^{-1}\\ 0& D^{-1}\end{array}\right)$. Swap-and-negate (both determinants are $1$):

   $$A^{-1}=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right),D^{-1}=\left(\begin{array}{cc}3& -2\\ -1& 1\end{array}\right).$$

   The corner block, left to right:

   $$A^{-1}B=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}2& 0\\ 1& 3\end{array}\right)=\left(\begin{array}{cc}1& -3\\ 1& 3\end{array}\right),A^{-1}B{D}^{-1}=\left(\begin{array}{cc}1& -3\\ 1& 3\end{array}\right)\left(\begin{array}{cc}3& -2\\ -1& 1\end{array}\right)=\left(\begin{array}{cc}6& -5\\ 0& 1\end{array}\right),$$

   so $-A^{-1}B{D}^{-1}=\left(\begin{array}{cc}-6& 5\\ 0& -1\end{array}\right)$ and

   $$M^{-1}=\left(\begin{array}{cccc}1& -1& -6& 5\\ 0& 1& 0& -1\\ 0& 0& 3& -2\\ 0& 0& -1& 1\end{array}\right).$$

   Check row $1$ of $M{M}^{-1}$: row $1$ of $M$ is $(1,1,2,0)$, and against the four columns of $M^{-1}$ it gives $1$, $-1+1=0$, $-6+0+6=0$, $5-1-4=0$. That is $(1,0,0,0)$, the first row of $I_4$. (Row $2$, $(0,1,1,3)$, gives $0$, $1$, $0+3-3=0$, $-1-2+3=0$ as well.)

   (c) Block elimination. From (b), $A^{-1}B=\left(\begin{array}{cc}1& -3\\ 1& 3\end{array}\right)$, so

   $$C{A}^{-1}B=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& -3\\ 1& 3\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 1& 3\end{array}\right),N/A=D-C{A}^{-1}B=\left(\begin{array}{cc}1& 2\\ 1& 3\end{array}\right)-\left(\begin{array}{cc}0& 0\\ 1& 3\end{array}\right)=\left(\begin{array}{cc}1& 2\\ 0& 0\end{array}\right).$$

   The criterion: since $A$ is invertible, $N=\left(\begin{array}{cc}I& 0\\ C{A}^{-1}& I\end{array}\right)\left(\begin{array}{cc}A& B\\ 0& N/A\end{array}\right)$. The first factor is block triangular with identity diagonal blocks, hence invertible, so $N$ is invertible exactly when the second factor is — that is, exactly when $A$ and $N/A$ are both invertible. Here $N/A$ has a zero row, so it is singular, and therefore $N$ is not invertible.

   Sanity check on $N$ itself: the change put a single $1$ in position $(4,2)$, so row $4$ of $N$ is $(0,1,1,3)$, identical to row $2$. Two equal rows means $R_4-R_2$ produces a zero row, so $N$ has fewer than $4$ pivots — singular by the Invertible Matrix Theorem, exactly as the Schur complement predicted.

9. 9.Critique · 8 min

   Each argument below reaches a false conclusion. For each: identify the exact step that fails and name the law of numbers that is being applied to matrices without justification; produce an explicit counterexample to the conclusion, with the verifying computation; and state the correct theorem that survives, with proof.

   1. (a)“Suppose $Q$ is both symmetric and orthogonal. Then $Q^T=Q$ and $Q^T=Q^{-1}$, so $Q=Q^{-1}$, hence $Q^2=I$. Taking square roots, $Q=I$. So the identity is the only symmetric orthogonal matrix.”
   2. (b)“Let $N$ be nilpotent, say $N^k=0$. Taking $k$-th roots of both sides gives $N=0$. So the only nilpotent matrix is the zero matrix.”
   3. (c)“If $P$ and $Q$ are idempotent $n\times n$ matrices then so is $PQ$, because $(PQ{)}^2=P^2{Q}^2=PQ$.”
   Show the solution to question 9

   (a) The failing step is “Taking square roots, $Q=I$.” Even for real numbers $x^2=1$ has two solutions, and for matrices $X^2=I$ has infinitely many: there is no square-root law. Everything before that step is correct, so what has actually been proved is that a symmetric orthogonal matrix satisfies $Q^2=I$.

   Counterexample: $Q=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. It is symmetric by inspection, and $Q^{T}Q=Q^2=\left(\begin{array}{cc}0\cdot 0+1\cdot 1& 0\cdot 1+1\cdot 0\\ 1\cdot 0+0\cdot 1& 1\cdot 1+0\cdot 0\end{array}\right)=I$, so it is orthogonal — yet $Q\ne I$. (It is the reflection in the line $y=x$.)

   Correct theorem: a symmetric matrix is orthogonal if and only if $Q^2=I$. Proof. If $Q^T=Q$ and $Q^{T}Q=I$ then $Q^2=Q^{T}Q=I$. Conversely if $Q^T=Q$ and $Q^2=I$ then $Q^{T}Q=Q^2=I$, which is the definition of orthogonal. $■$

   (b) The failing step is “Taking $k$-th roots of both sides gives $N=0$.” For numbers, $x^k=0$ forces $x=0$ because there are no zero divisors; matrices have zero divisors, so a power of a nonzero matrix can vanish and no root-extraction is available.

   Counterexample: $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\ne 0$, and $N^2=\left(\begin{array}{cc}0\cdot 0+1\cdot 0& 0\cdot 1+1\cdot 0\\ 0& 0\end{array}\right)=0$.

   Correct theorem: nilpotent matrices are far from being zero in general, but they are always singular — a nilpotent matrix is never invertible. Proof. If $N^k=0$ and $N$ were invertible, multiplying $N^k=0$ on the left by $(N^{-1}{)}^k$ gives $I=0$, which is false for $n\ge 1$. $■$

   (c) The failing step is “$(PQ{)}^2=P^2{Q}^2$.” By definition $(PQ{)}^2=PQPQ$, and rearranging the middle factors to $PPQQ$ uses commutativity, $QP=PQ$, which is not given. This is the same error as writing $(AB{)}^2=A^2{B}^2$.

   Counterexample: $P=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ and $Q=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$. Both are idempotent ($P^2=P$ by inspection; $Q^2=\frac{1}{4}\left(\begin{array}{cc}2& 2\\ 2& 2\end{array}\right)=Q$). Their product is $PQ=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$, and

   $$(PQ{)}^2=\frac{1}{4}\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)=\frac{1}{4}\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)=\frac{1}{2}PQ\ne PQ.$$

   So $PQ$ is not idempotent.

   Correct theorem: if $P$ and $Q$ are idempotent and commute, then $PQ$ is idempotent. Proof. $(PQ{)}^2=PQPQ=P(QP)Q=P(PQ)Q=P^2{Q}^2=PQ$, where the third equality is exactly the commuting hypothesis and the last uses idempotence. $■$

10. 10.Proof · 10 min

    Let $A$ be an $m\times n$ matrix, so that $A^{T}A$ is square of size $n\times n$. You may use that ${\mathbf{v}}^T\mathbf{v}=v_1^2+\cdots +v_m^2$ for any $\mathbf{v}\in {\mathbb{R}}^m$.

    1. (a)Prove that the equations $A\mathbf{x}=0$ and $A^{T}A\mathbf{x}=0$ have exactly the same solutions $\mathbf{x}\in {\mathbb{R}}^n$.
    2. (b)Deduce that $A^{T}A$ is invertible if and only if the columns of $A$ are linearly independent. Say which statements of the Invertible Matrix Theorem you use and to which matrix you apply them.
    3. (c)Let $A=\left(\begin{array}{cc}1& 2\\ 0& 1\\ 1& 0\end{array}\right)$. Verify (b) in both directions for this $A$. Then show that $A{A}^T$ is not invertible by exhibiting a nonzero $\mathbf{y}$ with $A{A}^T\mathbf{y}=0$, and explain why the same must happen for every $3\times 2$ matrix.
    Show the solution to question 10

    (a) Two inclusions. If $A\mathbf{x}=0$, then multiplying on the left by $A^T$ gives $A^{T}A\mathbf{x}=A^{T}0=0$. So every solution of the first equation solves the second.

    Conversely suppose $A^{T}A\mathbf{x}=0$. Multiply on the left by the $1\times n$ row ${\mathbf{x}}^T$: ${\mathbf{x}}^T{A}^{T}A\mathbf{x}=0$. By the reversal rule ${\mathbf{x}}^T{A}^T=(A\mathbf{x}{)}^T$, so with $\mathbf{v}=A\mathbf{x}\in {\mathbb{R}}^m$ this reads

    $${\mathbf{v}}^T\mathbf{v}=v_1^2+v_2^2+\cdots +v_m^2=0.$$

    A sum of squares of real numbers is zero only when every term is zero, so $\mathbf{v}=0$, that is $A\mathbf{x}=0$. The two solution sets coincide. $■$

    (b) Apply the theorem to the square matrix $A^{T}A$. Statements (a) and (d) of the Invertible Matrix Theorem, applied to the $n\times n$ matrix $A^{T}A$, say: $A^{T}A$ is invertible if and only if $A^{T}A\mathbf{x}=0$ has only the trivial solution. By part (a) that is the same as $A\mathbf{x}=0$ having only the trivial solution. Finally, $A\mathbf{x}=x_1{\mathbf{a}}_1+\cdots +x_n{\mathbf{a}}_n$, so “$A\mathbf{x}=0$ only for $\mathbf{x}=0$” is word for word the definition of the columns ${\mathbf{a}}_1,\dots ,{\mathbf{a}}_n$ being linearly independent — the (d)$\iff$(e) rewrite, which is valid for any shape of $A$ because it is a rewrite, not a theorem about square matrices. Chaining the three equivalences: $A^{T}A$ invertible $\iff$ columns of $A$ independent. $■$

    Note that the theorem is applied only to $A^{T}A$, which is square; $A$ itself may be any shape, and for $m\ne n$ it is never invertible.

    (c) Both directions, then the other product. The columns $(1,0,1{)}^T$ and $(2,1,0{)}^T$ are independent: neither is a multiple of the other (the second has a nonzero middle entry, the first does not). And

    $$A^{T}A=\left(\begin{array}{ccc}1& 0& 1\\ 2& 1& 0\end{array}\right)\left(\begin{array}{cc}1& 2\\ 0& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1+0+1& 2+0+0\\ 2+0+0& 4+1+0\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 2& 5\end{array}\right),$$

    with $\det =2\cdot 5-2\cdot 2=6\ne 0$, so it is invertible — independent columns and invertible $A^{T}A$, as (b) requires.

    For $A{A}^T$ (size $3\times 3$), first find $\mathbf{y}\ne 0$ with $A^T\mathbf{y}=0$: the equations are $y_1+y_3=0$ and $2y_1+y_2=0$. Taking $y_1=1$ gives $\mathbf{y}=(1,-2,-1{)}^T$. Then $A{A}^T\mathbf{y}=A(A^T\mathbf{y})=A0=0$ with $\mathbf{y}\ne 0$, so statement (d) fails for $A{A}^T$ and it is singular. As a direct check,

    $$A{A}^T=\left(\begin{array}{ccc}5& 2& 1\\ 2& 1& 0\\ 1& 0& 1\end{array}\right),A{A}^T\mathbf{y}=\left(\begin{array}{c}5-4-1\\ 2-2+0\\ 1+0-1\end{array}\right)=0.$$

    This is forced for every $3\times 2$ matrix $A$: $A^T$ is $2\times 3$, so $A^T\mathbf{y}=0$ is a homogeneous system of $2$ equations in $3$ unknowns. Its echelon form has at most $2$ pivots, hence at least one free variable, hence a nonzero solution $\mathbf{y}$; and then $A{A}^T\mathbf{y}=A(A^T\mathbf{y})=A0=0$ by the same two-step argument. So $A{A}^T$ is always singular when $A$ has more rows than columns, while $A^{T}A$ is invertible exactly when the columns are independent — which is why least squares works with $A^{T}A$ and never with $A{A}^T$.

11. 11.Proof · 10 min

    Let $A$ be an invertible $n\times n$ matrix and $N$ a nilpotent $n\times n$ matrix with $N^k=0$, and suppose $A$ and $N$ commute: $AN=NA$.

    1. (a)Prove that $A^{-1}$ also commutes with $N$: $A^{-1}N=N{A}^{-1}$.
    2. (b)Let $M=A^{-1}N$. Prove by induction that $M^j=(A^{-1}{)}^j{N}^j$ for every $j\ge 1$, and conclude that $M$ is nilpotent.
    3. (c)Prove that $A-N$ is invertible, with $(A-N{)}^{-1}=(I+M+M^2+\cdots +M^{k-1})A^{-1}$.
    4. (d)Apply this with $A=2I_3$ and $N=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right)$ to compute $(2I-N{)}^{-1}$ exactly, and check the first row of the product $(2I-N)(2I-N{)}^{-1}$.
    Show the solution to question 11

    (a) Multiply the commuting relation by $A^{-1}$ on both sides. Start from $AN=NA$. Multiplying on the left by $A^{-1}$ gives $N=A^{-1}NA$. Now multiply on the right by $A^{-1}$: $N{A}^{-1}=A^{-1}N(A{A}^{-1})=A^{-1}N$. $■$

    (b) Induction on $j$. The case $j=1$ is the definition of $M$. Suppose $M^j=(A^{-1}{)}^j{N}^j$. Then

    $$M^{j+1}=M^{j}M=(A^{-1}{)}^j{N}^j{A}^{-1}N.$$

    By (a), $A^{-1}$ commutes with $N$, hence with $N^j$ (move $A^{-1}$ past one factor of $N$ at a time, $j$ times): $N^j{A}^{-1}=A^{-1}{N}^j$. So

    $$M^{j+1}=(A^{-1}{)}^j{A}^{-1}{N}^{j}N=(A^{-1}{)}^{j+1}{N}^{j+1},$$

    which completes the induction. In particular $M^k=(A^{-1}{)}^k{N}^k=(A^{-1}{)}^k\cdot 0=0$, so $M$ is nilpotent. $■$ (Without the commuting hypothesis this fails: $M^2=A^{-1}N{A}^{-1}N$ cannot be regrouped.)

    (c) Factor out $A$ and use the finite geometric series. Since $A$ is invertible, $A-N=A(I-A^{-1}N)=A(I-M)$. Let $S=I+M+M^2+\cdots +M^{k-1}$. Multiplying out, the sum telescopes:

    $$(I-M)S=(I+M+\cdots +M^{k-1})-(M+M^2+\cdots +M^k)=I-M^k=I,$$

    and the same computation in the other order gives $S(I-M)=I$ (every factor here is a power of $M$, and powers of one matrix commute with each other). So $I-M$ is invertible with inverse $S$. Then $A-N=A(I-M)$ is a product of two invertible matrices, hence invertible, and by the reversal rule for inverses

    $$(A-N{)}^{-1}=(I-M{)}^{-1}{A}^{-1}=(I+M+M^2+\cdots +M^{k-1})A^{-1}.■$$

    (d) Specialise. $A=2I$ commutes with everything and $A^{-1}=\frac{1}{2}I$, so $M=\frac{1}{2}N$. Powers of $N$: $N^2=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$ (the single $1$ moves one place further from the diagonal) and $N^3=0$, so $k=3$ and $M^2=\frac{1}{4}{N}^2$. Therefore

    $$(2I-N{)}^{-1}=(I+\frac{1}{2}N+\frac{1}{4}{N}^2)\frac{1}{2}I=\frac{1}{2}\left(\begin{array}{ccc}1& \frac{1}{2}& \frac{1}{4}\\ 0& 1& \frac{1}{2}\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{2}& \frac{1}{4}& \frac{1}{8}\\ 0& \frac{1}{2}& \frac{1}{4}\\ 0& 0& \frac{1}{2}\end{array}\right).$$

    Check: $2I-N=\left(\begin{array}{ccc}2& -1& 0\\ 0& 2& -1\\ 0& 0& 2\end{array}\right)$, and its first row $(2,-1,0)$ against the three columns of the candidate inverse gives $2\cdot \frac{1}{2}=1$, $2\cdot \frac{1}{4}-\frac{1}{2}=0$, $2\cdot \frac{1}{8}-\frac{1}{4}=0$: the first row of $I$, as required.

12. 12.Sketch · 12 min

    Work in homogeneous coordinates, writing the point $(x,y)$ as $(x,y,1{)}^T$. Let $R$ be the $3\times 3$ matrix of the counterclockwise rotation by $90^{\circ }$ about the origin, and $T$ the $3\times 3$ matrix of the translation by $(2,0)$.

    1. (a)Write down $R$ and $T$, and compute both products $TR$ and $RT$. State in words which motion each product performs, in which order.
    2. (b)Let $U$ be the unit square with vertices $O=(0,0)$, $P=(1,0)$, $Q=(1,1)$, $S=(0,1)$. Compute the image of each vertex under $TR$ and under $RT$, then draw one set of axes showing $U$ and both image squares, labelling each image vertex with the name of the vertex it came from.
    3. (c)Each of $TR$ and $RT$ is a rotation by $90^{\circ }$ about some point. Find the fixed point of each by solving a linear system, and mark both centres on your sketch. Explain what the sketch shows about $TR$ versus $RT$.
    4. (d)Compute $(TR{)}^{-1}$ by the reversal rule, without row reduction, and verify that it fixes the centre you found for $TR$.
    Show the solution to question 12

    (a) The two matrices and their products. In homogeneous coordinates a linear map sits in the top-left $2\times 2$ block and a translation in the last column:

    $$R=\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right),T=\left(\begin{array}{ccc}1& 0& 2\\ 0& 1& 0\\ 0& 0& 1\end{array}\right).$$

    Multiplying (the rightmost factor acts first):

    $$TR=\left(\begin{array}{ccc}0& -1& 2\\ 1& 0& 0\\ 0& 0& 1\end{array}\right),RT=\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 2\\ 0& 0& 1\end{array}\right).$$

    For $TR$ the only new work is the last column: row $1$ of $T$ times column $3$ of $R$ is $1\cdot 0+0\cdot 0+2\cdot 1=2$. For $RT$, row $2$ of $R$ times column $3$ of $T$ is $1\cdot 2+0+0=2$. So $TR$ is “rotate, then translate”: $(x,y)\mapsto (2-y,x)$. And $RT$ is “translate, then rotate”: $(x,y)\mapsto (-y,x+2)$. The products differ, so the two orders are different motions.

    (b) Images of the vertices. Under $TR$, $(x,y)\mapsto (2-y,x)$:

    $$O=(0,0)\mapsto (2,0),P=(1,0)\mapsto (2,1),Q=(1,1)\mapsto (1,1),S=(0,1)\mapsto (1,0).$$

    Under $RT$, $(x,y)\mapsto (-y,x+2)$:

    $$O\mapsto (0,2),P\mapsto (0,3),Q\mapsto (-1,3),S\mapsto (-1,2).$$

    The sketch must show three unit squares on one grid. The original $U$ occupies $[0,1]\times [0,1]$. The $TR$-image occupies $[1,2]\times [0,1]$, immediately to the right of $U$ and sharing the edge $x=1$; its corner labels read $S$ at $(1,0)$, $O$ at $(2,0)$, $P$ at $(2,1)$, $Q$ at $(1,1)$ — the labels go round in the same counterclockwise sense as on $U$, because a rotation preserves orientation, but they have been turned a quarter turn. The $RT$-image occupies $[-1,0]\times [2,3]$, up and to the left, with $O$ at $(0,2)$, $P$ at $(0,3)$, $Q$ at $(-1,3)$, $S$ at $(-1,2)$. The two image squares do not overlap; the gap between them is the whole content of $TR\ne RT$.

    (c) Fixed points. A fixed point of $TR$ satisfies $(2-y,x)=(x,y)$, i.e. $x=2-y$ and $y=x$, so $x=y=1$: the centre is $(1,1)$, which is the vertex $Q$ — consistent with (b), where $Q$ stayed put. So $TR$ is the rotation by $90^{\circ }$ about $(1,1)$; the image square is $U$ spun a quarter turn about its top-right corner.

    A fixed point of $RT$ satisfies $(-y,x+2)=(x,y)$, i.e. $x=-y$ and $y=x+2$, so $y=-y+2$, $y=1$, $x=-1$: the centre is $(-1,1)$. Check: $RT$ sends $(-1,1)$ to $(-1,-1+2)=(-1,1)$. So $RT$ is the rotation by $90^{\circ }$ about $(-1,1)$, a point outside $U$, which is why the image landed far away.

    Mark both centres. The sketch shows that the two composites are the same kind of motion — a quarter turn — about different centres, two units apart. That is the geometry behind non-commutativity here: which operation runs first changes where the rotation centre ends up, not the amount of rotation.

    (d) Reverse the order and invert each factor. By the reversal rule $(TR{)}^{-1}=R^{-1}{T}^{-1}$. Undoing a translation by $(2,0)$ is translating by $(-2,0)$, and undoing a $90^{\circ }$ rotation is rotating by $-90^{\circ }$:

    $$T^{-1}=\left(\begin{array}{ccc}1& 0& -2\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),R^{-1}=R^T=\left(\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right)$$

    (the columns of $R$ are orthonormal, so $R$ is orthogonal and its inverse is its transpose), and

    $$(TR{)}^{-1}=R^{-1}{T}^{-1}=\left(\begin{array}{ccc}0& 1& 0\\ -1& 0& 2\\ 0& 0& 1\end{array}\right),$$

    the $(2,3)$ entry being $-1\cdot (-2)+0+0=2$. Check that it is the inverse: $(TR)(R^{-1}{T}^{-1})$ has first row $(0,-1,2)$ times the columns $(0,-1,0{)}^T$, $(1,0,0{)}^T$, $(0,2,1{)}^T$, giving $(1,0,-2+2)=(1,0,0)$; the other rows work the same way, so the product is $I_3$. As a map, $(TR{)}^{-1}$ sends $(x,y)\mapsto (y,2-x)$, and at the centre $(1,1)$ this gives $(1,2-1)=(1,1)$: the inverse fixes the same point, as it must — undoing a rotation about $(1,1)$ is rotating the other way about $(1,1)$.