Corrected exercises: determinants, cofactor expansion and row operations (MATH 133)

a) $(-1)^{2+3} = (-1)^{5} = -1$, $(-1)^{3+2} = -1$, $(-1)^{4+1} = -1$ and $(-1)^{4+4} = +1$. On the figure, the cell in row $2$ and column $3$ carries a minus, and so does its mirror image in row $3$ and column $2$: positions symmetric about the diagonal always share their sign, since $i + j = j + i$. The whole diagonal is $+$, because $i + i$ is even. The practical consequence is the one that costs marks: row $2$ of ANY matrix starts with a minus, so the chant $+, -, +$ is only right on rows $1$ and $3$ of a $3 \times 3$. Before expanding along a line, write its signs above it from the checkerboard, never from memory.

b) Delete row $1$ and column $2$: $M_{12} = \det\begin{pmatrix} 0 & 1 \\ 5 & -2 \end{pmatrix} = 0 \times (-2) - 1 \times 5 = -5$. Position $(1,2)$ is odd, so $C_{12} = -M_{12} = 5$. Delete row $2$ and column $1$: $M_{21} = \det\begin{pmatrix} -1 & 3 \\ 0 & -2 \end{pmatrix} = (-1)(-2) - 3 \times 0 = 2$, and $C_{21} = -2$. The minor and the cofactor are never two different computations, they are one determinant and one sign. But a question that asks for a cofactor and gets the minor back is marked wrong, and so is every expansion built on it.

c) Along row $1$, signs $+, -, +$: $C_{11} = +\det\begin{pmatrix} 4 & 1 \\ 0 & -2 \end{pmatrix} = -8$, $C_{12} = 5$ from b), $C_{13} = +\det\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} = 0 - 20 = -20$. Then $\det A = 2(-8) + (-1)(5) + 3(-20) = -16 - 5 - 60 = -81$. Write the three products with their brackets before adding: the entry $-1$ and the cofactor $5$ carry one sign each, and doing both in your head is where the slip happens.

d) Column $2$ holds $-1$, $4$, $0$ with signs $-, +, -$. The zero kills its whole term, so only two $2 \times 2$ determinants remain: $C_{22} = +\det\begin{pmatrix} 2 & 3 \\ 5 & -2 \end{pmatrix} = -4 - 15 = -19$, and $\det A = (-1)C_{12} + 4C_{22} + 0 = (-1)(5) + 4(-19) = -5 - 76 = -81$. Same value, as the expansion theorem guarantees: every row and every column gives the same determinant. Row $2$, row $3$ and column $1$ also contain exactly one zero and would cost the same two $2 \times 2$ determinants. Row $1$ and column $3$ contain none and cost three. On a $3 \times 3$ the saving is one small determinant; on a $4 \times 4$ it is a whole $3 \times 3$, which is the point of the next exercise.

e) He gets $2(-8) + (-1)(-5) + 3(-20) = -16 + 5 - 60 = -71$, which is $10$ above the true $-81$. The first and third terms are right by luck: positions $(1,1)$ and $(1,3)$ carry a plus, so there the minor IS the cofactor. The middle term is wrong: it should be $(-1) \times C_{12} = (-1)(+5) = -5$, and he wrote $(-1) \times M_{12} = +5$. The gap is exactly twice that term, $2 \times 5 = 10$. The error is invisible from inside the computation, which is why the only reliable check is a second expansion along another line, as in d): two different answers mean a sign was lost somewhere.

a) Rows $1$ to $4$ contain $1$, $2$, $1$ and $2$ zeros. Columns $1$ to $4$ contain $1$, $3$, $1$ and $1$. Column $2$ has a single non-zero entry, the $5$ in position $(3,2)$, so its expansion has ONE term. Row $1$ has three non-zero entries, so it would need three $3 \times 3$ determinants, that is up to nine $2 \times 2$ ones. Rows $2$ and $4$, with two zeros each, would need two. The count takes ten seconds and decides whether the question takes three minutes or fifteen.

b) Position $(3,2)$: $3 + 2 = 5$ is odd, so $C_{32} = -M_{32}$. Hence $\det A = 0 \cdot C_{12} + 0 \cdot C_{22} + 5 \, C_{32} + 0 \cdot C_{42} = -5 \, M_{32}$, where $M_{32}$ is the determinant of $A$ with row $3$ and column $2$ removed: $M_{32} = \det\begin{pmatrix} 3 & 2 & -1 \\ 1 & 0 & 4 \\ 0 & 3 & 2 \end{pmatrix}$. Write the minus sign on this line, next to the $5$, before touching the $3 \times 3$: it is the sign most often lost, because the entry $5$ is positive and nothing in it warns you.

c) Row $2$ of the $3 \times 3$ is $(1, 0, 4)$, with signs $-, +, -$. $C_{21} = -\det\begin{pmatrix} 2 & -1 \\ 3 & 2 \end{pmatrix} = -(4 + 3) = -7$ and $C_{23} = -\det\begin{pmatrix} 3 & 2 \\ 0 & 3 \end{pmatrix} = -(9 - 0) = -9$. So $M_{32} = 1(-7) + 0 + 4(-9) = -7 - 36 = -43$ and $\det A = -5 \times (-43) = 215$. Two minus signs meet here, one from the checkerboard at $(3,2)$ and one from the value of the minor. Dropping either gives $-215$, and the marker sees a right method with a wrong sign, which usually costs half the question.

d) Column $1$ of the $3 \times 3$ is $(3, 1, 0)$, signs $+, -, +$: $3\det\begin{pmatrix} 0 & 4 \\ 3 & 2 \end{pmatrix} - 1\det\begin{pmatrix} 2 & -1 \\ 3 & 2 \end{pmatrix} + 0 = 3(0 - 12) - (4 + 3) = -36 - 7 = -43$. The two expansions agree. On an exam this check costs two $2 \times 2$ determinants, one of which you already computed in c), and it catches the sign errors that no amount of re-reading finds.

e) The determinant is $0$. Expand along the zero column: every term is $0 \times C_{i3}$, so the sum is $0$, whatever the other twelve entries are. The same argument applies to a row of zeros. It is the first case of a general fact that the rest of the chapter will use constantly: as soon as row operations produce a zero row, the determinant is known to be $0$ and the computation stops.

a) $\det T = 2 \times (-1) \times 3 \times (-2) = 12$. Justification: expand along column $1$, whose only non-zero entry is the $2$ at position $(1,1)$, sign $+$. That gives $2\det\begin{pmatrix} -1 & 4 & 8 \\ 0 & 3 & 6 \\ 0 & 0 & -2 \end{pmatrix}$, and the new matrix is again triangular. Expanding along its first column gives $2 \times (-1) \times \det\begin{pmatrix} 3 & 6 \\ 0 & -2 \end{pmatrix} = 2 \times (-1) \times (-6) = 12$. The entries above the diagonal, $7$, $-3$, $5$, $4$, $8$, $6$, never enter the answer: each time they sit in a row that is deleted before they could be used.

b) $R_2 \to R_2 - 2R_1$ gives $(0, 0, 3, 3)$. $R_3 \to R_3 + R_1$ gives $(0, 2, 2, 2)$. $R_4 \to R_4 - 3R_1$ gives $(0, -2, 4, 5)$. Each is a replacement, a row minus a multiple of ANOTHER row, so the determinant does not move: $\det B = \det\begin{pmatrix} 1 & 2 & -1 & 0 \\ 0 & 0 & 3 & 3 \\ 0 & 2 & 2 & 2 \\ 0 & -2 & 4 & 5 \end{pmatrix}$. Write the equality sign between the two determinants on the copy. A chain of matrices joined by arrows, with no equality and no factor, is exactly what the marker cannot give method marks for.

c) A pivot is used to clear the entries below it, by subtracting multiples of its row. A $0$ cannot clear a $2$ or a $-2$, whatever multiple you take. So the rows are exchanged: $R_2 \leftrightarrow R_3$, and the ledger records $\times(-1)$. Now $\det B = -\det\begin{pmatrix} 1 & 2 & -1 & 0 \\ 0 & 2 & 2 & 2 \\ 0 & 0 & 3 & 3 \\ 0 & -2 & 4 & 5 \end{pmatrix}$. A swap is never the mistake; forgetting to record it is, and it produces $+6$ instead of $-6$ with every line of arithmetic correct.

d) $R_4 \to R_4 + R_2$ gives $(0, 0, 6, 7)$, then $R_4 \to R_4 - 2R_3$ gives $(0, 0, 0, 1)$. Both are replacements. The matrix is now upper triangular with diagonal $1, 2, 3, 1$, whose product is $6$. With the single swap of c), $\det B = -6$. The whole computation used integers only because the pivot $2$ divides the entry $-2$ below it and the pivot $3$ divides the $6$: choosing the swap that brings a convenient pivot up is part of the method, not luck.

e) For $n = 4$ the full expansion has $4! = 24$ products of four entries, about $70$ multiplications; the reduction above took about twenty. The gap then explodes: at $n = 10$, $10! = 3\,628\,800$ products, while elimination needs roughly $n^{3}/3$, a few hundred operations. That is why a computer never expands a large determinant along a row, and why on paper you expand only along a line that the zeros have already made cheap.

a) Row $1$ is $4(1, -2, 3)$ and row $3$ is $5(2, 1, 3)$. Pulling each factor out multiplies the outside by it: $\det M = 4 \times 5 \times \det\begin{pmatrix} 1 & -2 & 3 \\ 3 & 1 & -2 \\ 2 & 1 & 3 \end{pmatrix} = 20\det M'$. Write the $20$ on the same line as the new determinant and keep it there until the last line. Pulling $4$ out of row $1$ does not allow pulling a second $4$ out of another row unless that row also has it: each factor belongs to one row.

b) In $M'$, $R_2 \to R_2 - 3R_1$ gives $(0, 7, -11)$ and $R_3 \to R_3 - 2R_1$ gives $(0, 5, -3)$: two replacements, no change. Expanding along column $1$, whose only non-zero entry is the pivot $1$: $\det M' = 1 \times \det\begin{pmatrix} 7 & -11 \\ 5 & -3 \end{pmatrix} = -21 + 55 = 34$. So $\det M = 20 \times 34 = 680$. The factor $20$ was the whole difficulty; the arithmetic after it was on small integers precisely because the factors were removed first.

c) $R_2 \to 4R_2 - 3R_1$ is NOT a replacement. A replacement changes a row by adding a multiple of another row to it, and keeps the row itself with coefficient $1$. Here $R_2$ is first multiplied by $4$, which multiplies the determinant by $4$, and only then is $-3R_1$ added, which changes nothing. Likewise $R_3 \to 2R_3 - 5R_1$ multiplies it by $2$. His matrix is $\begin{pmatrix} 4 & -8 & 12 \\ 0 & 28 & -44 \\ 0 & 50 & -30 \end{pmatrix}$, whose determinant is $4(28 \times (-30) - (-44) \times 50) = 4 \times 1360 = 5440$, correctly computed, and equal to $8\det M$. So $\det M = 5440 / 8 = 680$. The fraction-free trick is perfectly legitimate, provided the ledger says $\times 4$ and $\times 2$ on the lines where it is used.

d) The columns of $K$ are $3(1, 2, 3)$, $5(2, -1, 3)$ and $(1, 2, 4)$. Since $\det(K^{T}) = \det K$ and the columns of $K$ are the rows of $K^{T}$, every row rule is also a column rule: $\det K = 15\det\begin{pmatrix} 1 & 2 & 1 \\ 2 & -1 & 2 \\ 3 & 3 & 4 \end{pmatrix}$. In this matrix $R_2 \to R_2 - 2R_1$ gives $(0, -5, 0)$ and $R_3 \to R_3 - 3R_1$ gives $(0, -3, 1)$; expanding along column $1$, $\det = \det\begin{pmatrix} -5 & 0 \\ -3 & 1 \end{pmatrix} = -5$. Hence $\det K = 15 \times (-5) = -75$. Rows are not special: a factor, a swap or a replacement on columns affects the determinant exactly as it does on rows.

a) This is $A$ with rows $1$ and $2$ exchanged: one swap, $\times(-1)$, so the determinant is $-4$. Nothing else changed, and nothing else needs to be said.

b) Row $2$ has been multiplied by $3$: $\times 3$, so the determinant is $3 \times 4 = 12$. The factor $3$ appears once, because it multiplies ONE row. The answer $3^{3} \times 4 = 108$ confuses this matrix with $3A$, where every row is multiplied by $3$.

c) Row $2$ has been replaced by $R_2 + 2R_1$: a replacement, so the determinant is still $4$. The row that changed kept its coefficient $1$, which is the test that separates a replacement from part b). Writing $2 \times 4 = 8$ is the typical slip: the $2$ multiplies the OTHER row, and adding a multiple of another row never scales anything.

d) Build it from $A$ in three steps. Swap $R_1 \leftrightarrow R_3$: the rows are now $(g, h, i)$, $(d, e, f)$, $(a, b, c)$ and the ledger reads $\times(-1)$, so $-4$. Multiply the new $R_2$ by $2$: $\times 2$, so $-8$. Replace $R_1 \to R_1 - 3R_3$, which turns $(g, h, i)$ into $(g - 3a, h - 3b, i - 3c)$ since $R_3$ is now $(a, b, c)$: replacement, $\times 1$. The determinant is $-8$. The order of the chain matters only for the bookkeeping, not for the result: any order of the same operations gives $(-1) \times 2 \times 1 \times 4$. What matters is that each operation is named and each factor is written.

e) $2A$ multiplies EVERY row by $2$: three rows, so $\det(2A) = 2^{3} \times 4 = 32$, never $2 \times 4$. Likewise $\det(-A) = (-1)^{3} \times 4 = -4$. Transposing changes nothing, $\det(A^{T}) = \det A = 4$, so $\det(2A^{T}) = 2^{3} \times 4 = 32$. The general rule is $\det(kA) = k^{n}\det A$ for an $n \times n$ matrix, and the exponent is the size: for a $4 \times 4$, $\det(-A) = (-1)^{4}\det A = +4$. Negating a matrix changes the sign of its determinant only in odd size, a fact that is asked every year as a true or false question.

a) $R_1 \to R_1 + R_2 + R_3$ is two replacements in a row, so the determinant does not change. The new first row is $(x + 2, x + 2, x + 2)$, because each column of $M(x)$ contains one $x$ and two $1$s. Pulling the common factor out of that row gives $d(x) = (x + 2)\det\begin{pmatrix} 1 & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{pmatrix}$. This is the gesture to look for on any matrix whose rows or columns all have the same sum: summing them first extracts that sum as a factor.

b) In the remaining determinant, $R_2 \to R_2 - R_1$ gives $(0, x - 1, 0)$ and $R_3 \to R_3 - R_1$ gives $(0, 0, x - 1)$. The matrix is upper triangular with diagonal $1$, $x - 1$, $x - 1$. Hence $d(x) = (x + 2)(x - 1)^{2}$. No cubic was ever solved: the row operations delivered the factors one by one, and each factor came with its reason.

c) Along row $1$, signs $+, -, +$: $d(x) = x\det\begin{pmatrix} x & 1 \\ 1 & x \end{pmatrix} - 1\det\begin{pmatrix} 1 & 1 \\ 1 & x \end{pmatrix} + 1\det\begin{pmatrix} 1 & x \\ 1 & 1 \end{pmatrix}$ $= x(x^{2} - 1) - (x - 1) + (1 - x) = x^{3} - 3x + 2$. And $(x + 2)(x - 1)^{2} = (x + 2)(x^{2} - 2x + 1) = x^{3} - 3x + 2$. The two routes agree. A quick numerical check is also available: at $x = 0$, $d(0) = 2$ and $(0 + 2)(0 - 1)^{2} = 2$.

d) $d(x) = 0$ exactly when $x = 1$ or $x = -2$. At $x = 1$ the three rows are all $(1, 1, 1)$, so $R_2 \to R_2 - R_1$ gives a zero row. At $x = -2$ the rows are $(-2, 1, 1)$, $(1, -2, 1)$, $(1, 1, -2)$, which add up to $(0, 0, 0)$, so $R_1 \to R_1 + R_2 + R_3$ gives a zero row. In both cases the determinant is $0$ because a replacement, which does not change it, produced a row of zeros. The question said every value: listing $1$ and $-2$ without saying that $d$ is non-zero everywhere else is half an answer, and the factored form is the proof that there is nothing else.

e) Near $x = -2$ the factor $x + 2$ changes sign while $(x - 1)^{2}$ stays positive, so $d$ changes sign and the graph crosses. Near $x = 1$, $(x - 1)^{2}$ is never negative, so $d$ keeps the sign of $x + 2 = 3 > 0$ on both sides and the graph only touches. The student's division is harmless: $x^{3} - 3x + 2 = (x - 1)(x^{2} + x - 2)$ and $x^{2} + x - 2 = (x - 1)(x + 2)$ still has $1$ as a root, so he finds $1$ twice and $-2$. The real danger is the other slip, CANCELLING $x - 1$ from an equation such as $(x - 1)^{2}(x + 2) = 0$ as if it were non-zero, which deletes the root $x = 1$ altogether.