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1. 1.Computation · 6 min

   Let $D$ assign a number to each list of three row vectors in ${\mathbb{R}}^3$, and suppose $D$ is multilinear, alternating and normalised. No determinant formula may be used anywhere in this question: every step must cite one of the three defining properties or a consequence proved from them (a swap changes the sign; adding a multiple of one row to another changes nothing).

   1. (a)Compute $D(3{\mathbf{e}}_2,{\mathbf{e}}_1+{\mathbf{e}}_3,{\mathbf{e}}_3)$.
   2. (b)Suppose $D({\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3)=5$. Compute $D({\mathbf{a}}_1+2{\mathbf{a}}_2,{\mathbf{a}}_2,{\mathbf{a}}_3-{\mathbf{a}}_1)$ and $D({\mathbf{a}}_3,{\mathbf{a}}_1,{\mathbf{a}}_2)$.
   3. (c)With the same ${\mathbf{a}}_i$, compute $D(-2{\mathbf{a}}_1,-2{\mathbf{a}}_2,-2{\mathbf{a}}_3)$, and say in one sentence why the answer is not $-10$.
   Show the solution to question 1

   (a) Expand the slot that is a sum. By linearity in the second slot,

   $$D(3{\mathbf{e}}_2,{\mathbf{e}}_1+{\mathbf{e}}_3,{\mathbf{e}}_3)=D(3{\mathbf{e}}_2,{\mathbf{e}}_1,{\mathbf{e}}_3)+D(3{\mathbf{e}}_2,{\mathbf{e}}_3,{\mathbf{e}}_3).$$

   The second term has two equal rows, so it is $0$ because $D$ is alternating. In the first term pull the scalar out of slot one by linearity: $3D({\mathbf{e}}_2,{\mathbf{e}}_1,{\mathbf{e}}_3)$. Swapping the first two rows changes the sign, so this is $-3D({\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3)=-3\cdot 1=-3$ by normalisation.

   Sanity check: the rows are $(0,3,0)$, $(1,0,1)$, $(0,0,1)$, and the answer $-3$ is what the volume picture predicts — a unit box stretched by $3$ in one direction and with two edges listed in the wrong order.

   (b) Both changes are row replacements. Expand slot one: $D({\mathbf{a}}_1+2{\mathbf{a}}_2,{\mathbf{a}}_2,{\mathbf{a}}_3-{\mathbf{a}}_1)=D({\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3-{\mathbf{a}}_1)+2D({\mathbf{a}}_2,{\mathbf{a}}_2,{\mathbf{a}}_3-{\mathbf{a}}_1)$, and the second term dies because two rows are equal. Now expand slot three of what is left: $D({\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3)-D({\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_1)=5-0=5$. (This is the corollary that adding a multiple of one row to a different row is invisible, applied twice.)

   For the cyclic shift, swap slots one and two, then slots two and three:

   $$D({\mathbf{a}}_3,{\mathbf{a}}_1,{\mathbf{a}}_2)=-D({\mathbf{a}}_1,{\mathbf{a}}_3,{\mathbf{a}}_2)=+D({\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3)=5.$$

   Two swaps, two sign changes, so a cyclic shift of three rows leaves $D$ alone.

   (c) One scalar per row. Multilinearity lets a scalar be pulled out of one slot at a time, so the factor $-2$ comes out three times:

   $$D(-2{\mathbf{a}}_1,-2{\mathbf{a}}_2,-2{\mathbf{a}}_3)=(-2{)}^{3}D({\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3)=-8\cdot 5=-40.$$

   It is not $-10$ because $D$ is linear in each row separately, not in the whole matrix: scaling the entire matrix by $c$ scales every one of the $n$ rows, and each row contributes its own factor of $c$.

2. 2.Computation · 7 min

   Work directly with the permutation definition $\det (A)={\sum }_{\sigma \in S_n}\mathrm{sgn}(\sigma ){\prod }_i{a}_{i\sigma (i)}$ throughout. Cofactor expansion may be used only as a final check.

   1. (a)For $\sigma =41352$ in $S_5$, list every inversion, state $\mathrm{sgn}(\sigma )$, and write down the term $\sigma$ contributes to the determinant of a general $5\times 5$ matrix $A=(a_{ij})$. Then confirm the sign by turning $12345$ into $\sigma$ with a sequence of transpositions.
   2. (b)Compute $\det \left(\begin{array}{cccc}0& 2& 0& 3\\ 1& 0& 0& 0\\ 0& 0& 4& 0\\ 0& 5& 0& 1\end{array}\right)$ by identifying every permutation whose term survives, signing each, and adding.
   3. (c)Without listing them, explain why exactly $12$ of the $24$ terms in the permutation sum for a general $4\times 4$ matrix carry a plus sign.
   Show the solution to question 2

   (a) Count the pairs that are out of order. Reading $41352$ left to right: the $4$ precedes the smaller entries $1$, $3$ and $2$, giving the inversions $(4,1)$, $(4,3)$, $(4,2)$; the $1$ precedes nothing smaller; the $3$ precedes the $2$, giving $(3,2)$; the $5$ precedes the $2$, giving $(5,2)$. That is $\mathrm{inv}(\sigma )=5$, so $\mathrm{sgn}(\sigma )=(-1{)}^5=-1$.

   Row $1$ takes column $4$, row $2$ takes column $1$, row $3$ takes column $3$, row $4$ takes column $5$, row $5$ takes column $2$, so the term is

   $$-a_{14}{a}_{21}{a}_{33}{a}_{45}{a}_{52}.$$

   Transposition check, starting from $12345$: exchange the entries $1$ and $4$ to get $42315$; exchange $2$ and $1$ to get $41325$; exchange $2$ and $5$ to get $41352$. Three transpositions, an odd number, so the sign is $-1$. ✓

   (b) Let the zeros choose the permutation. A term survives only if every factor is nonzero. Row $2$ has its only nonzero entry in column $1$, so $\sigma (2)=1$; row $3$ forces $\sigma (3)=3$. Rows $1$ and $4$ must then share columns $2$ and $4$, and both ways are available: $\sigma (1)=2,\sigma (4)=4$, or $\sigma (1)=4,\sigma (4)=2$. So exactly two permutations survive:

   $\sigma =2134$ has the single inversion $(2,1)$, sign $-1$, and product $a_{12}{a}_{21}{a}_{33}{a}_{44}=2\cdot 1\cdot 4\cdot 1=8$, contributing $-8$.

   $\sigma =4132$ has inversions $(4,1),(4,3),(4,2),(3,2)$, four of them, sign $+1$, and product $a_{14}{a}_{21}{a}_{33}{a}_{42}=3\cdot 1\cdot 4\cdot 5=60$, contributing $+60$.

   Hence $\det =-8+60=52$.

   Sanity check by expanding along row $2$, whose only nonzero entry is the $1$ at position $(2,1)$ with cofactor sign $(-1{)}^{2+1}=-1$. The minor deletes row $2$ and column $1$, leaving $\left(\begin{array}{ccc}2& 0& 3\\ 0& 4& 0\\ 5& 0& 1\end{array}\right)$; expanding that along its middle row gives $4(2\cdot 1-3\cdot 5)=-52$. So $\det =-1\cdot (-52)=52$. ✓

   (c) Pair each permutation with a partner of the opposite sign. Let $\tau$ be the transposition that exchanges $1$ and $2$, and send each $\sigma \in S_4$ to $\sigma \tau$. This map is a bijection of $S_4$ onto itself, because applying it twice returns $\sigma$. By the lemma that a transposition is odd, $\mathrm{sgn}(\sigma \tau )=-\mathrm{sgn}(\sigma )$, so every even permutation is paired with an odd one and vice versa. A bijection between the even and the odd permutations means there are equally many of each: $24/2=12$ terms with a plus sign and $12$ with a minus sign. (The same pairing is what proves the permutation sum is alternating.)

3. 3.Computation · 7 min

   Cofactor expansion only in this question — no elimination. Before each expansion, say which row or column you are using and why, and write the checkerboard sign of every position you expand through.

   1. (a)Compute $\det \left(\begin{array}{cccc}3& 0& 2& 0\\ 1& 4& -1& 2\\ 0& 0& 5& 0\\ 2& 1& 3& -1\end{array}\right)$, and confirm the value with a second, different expansion.
   2. (b)Find every real $t$ for which $\left(\begin{array}{ccc}t& 3& 0\\ 1& t& 0\\ 2& 5& t-4\end{array}\right)$ is singular.
   3. (c)State $\det \left(\begin{array}{cccc}2& 7& -3& 9\\ 0& -1& 4& 8\\ 0& 0& 3& -6\\ 0& 0& 0& 5\end{array}\right)$ without computing any minor, and explain what repeated expansion along the first column would do.
   Show the solution to question 3

   (a) Expand along the line with the most zeros. Row $3$ is $(0,0,5,0)$, so only the $(3,3)$ position survives, and its sign is $(-1{)}^{3+3}=+$. Deleting row $3$ and column $3$ leaves

   $$M_{33}=\det \left(\begin{array}{ccc}3& 0& 0\\ 1& 4& 2\\ 2& 1& -1\end{array}\right).$$

   Its first row has two zeros, so expand along it: $3\det \left(\begin{array}{cc}4& 2\\ 1& -1\end{array}\right)=3(-4-2)=-18$. Hence $\det =5\cdot (-18)=-90$.

   Second expansion, along column $2$ of the original, whose entries are $0,4,0,1$. The signs at positions $(2,2)$ and $(4,2)$ are $(-1{)}^4=+$ and $(-1{)}^6=+$, so

   $$\det =4\det \left(\begin{array}{ccc}3& 2& 0\\ 0& 5& 0\\ 2& 3& -1\end{array}\right)+1\cdot \det \left(\begin{array}{ccc}3& 2& 0\\ 1& -1& 2\\ 0& 5& 0\end{array}\right).$$

   The first minor, expanded along its middle row, is $5(3\cdot (-1)-0\cdot 2)=-15$. The second, expanded along its bottom row where only the $5$ at position $(3,2)$ survives with sign $-$, is $-5(3\cdot 2-0\cdot 1)=-30$. Total: $4(-15)+(-30)=-90$. ✓

   (b) Expand along column 3. Only the $(3,3)$ entry is nonzero there, with sign $+$, so

   $$\det =(t-4)\det \left(\begin{array}{cc}t& 3\\ 1& t\end{array}\right)=(t-4)(t^2-3).$$

   By the singularity criterion the matrix is singular exactly when this vanishes:

   $$t=4,t=\sqrt{3},t=-\sqrt{3}.$$

   Sanity check at $t=4$: the third column becomes entirely zero, so the determinant is $0$ by inspection. ✓

   (c) Triangular means product of the diagonal. The matrix is upper triangular, so $\det =2\cdot (-1)\cdot 3\cdot 5=-30$.

   Expanding along the first column shows why: the only nonzero entry is the $2$ at $(1,1)$, with sign $+$, so $\det =2M_{11}$, and the minor is again upper triangular with diagonal $-1,3,5$. Repeating peels off one diagonal entry at a time, each with sign $+$, until a $1\times 1$ matrix is left. The entries above the diagonal — $7,-3,9,4,8,-6$ — never appear, because each expansion discards the row and column that contain them.

4. 4.Computation · 6 min

   Let $A$ and $B$ be $3\times 3$ matrices with $\det (A)=-2$ and $\det (B)=5$. Name the theorem behind every step.

   1. (a)Compute $\det \left(3A^{-1}{B}^{T}A\right)$.
   2. (b)Compute $\det \left(A^2{B}^{-1}\right)$ and $\det (-B)$.
   3. (c)Is there an invertible $P$ with $P^{-1}AP=B$? Is there an invertible $P$ with $P^{-1}AP=-A^T$? Justify each answer in one or two lines.
   Show the solution to question 4

   (a) Scalar first, then multiply. The scalar $3$ multiplies a $3\times 3$ matrix, so by the scaling theorem it contributes $3^3=27$. Multiplicativity splits the rest into factors, transpose invariance gives $\det (B^T)=\det (B)=5$, and the inverse corollary gives $\det (A^{-1})=1/\det (A)=-\frac{1}{2}$:

   $$\det \left(3A^{-1}{B}^{T}A\right)=27\cdot \left(-\frac{1}{2}\right)\cdot 5\cdot (-2)=27\cdot 5=135.$$

   Sanity check on the sign: two negative factors appear, $\det (A^{-1})$ and $\det (A)$, so the product is positive. ✓

   (b) By multiplicativity and the powers corollary, $\det (A^2{B}^{-1})=\det (A{)}^2\cdot \frac{1}{\det (B)}=\frac{4}{5}$.

   For $-B=(-1)B$, the scaling theorem gives $(-1{)}^3\det (B)=-5$. (For a $4\times 4$ matrix the answer would have been $+5$; the sign depends on the size.)

   (c) Similar matrices have equal determinants. If $P^{-1}AP=B$ then $\det (B)=\det (P^{-1})\det (A)\det (P)=\det (A)$, but $5\ne -2$. So no such $P$ exists.

   For the second question compute the target's determinant: $\det (-A^T)=(-1{)}^3\det (A^T)=-\det (A)=2$, using scaling and transpose invariance. Since $2\ne -2$, no invertible $P$ can make $A$ similar to $-A^T$ either. (Note that $A$ is always similar to $A^T$ in the sense that they share a determinant; the minus sign is what breaks it here, and only because the size is odd.)

5. 5.Critique · 8 min

   Each of the three arguments below is wrong. For each one, quote the exact sentence that fails, say which theorem or hypothesis it abuses, and then give the correct conclusion — with a proof if the claim can be rescued, or a counterexample if it cannot.

   1. (a)“Let $A$ be $3\times 3$ with $\det (A)=4$. Then $\det (2A)=2\cdot 4=8$. As a check, $2A=A+A$, and multilinearity gives $\det (A+A)=\det (A)+\det (A)=8$, so the two computations agree.”
   2. (b)“Let $A$ be a real $2\times 2$ matrix with $A^2=-I$. Taking determinants, $\det (A{)}^2=\det (-I)=-1$. A real number cannot have a negative square, so no such matrix exists.”
   3. (c)“Let $A$ be $n\times n$ with $A^3=A$. Taking determinants, $\det (A{)}^3=\det (A)$, so dividing by $\det (A)$ gives $\det (A{)}^2=1$. Hence $\det (A)=\pm 1$ and $A$ is invertible.”
   Show the solution to question 5

   (a) Two errors that happen to agree with each other. The sentence “$\det (2A)=2\cdot 4=8$” is wrong: $2A$ scales each of the three rows by $2$, so by the scaling theorem $\det (2A)=2^3\cdot 4=32$. The sentence “multilinearity gives $\det (A+A)=\det (A)+\det (A)$” is also wrong, and for a related reason: multilinearity is linearity in one row at a time, while $A+A$ changes every row at once. The determinant is not additive. A counterexample to additivity is $A=B=I_3$, where $\det (A+B)=\det (2I)=8$ but $\det (A)+\det (B)=2$.

   Correct conclusion: $\det (A+A)=\det (2A)=32$. The student's “check” only appeared to work because both mistakes drop the same factor.

   (b) The size of the identity matters. The sentence “$\det (-I)=-1$” is false for $2\times 2$ matrices: $-I=(-1)I$ and the scaling theorem gives $(-1{)}^2\det (I)=+1$. So the correct equation is $\det (A{)}^2=1$, which has the real solutions $\det (A)=\pm 1$, and there is no contradiction.

   Such matrices do exist. Take the rotation through a right angle, $R=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Multiplying out,

   $$R^2=\left(\begin{array}{cc}0\cdot 0+(-1)\cdot 1& 0\cdot (-1)+(-1)\cdot 0\\ 1\cdot 0+0\cdot 1& 1\cdot (-1)+0\cdot 0\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=-I,$$

   and $\det (R)=0-(-1)=1$, consistent with $\det (R{)}^2=1$.

   The argument becomes valid for an odd size. For $3\times 3$ real matrices $\det (-I)=(-1{)}^3=-1$, so $\det (A{)}^2=-1$ really is impossible and no real $3\times 3$ matrix satisfies $A^2=-I$.

   (c) Dividing by something that may be zero. The sentence “dividing by $\det (A)$ gives $\det (A{)}^2=1$” presupposes $\det (A)\ne 0$, which is exactly what the argument is trying to prove. Without dividing, $\det (A{)}^3-\det (A)=\det (A)(\det (A{)}^2-1)=0$, so the honest conclusion is

   $$\det (A)\in \{-1,0,1\}.$$

   The value $0$ genuinely occurs: $A=0$ satisfies $A^3=A$ and is not invertible, and so does the projection $A=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$, for which $A^3=A$ and $\det (A)=0$. So “hence $A$ is invertible” is false. What survives is: if $A^3=A$ and $A$ happens to be invertible, then $\det (A)=\pm 1$ — but invertibility must be supplied separately, by the singularity criterion, not deduced from $A^3=A$.

6. 6.Computation · 9 min

   Compute each determinant by elimination. Use row replacements and swaps only, except where a part explicitly asks you to scale a row, and keep a running record in the margin of every factor that must be restored at the end.

   1. (a)$\det \left(\begin{array}{cccc}0& 1& 2& -1\\ 2& 4& 0& 6\\ 1& 3& 1& 2\\ -1& 1& 3& 0\end{array}\right)$, then confirm the value by a cofactor expansion along the first row.
   2. (b)$\det \left(\begin{array}{ccc}2& 4& 6\\ 3& 1& 2\\ 5& 0& 4\end{array}\right)$: begin by factoring the common factor out of the first row, and say exactly where that factor re-enters the answer.
   3. (c)$\det \left(\begin{array}{cccc}1& 2& -1& 3\\ 2& 4& 1& 1\\ -1& -2& 4& -8\\ 3& 6& 0& 5\end{array}\right)$. Explain what happens at the second pivot, and then give a one-line reason, needing no elimination at all, why the answer had to come out as it did.
   Show the solution to question 6

   (a) Swap first, then clear. The $(1,1)$ entry is $0$, so swap rows $1$ and $3$ and record a factor $-1$:

   $$\det =-\det \left(\begin{array}{cccc}1& 3& 1& 2\\ 2& 4& 0& 6\\ 0& 1& 2& -1\\ -1& 1& 3& 0\end{array}\right).$$

   Clear column $1$ with $R_2\leftarrow R_2-2R_1$ and $R_4\leftarrow R_4+R_1$, both free:

   $$\left(\begin{array}{cccc}1& 3& 1& 2\\ 0& -2& -2& 2\\ 0& 1& 2& -1\\ 0& 4& 4& 2\end{array}\right).$$

   The pivot $-2$ is in place. Clear below it with $R_3\leftarrow R_3+\frac{1}{2}{R}_2$, giving $(0,0,1,0)$, and $R_4\leftarrow R_4+2R_2$, giving $(0,0,0,6)$:

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

   The triangular product is $1\cdot (-2)\cdot 1\cdot 6=-12$, and restoring the swap factor gives $\det =-(-12)=12$.

   Cofactor check along the original first row $(0,1,2,-1)$, with signs $+,-,+,-$. The $0$ contributes nothing. Position $(1,2)$: $-1\cdot \det \left(\begin{array}{ccc}2& 0& 6\\ 1& 1& 2\\ -1& 3& 0\end{array}\right)=-[2(0-6)-0+6(3+1)]=-(-12+24)=-12$. Position $(1,3)$: $+2\det \left(\begin{array}{ccc}2& 4& 6\\ 1& 3& 2\\ -1& 1& 0\end{array}\right)=2[2(0-2)-4(0+2)+6(1+3)]=2(-4-8+24)=24$. Position $(1,4)$: $-(-1)\det \left(\begin{array}{ccc}2& 4& 0\\ 1& 3& 1\\ -1& 1& 3\end{array}\right)=+[2(9-1)-4(3+1)+0]=16-16=0$. Total $-12+24+0=12$. ✓

   (b) Factor, then remember the factor. Row $1$ is $2\cdot (1,2,3)$, and scaling a row by $c$ scales the determinant by $c$, so

   $$\det \left(\begin{array}{ccc}2& 4& 6\\ 3& 1& 2\\ 5& 0& 4\end{array}\right)=2\det \left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\\ 5& 0& 4\end{array}\right).$$

   The factor $2$ now sits in front and must multiply whatever the elimination produces. Clear column $1$ with $R_2\leftarrow R_2-3R_1$ and $R_3\leftarrow R_3-5R_1$, giving rows $(0,-5,-7)$ and $(0,-10,-11)$; then $R_3\leftarrow R_3-2R_2$ gives $(0,0,3)$. The triangular product is $1\cdot (-5)\cdot 3=-15$, so the determinant is $2\cdot (-15)=-30$.

   Sanity check by expanding the original along row $1$: $2(4-0)-4(12-10)+6(0-5)=8-8-30=-30$. ✓ Had the factor been forgotten, the answer would have been $-15$: the standard error.

   (c) The second pivot is missing. Clear column $1$ with $R_2\leftarrow R_2-2R_1$, $R_3\leftarrow R_3+R_1$ and $R_4\leftarrow R_4-3R_1$:

   $$\left(\begin{array}{cccc}1& 2& -1& 3\\ 0& 0& 3& -5\\ 0& 0& 3& -5\\ 0& 0& 3& -4\end{array}\right).$$

   Every entry of column $2$ below the first row is $0$, so there is no lower row to swap into the pivot position: column $2$ has no pivot. Going on anyway, $R_3\leftarrow R_3-R_2$ produces a row of zeros, and the determinant is $0$ — the recipe says to stop as soon as that happens.

   The one-line reason: column $2$ of the original matrix is exactly twice column $1$, so the columns are linearly dependent and the determinant is $0$ by inspection (columns, by transpose invariance). No arithmetic was ever needed.

7. 7.Computation · 9 min

   Let $A=\left(\begin{array}{ccc}1& 0& 2\\ 2& 1& 0\\ 0& 3& 1\end{array}\right)$. The matrix is deliberately not symmetric, so a transposition slip will show.

   1. (a)Compute all nine cofactors $C_{ij}$, write down the cofactor matrix $C$, and write down $\mathrm{adj}(A)$.
   2. (b)Compute $\det (A)$ by expanding along row $1$. Then compute the second row of the product $A\mathrm{adj}(A)$ entry by entry, and say which theorem accounts for the diagonal entry and which lemma accounts for the two zeros.
   3. (c)Write down $(A^{-1}{)}_{31}$ and $(A^{-1}{)}_{13}$, stating which cofactor each one comes from.
   4. (d)If $B$ is any $3\times 3$ matrix with $\det (B)=13$, find $\det (\mathrm{adj}(B))$ with justification.
   Show the solution to question 7

   (a) Nine $2\times 2$ determinants, each with its checkerboard sign. Along row $1$:

   $$C_{11}=+\det \left(\begin{array}{cc}1& 0\\ 3& 1\end{array}\right)=1,C_{12}=-\det \left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)=-2,C_{13}=+\det \left(\begin{array}{cc}2& 1\\ 0& 3\end{array}\right)=6.$$

   Along row $2$:

   $$C_{21}=-\det \left(\begin{array}{cc}0& 2\\ 3& 1\end{array}\right)=-(0-6)=6,C_{22}=+\det \left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)=1,C_{23}=-\det \left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)=-3.$$

   Along row $3$:

   $$C_{31}=+\det \left(\begin{array}{cc}0& 2\\ 1& 0\end{array}\right)=-2,C_{32}=-\det \left(\begin{array}{cc}1& 2\\ 2& 0\end{array}\right)=-(0-4)=4,C_{33}=+\det \left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)=1.$$

   So the cofactor matrix and its transpose are

   $$C=\left(\begin{array}{ccc}1& -2& 6\\ 6& 1& -3\\ -2& 4& 1\end{array}\right),\mathrm{adj}(A)=C^T=\left(\begin{array}{ccc}1& 6& -2\\ -2& 1& 4\\ 6& -3& 1\end{array}\right).$$

   (b) Determinant, then the product. Along row $1$: $\det (A)=1\cdot C_{11}+0\cdot C_{12}+2\cdot C_{13}=1+0+12=13$. (Along row $3$ as a check: $0\cdot C_{31}+3\cdot C_{32}+1\cdot C_{33}=12+1=13$. ✓)

   The second row of $A$ is $(2,1,0)$. Dotting it with the three columns of $\mathrm{adj}(A)$:

   $$2\cdot 1+1\cdot (-2)+0\cdot 6=0,2\cdot 6+1\cdot 1+0\cdot (-3)=13,2\cdot (-2)+1\cdot 4+0\cdot 1=0.$$

   So the second row of $A\mathrm{adj}(A)$ is $(0,13,0)=\det (A)\cdot (0,1,0)$, as the adjugate identity promises. The middle entry is ${\sum }_j{a}_{2j}{C}_{2j}$, which is Laplace expansion along row $2$ and therefore equals $\det (A)$. The outer entries are ${\sum }_j{a}_{2j}{C}_{1j}$ and ${\sum }_j{a}_{2j}{C}_{3j}$ — row $2$ against the cofactors of a different row — and the alien cofactor lemma says those sums are $0$.

   (c) Remember the transpose. $A^{-1}=\frac{1}{13}\mathrm{adj}(A)$, and $(\mathrm{adj}A{)}_{ij}=C_{ji}$. Hence

   $$(A^{-1}{)}_{31}=\frac{C_{13}}{13}=\frac{6}{13},(A^{-1}{)}_{13}=\frac{C_{31}}{13}=-\frac{2}{13}.$$

   The indices reverse; using $C_{31}$ for the $(3,1)$ entry would give $-\frac{2}{13}$ in the wrong place. Sanity check: the third row of $A^{-1}$ is $\frac{1}{13}(6,-3,1)$, and against the first column $(1,2,0)$ of $A$ it gives $\frac{6-6+0}{13}=0$, as the $(3,1)$ entry of $A^{-1}A=I$ must. ✓

   (d) Rescale the inverse. Since $\det (B)=13\ne 0$, the adjugate identity gives $\mathrm{adj}(B)=13B^{-1}$. The scalar $13$ multiplies a $3\times 3$ matrix, so

   $$\det (\mathrm{adj}(B))=13^3\det (B^{-1})=13^3\cdot \frac{1}{13}=13^2=169.$$

   In general $\det (\mathrm{adj}B)=\det (B{)}^{n-1}$ for an invertible $n\times n$ matrix. Checking on part (a): $\det (\mathrm{adj}A)$ should be $169$, and expanding $\mathrm{adj}(A)$ along its first row gives $1(1+12)-6(-2-24)+(-2)(6-6)=13+156+0=169$. ✓

8. 8.Computation · 8 min

   Use Cramer's rule where the parts ask for it, writing $A_i(\mathbf{b})$ for the coefficient matrix with its $i$-th column replaced by the right-hand side.

   1. (a)For the system $2x-y+z=5,x+2y-z=-3,3x+y+2z=6$, find $y$ alone by Cramer's rule. Then find $x$ the same way and check both against the first equation.
   2. (b)Consider $kx+y=1,x+ky=1$, where $k$ is a real parameter. For which $k$ does Cramer's rule apply? Solve for $x$ and $y$ in those cases, simplify, and then explain why the simplified formula still may not be used at the excluded values. Finally, decide what actually happens to the system at each excluded value.
   Show the solution to question 8

   (a) Two determinants for one unknown. The coefficient matrix is $A=\left(\begin{array}{ccc}2& -1& 1\\ 1& 2& -1\\ 3& 1& 2\end{array}\right)$ and $\mathbf{b}=(5,-3,6)$. Expanding $\det (A)$ along row $1$:

   $$\det (A)=2(4+1)-(-1)(2+3)+1(1-6)=10+5-5=10.$$

   It is nonzero, so the rule applies. Replace column $2$ by $\mathbf{b}$ and expand along row $1$:

   $$\det (A_2(\mathbf{b}))=\det \left(\begin{array}{ccc}2& 5& 1\\ 1& -3& -1\\ 3& 6& 2\end{array}\right)=2(-6+6)-5(2+3)+1(6+9)=0-25+15=-10.$$

   Hence $y=\frac{-10}{10}=-1$.

   For $x$ replace column $1$:

   $$\det (A_1(\mathbf{b}))=\det \left(\begin{array}{ccc}5& -1& 1\\ -3& 2& -1\\ 6& 1& 2\end{array}\right)=5(4+1)-(-1)(-6+6)+1(-3-12)=25+0-15=10,$$

   so $x=\frac{10}{10}=1$. The first equation then gives $z=5-2x+y=5-2-1=2$, and as a check the second equation reads $1-2-2=-3$ ✓ and the third $3-1+4=6$ ✓.

   (b) The rule needs a nonzero determinant. Here $A=\left(\begin{array}{cc}k& 1\\ 1& k\end{array}\right)$ with $\det (A)=k^2-1$, so Cramer's rule applies exactly when $k\ne \pm 1$. In that case

   $$x=\frac{\det \left(\begin{array}{cc}1& 1\\ 1& k\end{array}\right)}{k^2-1}=\frac{k-1}{k^2-1},y=\frac{\det \left(\begin{array}{cc}k& 1\\ 1& 1\end{array}\right)}{k^2-1}=\frac{k-1}{k^2-1}.$$

   Since $k^2-1=(k-1)(k+1)$ and $k\ne 1$, both simplify to $x=y=\frac{1}{k+1}$.

   The cancellation of $k-1$ was legal only because $k\ne 1$; at $k=1$ the unsimplified expression is $\frac{0}{0}$, which is not a number, and the rule's hypothesis $\det (A)\ne 0$ fails. At $k=-1$ the simplified formula would divide by zero. So the formula $\frac{1}{k+1}$ is valid for every $k$ other than $\pm 1$ and says nothing at either excluded value — Cramer's rule reports only that the solution is not unique there, never which way it fails.

   What actually happens must be read from the equations. At $k=1$ both equations are $x+y=1$: consistent, with infinitely many solutions $(t,1-t)$. At $k=-1$ the equations are $-x+y=1$ and $x-y=1$; adding them gives $0=2$, so the system is inconsistent and has no solution. Same zero determinant, opposite outcomes.

9. 9.Computation · 9 min

   Recognise the structure before you compute. In each part, name the theorem that lets you shortcut the work, and check at least one answer by an independent method.

   1. (a)$\det \left(\begin{array}{ccccc}1& 2& 0& 0& 0\\ 3& 4& 0& 0& 0\\ 5& -7& 2& 1& 0\\ 9& 0& 0& 3& 1\\ 4& 8& 1& 0& 2\end{array}\right)$
   2. (b)$\det \left(\begin{array}{ccc}1& -1& 1\\ 1& 3& 9\\ 1& 4& 16\end{array}\right)$
   3. (c)$\det \left(\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 3& 4\\ 1& 4& 9& 16\\ 1& 8& 27& 64\end{array}\right)$ — the powers here run down the columns, not along the rows; say why the shortcut still applies.
   4. (d)Find every real $a$ for which $\left(\begin{array}{ccc}1& a& a^2\\ 1& -2& 4\\ 1& a+1& (a+1{)}^2\end{array}\right)$ is singular.
   Show the solution to question 9

   (a) Block lower triangular. The upper-right $2\times 3$ block is zero, so the matrix has the form $\left(\begin{array}{cc}A& 0\\ C& D\end{array}\right)$ with $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$ and $D=\left(\begin{array}{ccc}2& 1& 0\\ 0& 3& 1\\ 1& 0& 2\end{array}\right)$. By the block triangular theorem $\det =\det (A)\det (D)$.

   $\det (A)=4-6=-2$. For $D$, expand along row $1$: $2(6-0)-1(0-1)+0=12+1=13$. So $\det =-2\cdot 13=-26$.

   The six entries of the lower-left block, $5,-7,9,0,4,8$, play no part, exactly as the theorem promises.

   (b) Vandermonde with nodes $-1,3,4$. Each row is $(1,x,x^2)$, so by the Vandermonde theorem

   $$\det =(x_2-x_1)(x_3-x_1)(x_3-x_2)=(3+1)(4+1)(4-3)=4\cdot 5\cdot 1=20.$$

   Check by elimination: $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-R_1$ give rows $(0,4,8)$ and $(0,5,15)$; then $R_3\leftarrow R_3-\frac{5}{4}{R}_2$ gives $(0,0,5)$. The triangular product is $1\cdot 4\cdot 5=20$. ✓

   (c) Transpose invariance. Column $j$ is $(1,x_j,x_j^2,x_j^3)$ with $x_j=1,2,3,4$, so the matrix is $V^T$ for the Vandermonde matrix $V$ on those nodes. Since $\det (V^T)=\det (V)$,

   $$\det =\prod _{i<j}(x_j-x_i)=(2-1)(3-1)(4-1)\cdot (3-2)(4-2)\cdot (4-3)=1\cdot 2\cdot 3\cdot 1\cdot 2\cdot 1=12.$$

   Sanity check on the sign: the nodes increase left to right, so every factor $x_j-x_i$ with $i<j$ is positive and the determinant must be positive. ✓

   (d) A Vandermonde with a parameter. The rows are $(1,x,x^2)$ for the nodes $x_1=a$, $x_2=-2$, $x_3=a+1$, so

   $$\det =(x_2-x_1)(x_3-x_1)(x_3-x_2)=(-2-a)((a+1)-a)((a+1)+2)=-(a+2)\cdot 1\cdot (a+3).$$

   This vanishes exactly when $a=-2$ or $a=-3$. The middle factor is the constant $1$: the nodes $a$ and $a+1$ can never coincide, so the only way to repeat a node is for $-2$ to equal one of them.

   Sanity check: at $a=-2$ rows $1$ and $2$ both read $(1,-2,4)$; at $a=-3$ rows $2$ and $3$ both read $(1,-2,4)$. Two equal rows, determinant $0$ by inspection. ✓

10. 10.Proof · 11 min

    Let $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$, where $A$ is $p\times p$ and invertible, $D$ is $q\times q$, and $B$, $C$ have the matching rectangular shapes. The block triangular theorem needs a zero block; this question removes that restriction.

    1. (a)Verify by block multiplication that $M=\left(\begin{array}{cc}A& 0\\ C& I_q\end{array}\right)\left(\begin{array}{cc}{I}_p& A^{-1}B\\ 0& D-C{A}^{-1}B\end{array}\right)$.
    2. (b)Deduce that $\det (M)=\det (A)\det (D-C{A}^{-1}B)$, naming every theorem you use.
    3. (c)Use the formula to compute $\det \left(\begin{array}{cccc}1& 0& 1& 2\\ 0& 1& 3& 1\\ 1& 1& 0& 0\\ 2& 0& 0& 1\end{array}\right)$ with $p=q=2$, and check that the tempting guess $\det (A)\det (D)-\det (B)\det (C)$ gives a different number.
    4. (d)Now suppose $p=q$ and that $A$ commutes with $C$, that is $AC=CA$. Prove that $\det (M)=\det (AD-CB)$.
    Show the solution to question 10

    (a) Multiply the blocks as if they were numbers, keeping their order. Write $S=D-C{A}^{-1}B$ for the lower-right block of the second factor. The product's four blocks are

    $$\left(\begin{array}{cc}A\cdot I_p+0\cdot 0& A\cdot A^{-1}B+0\cdot S\\ C\cdot I_p+I_q\cdot 0& C\cdot A^{-1}B+I_q\cdot S\end{array}\right)=\left(\begin{array}{cc}A& B\\ C& C{A}^{-1}B+D-C{A}^{-1}B\end{array}\right)=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right).$$

    The shapes are compatible throughout ($A^{-1}B$ is $p\times q$, $C{A}^{-1}B$ is $q\times q$), so the factorisation is valid. $■$

    (b) Multiplicativity, then two block triangular determinants. By the multiplicativity theorem, $\det (M)$ is the product of the determinants of the two factors. The first factor has a zero block in its upper-right corner, so by the block triangular theorem its determinant is $\det (A)\det (I_q)=\det (A)$. The second factor has a zero block in its lower-left corner, so its determinant is $\det (I_p)\det (S)=\det (S)$. Hence

    $$\det (M)=\det (A)\det (D-C{A}^{-1}B).■$$

    (c) Read off the blocks. Here $A=I_2$, $B=\left(\begin{array}{cc}1& 2\\ 3& 1\end{array}\right)$, $C=\left(\begin{array}{cc}1& 1\\ 2& 0\end{array}\right)$, $D=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$. Since $A^{-1}=I$,

    $$C{A}^{-1}B=CB=\left(\begin{array}{cc}1+3& 2+1\\ 2+0& 4+0\end{array}\right)=\left(\begin{array}{cc}4& 3\\ 2& 4\end{array}\right),D-CB=\left(\begin{array}{cc}-4& -3\\ -2& -3\end{array}\right),$$

    and $\det (D-CB)=12-6=6$. With $\det (A)=1$ the formula gives $\det (M)=6$.

    Check by elimination on the full matrix: $R_3\leftarrow R_3-R_1$ gives $(0,1,-1,-2)$, $R_4\leftarrow R_4-2R_1$ gives $(0,0,-2,-3)$, then $R_3\leftarrow R_3-R_2$ gives $(0,0,-4,-3)$. Finally $R_4\leftarrow R_4-\frac{1}{2}{R}_3$ gives $(0,0,0,-\frac{3}{2})$. The diagonal is $1,1,-4,-\frac{3}{2}$ with product $6$. ✓

    The false formula: $\det (B)=1-6=-5$, $\det (C)=0-2=-2$, $\det (D)=0$, so $\det (A)\det (D)-\det (B)\det (C)=0-(-5)(-2)=-10\ne 6$.

    (d) Push $\det (A)$ inside. With $p=q$ all four blocks are square of the same size, so $\det (A)\det (S)=\det (AS)$ by multiplicativity. Compute

    $$AS=A(D-C{A}^{-1}B)=AD-AC{A}^{-1}B=AD-CA{A}^{-1}B=AD-CB,$$

    where the third equality replaces $AC$ by $CA$ using the hypothesis, and the last uses $A{A}^{-1}=I$. Therefore, by (b),

    $$\det (M)=\det (A)\det (S)=\det (AS)=\det (AD-CB).■$$

    The commuting hypothesis is exactly what lets $A$ slide past $C$ to cancel $A^{-1}$; without it the expression $AC{A}^{-1}B$ does not simplify and no such formula holds. Note that in part (c), $A=I$ commutes with everything, and indeed $\det (AD-CB)=\det (D-CB)=6$ there. ✓

11. 11.Sketch · 10 min

    Let $T(\mathbf{x})=A\mathbf{x}$ with $A=\left(\begin{array}{cc}2& 1\\ 1& -1\end{array}\right)$, and let $S$ be the unit square with corners $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$. A drawing with no computation beside it earns nothing: every feature drawn must be justified.

    1. (a)Sketch $S$ and its image $T(S)$ on the same axes. Label where each of the four corners goes, and mark $T({\mathbf{e}}_1)$ and $T({\mathbf{e}}_2)$ as arrows from the origin.
    2. (b)Compute $\det (A)$. State the area of $T(S)$ and whether $T$ preserves orientation, and point to the feature of your drawing that shows the sign of the determinant.
    3. (c)Verify the area of $T(S)$ without any determinant, by shearing the parallelogram into a rectangle: replace the second edge by the component of it perpendicular to the first, and say why that does not change the area.
    4. (d)Without drawing, give the area of the image of $S$ under $A^2$ and say whether that image is traversed anticlockwise. Then confirm by computing $A^2$ and its determinant directly.
    Show the solution to question 11

    (a) The columns of $A$ are the images of the basis vectors. $T({\mathbf{e}}_1)$ is the first column, $(2,1)$, and $T({\mathbf{e}}_2)$ is the second, $(1,-1)$. Linearity sends the corner $(1,1)={\mathbf{e}}_1+{\mathbf{e}}_2$ to $(2,1)+(1,-1)=(3,0)$, and the origin stays fixed. So

    $$(0,0)\mapsto (0,0),(1,0)\mapsto (2,1),(1,1)\mapsto (3,0),(0,1)\mapsto (1,-1).$$

    The drawing shows the unit square in the first quadrant and, overlapping it, a parallelogram with vertices $(0,0)$, $(2,1)$, $(3,0)$, $(1,-1)$ — a long thin shape leaning to the right, with one vertex below the $x$-axis and its far vertex on the $x$-axis at $(3,0)$. Two arrows from the origin, to $(2,1)$ and to $(1,-1)$, are its edges. Draw the edges of $T(S)$ as the images of the edges of $S$: the bottom edge of $S$ becomes the arrow to $(2,1)$, the left edge becomes the arrow to $(1,-1)$, and the other two edges are parallel copies of these.

    (b) Determinant, area, orientation. $\det (A)=2\cdot (-1)-1\cdot 1=-3$. By the volume scaling theorem the area of $T(S)$ is $|\det (A)|\cdot \mathrm{area}(S)=3\cdot 1=3$. The determinant is negative, so $T$ reverses orientation.

    The sign is visible in the drawing as the direction of travel. Walking the corners of $S$ in the order $(0,0)\to (1,0)\to (1,1)\to (0,1)$ goes anticlockwise. Their images in the same order, $(0,0)\to (2,1)\to (3,0)\to (1,-1)$, go clockwise: from $(2,1)$ the path turns right to reach $(3,0)$, then right again to $(1,-1)$. Equivalently, the shorter turn from the arrow $T({\mathbf{e}}_1)=(2,1)$ to the arrow $T({\mathbf{e}}_2)=(1,-1)$ is clockwise, whereas from ${\mathbf{e}}_1$ to ${\mathbf{e}}_2$ it was anticlockwise. That reversal is what $\det (A)<0$ records.

    (c) Shear, then measure a rectangle. Let $\mathbf{u}=(2,1)$ and $\mathbf{v}=(1,-1)$. The component of $\mathbf{v}$ along $\mathbf{u}$ has coefficient

    $$c=\frac{\mathbf{u}\cdot \mathbf{v}}{\Vert \mathbf{u}{\Vert }^2}=\frac{2-1}{4+1}=\frac{1}{5},$$

    so the perpendicular component is $\mathbf{w}=\mathbf{v}-\frac{1}{5}\mathbf{u}=(1-\frac{2}{5},-1-\frac{1}{5})=(\frac{3}{5},-\frac{6}{5})$. Check: $\mathbf{u}\cdot \mathbf{w}=\frac{6}{5}-\frac{6}{5}=0$. ✓

    Replacing the edge $\mathbf{v}$ by $\mathbf{v}-c\mathbf{u}$ slides the far edge of the parallelogram parallel to the base $\mathbf{u}$; the base and the perpendicular height are untouched, so the area is unchanged. (In matrix language this is a row replacement, which leaves the determinant alone.) The sheared shape is a rectangle with sides $\Vert \mathbf{u}\Vert =\sqrt{5}$ and $\Vert \mathbf{w}\Vert =\sqrt{\frac{9}{25}+\frac{36}{25}}=\sqrt{\frac{45}{25}}=\frac{3}{\sqrt{5}}$, so

    $$\mathrm{area}=\sqrt{5}\cdot \frac{3}{\sqrt{5}}=3,$$

    in agreement with (b).

    (d) Multiplicativity does the work. $\det (A^2)=\det (A{)}^2=9$, so the image of $S$ under $A^2$ has area $9$, and since $9>0$ orientation is preserved: two reversals make a preservation, and the image is traversed anticlockwise. Directly,

    $$A^2=\left(\begin{array}{cc}2\cdot 2+1\cdot 1& 2\cdot 1+1\cdot (-1)\\ 1\cdot 2+(-1)\cdot 1& 1\cdot 1+(-1)(-1)\end{array}\right)=\left(\begin{array}{cc}5& 1\\ 1& 2\end{array}\right),\det (A^2)=10-1=9.✓$$

12. 12.Proof · 12 min

    You may quote the theorem that the parallelogram spanned by the rows $\mathbf{u},\mathbf{v}$ of a $2\times 2$ matrix has area $|\det |$, and the decomposition $\mathbf{v}=c\mathbf{u}+\mathbf{w}$ with $\mathbf{u}\cdot \mathbf{w}=0$ used in its proof. Everything else must be argued.

    1. (a)Let $P=(p_1,p_2)$, $Q=(q_1,q_2)$ and $R=(r_1,r_2)$ be points in the plane. Prove that the triangle $PQR$ has area $$\frac{1}{2}\left|\det \left(\begin{array}{ccc}1& p_1& p_2\\ 1& q_1& q_2\\ 1& r_1& r_2\end{array}\right)\right|.$$
    2. (b)Deduce that $P$, $Q$, $R$ are collinear if and only if that $3\times 3$ determinant is zero.
    3. (c)Prove that for any $\mathbf{u},\mathbf{v}\in {\mathbb{R}}^2$, $\left|\det \left(\begin{array}{c}\mathbf{u}\\ \mathbf{v}\end{array}\right)\right|\le \Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert$, and determine exactly when equality holds.
    4. (d)Apply (a) to $P=(-1,2)$, $Q=(3,4)$, $R=(1,7)$, check the answer by a method that uses no determinant, and confirm that the inequality in (c) holds for the edge vectors $Q-P$ and $R-P$.
    Show the solution to question 12

    (a) Reduce the $3\times 3$ to the edge-vector $2\times 2$. Subtract row $1$ from row $2$ and from row $3$. These are row replacements, so the determinant is unchanged:

    $$\det \left(\begin{array}{ccc}1& p_1& p_2\\ 1& q_1& q_2\\ 1& r_1& r_2\end{array}\right)=\det \left(\begin{array}{ccc}1& p_1& p_2\\ 0& q_1-p_1& q_2-p_2\\ 0& r_1-p_1& r_2-p_2\end{array}\right).$$

    Expand along the first column, where only the $(1,1)$ entry survives, with sign $+$:

    $$=\det \left(\begin{array}{cc}{q}_1-p_1& q_2-p_2\\ r_1-p_1& r_2-p_2\end{array}\right),$$

    the determinant whose rows are the edge vectors $\mathbf{u}=Q-P$ and $\mathbf{v}=R-P$. By the quoted theorem its absolute value is the area of the parallelogram spanned by $\mathbf{u}$ and $\mathbf{v}$, placed with its corner at $P$. The diagonal from $P$ to $P+\mathbf{u}+\mathbf{v}$ cuts that parallelogram into two congruent triangles, one of which is $PQR$ (its vertices are $P$, $P+\mathbf{u}=Q$, $P+\mathbf{v}=R$). So the triangle has half the area, which is the claim. $■$

    (b) Zero area means a flat triangle. By (a), the determinant is zero exactly when the triangle has area zero. If $P,Q,R$ lie on one line then the edge vectors $\mathbf{u}$ and $\mathbf{v}$ are parallel (or one of them is $0$, if two points coincide), so the rows of the $2\times 2$ matrix are linearly dependent and its determinant is $0$ by inspection. Conversely, if the $2\times 2$ determinant is $0$ then by the singularity criterion its rows are dependent: either one of $\mathbf{u},\mathbf{v}$ is zero, or one is a multiple of the other. In every case $R-P$ is a multiple of $Q-P$ (or $Q=P$ and any three points including a repeated one are collinear), so $R$ lies on the line through $P$ and $Q$. $■$

    (c) The height is at most the slant edge. If $\mathbf{u}=0$ both sides are $0$ and the inequality holds with equality. Otherwise decompose $\mathbf{v}=c\mathbf{u}+\mathbf{w}$ with $\mathbf{u}\cdot \mathbf{w}=0$. From the proof of the area theorem, $|\det |=\Vert \mathbf{u}\Vert \Vert \mathbf{w}\Vert$. Because $\mathbf{u}\perp \mathbf{w}$, expanding $\Vert \mathbf{v}{\Vert }^2=(c\mathbf{u}+\mathbf{w})\cdot (c\mathbf{u}+\mathbf{w})$ kills the cross term and gives

    $$\Vert \mathbf{v}{\Vert }^2=c^2\Vert \mathbf{u}{\Vert }^2+\Vert \mathbf{w}{\Vert }^2\ge \Vert \mathbf{w}{\Vert }^2,$$

    so $\Vert \mathbf{w}\Vert \le \Vert \mathbf{v}\Vert$ and therefore $|\det |=\Vert \mathbf{u}\Vert \Vert \mathbf{w}\Vert \le \Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert$.

    Equality holds exactly when $c^2\Vert \mathbf{u}{\Vert }^2=0$, that is $c=0$ (given $\mathbf{u}\ne 0$), which by the formula $c=\frac{\mathbf{u}\cdot \mathbf{v}}{\Vert \mathbf{u}{\Vert }^2}$ means $\mathbf{u}\cdot \mathbf{v}=0$. Together with the degenerate case: equality if and only if $\mathbf{u}\perp \mathbf{v}$ or one of the vectors is zero. Geometrically, a parallelogram with given side lengths has the largest area when it is a rectangle. $■$

    (d) Numbers. With $P=(-1,2)$, $Q=(3,4)$, $R=(1,7)$, the row replacements of (a) give edge vectors $\mathbf{u}=Q-P=(4,2)$ and $\mathbf{v}=R-P=(2,5)$, so

    $$\det \left(\begin{array}{ccc}1& -1& 2\\ 1& 3& 4\\ 1& 1& 7\end{array}\right)=\det \left(\begin{array}{cc}4& 2\\ 2& 5\end{array}\right)=20-4=16,\mathrm{area}=\frac{1}{2}\cdot 16=8.$$

    Determinant-free check with the bounding box: the vertices lie in the rectangle $[-1,3]\times [2,7]$, of area $4\cdot 5=20$. Cutting off the three right triangles outside $PQR$ — legs $4$ and $2$ along $PQ$, legs $2$ and $3$ along $QR$, legs $2$ and $5$ along $RP$ — removes $4+3+5=12$, leaving $8$. ✓

    For (c): $\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert =\sqrt{20}\sqrt{29}=\sqrt{580}$, and $16\le \sqrt{580}$ because $16^2=256<580$. Strict inequality is expected, since $\mathbf{u}\cdot \mathbf{v}=8+10=18\ne 0$: the edges are not perpendicular. ✓