Determinants

(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 $c$.

(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.)

(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.

(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.)

(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 $A^3=A$.

(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.

(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 $1(1+12)-6(-2-24)+(-2)(6-6)=13+156+0=169$. ✓

(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.

(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. ✓

(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. ✓

(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.