Corrected exercises: determinants, inverses and Cramer's rule (MATH 133)

$\det(AB) = \det A \, \det B = 2 \times (-3) = -6$, and $\det(BA) = \det B \, \det A = -6$ as well, because the two determinants are real NUMBERS and numbers commute even when matrices do not. Nothing follows about $AB$ and $BA$ themselves: two different matrices can share a determinant exactly as two different parallelograms can share an area. Writing $AB = BA$ here costs the part. The determinant forgets almost everything about a matrix and keeps one number, which is precisely why it is so easy to compute with and so dangerous to over-read.

Take the factors one at a time and write one line for each. $\det(A^{T}) = \det A = 2$: transposing swaps rows and columns, and expanding $A^{T}$ along a row is expanding $A$ along a column. $\det(B^{-1}) = \dfrac{1}{\det B} = -\dfrac{1}{3}$: from $BB^{-1} = I$ you get $\det B \, \det(B^{-1}) = \det I = 1$. So $\det(A^{T}B^{-1}) = 2 \times \left(-\dfrac{1}{3}\right) = -\dfrac{2}{3}$. The inverse DIVIDES by $\det B$, it does not change its sign: writing $\det(B^{-1}) = 3$ is the second most common slip of the chapter.

One line per factor before multiplying, it is what earns the method marks. The scalar $2$ multiplies a $3 \times 3$ matrix, so it contributes $2^{3} = 8$, not $2$. Then $\det(A^{-1}) = \dfrac{1}{2}$, $\det(B^{T}) = -3$ and $\det(A^{3}) = (\det A)^{3} = 8$. Multiply: $8 \times \dfrac{1}{2} \times (-3) \times 8 = -96$. Two traps sit in this single line. Writing $2$ instead of $2^{3}$ gives $-24$. Reading $A^{3}$ as $3A$ gives $\det(3A) = 27 \times 2 = 54$ instead of $8$. Notice also what you did NOT need: the order of the factors. The matrix $A^{-1}B^{T}A^{3}$ cannot be simplified to $B^{T}A^{2}$, since the factors do not commute, but its determinant is a product of four numbers and those do.

$\det(B^{-1}AB) = \det(B^{-1}) \, \det A \, \det B = \dfrac{1}{\det B} \times 2 \times \det B = 2$. The matrices $B^{-1}$ and $B$ cannot be cancelled inside the product, because $B^{-1}AB$ is in general a different matrix from $A$; their DETERMINANTS are numbers and do cancel. So every matrix of the form $B^{-1}AB$ has the determinant of $A$, whichever invertible $B$ is chosen. Geometrically, $B^{-1}AB$ is the transformation $A$ described in another coordinate system, and changing the coordinates does not change the factor by which volumes are scaled.

$\det(3A) = 3^{3} \times 2 = 54$, so $\det\big((3A)^{-1}\big) = \dfrac{1}{54}$. On the other side, $\det(3A^{-1}) = 3^{3} \times \det(A^{-1}) = 27 \times \dfrac{1}{2} = \dfrac{27}{2}$. The two expressions look alike and differ by a factor of $729 = 27^{2}$. In $(3A)^{-1}$ the scalar is INSIDE the inverse and gets inverted with the matrix, since $(3A)^{-1} = \dfrac{1}{3}A^{-1}$; in $3A^{-1}$ it is outside and is not. Reading the parentheses is part of the question, and a marker reads them too.

For a $3 \times 3$ matrix, $\det(2A) = 2^{3}\det A = 8 \times 5 = 40$. For $-A$ the scalar is $-1$, so $\det(-A) = (-1)^{3} \times 5 = -5$. For $-2A^{T}$, transposing changes nothing and the scalar $-2$ contributes $(-2)^{3} = -8$, so $\det(-2A^{T}) = -8 \times 5 = -40$. The answer $\det(2A) = 10$, that is $2\det A$, is the single most common wrong line on this topic: it treats the scalar as if it multiplied one row only.

With $n = 4$: $\det(2A) = 2^{4} \times 5 = 80$, $\det(-A) = (-1)^{4} \times 5 = 5$ and $\det(-2A^{T}) = (-2)^{4} \times 5 = 80$. The determinant of $A$ is still $5$ and yet every answer changed, because the size of the matrix is a hidden input of every scalar rule. The sign of $\det(-A)$ is the sharpest symptom: negating a matrix reverses the sign of its determinant when $n$ is odd and leaves it alone when $n$ is even. So never apply a scalar rule before writing $n$ on the line.

$\det(2B^{-1}) = 2^{3}\det(B^{-1}) = \dfrac{8}{\det B}$. Setting this equal to $4$ gives $\det B = 2$. Then $\det(2B) = 8 \times 2 = 16$ and its inverse has determinant $\dfrac{1}{16}$. A student who forgets the cube writes $\dfrac{2}{\det B} = 4$ and finds $\det B = \dfrac{1}{2}$, wrong by a factor of $4$, and the second answer inherits the error. Substituting back is a five-second check: $8/2 = 4$.

Multiplying ONE row by $2$ is an elementary row operation and it multiplies the determinant by $2$: $\det E = 2 \times (-4) = -8$. Multiplying the whole matrix multiplies all three rows: $\det(2D) = 2^{3} \times (-4) = -32$. Both operations are described as multiplying by $2$, and they differ by a factor of $2^{2} = 4$, one factor of $2$ for each row the row operation did not touch. In the language of the figure: stretching a cube in one direction doubles its volume, stretching it in all three directions multiplies it by $8$.

$kA = (kI)A$, where $kI$ is the diagonal matrix with $k$ in every diagonal position. The product rule gives $\det(kA) = \det(kI)\,\det A$, and the determinant of a diagonal matrix is the product of its diagonal entries, so $\det(kI) = k \cdot k \cdots k = k^{n}$. Hence $\det(kA) = k^{n}\det A$. The proof is worth keeping because it also says what $n$ is: the number of diagonal entries of $kI$, that is the SIZE of the matrix, never the number of times $k$ is written in the expression.

Rows $1$ and $3$ share their first two entries, so $R_3 \to R_3 - R_1$ creates two zeros, and adding a multiple of one row to another leaves the determinant unchanged. The new third row is $(0, 0, 2 - k)$. Expand along that row, where only the last entry survives, with sign $(-1)^{3+3} = +1$: $\det(A_k) = (2 - k)\det\begin{pmatrix} 1 & 1 \\ -1 & k \end{pmatrix} = (2 - k)(k + 1)$. Keeping the factored form is the whole point: expanding to $-k^{2} + k + 2$ and factoring again wastes a line and invites a sign error. Check with $k = 0$ on the ORIGINAL matrix, expanded along its first row: $1(0 \cdot 2 - 1 \cdot 1) - 1\big((-1) \cdot 2 - 1 \cdot 1\big) + 0 = -1 + 3 = 2$, and $(2 - 0)(0 + 1) = 2$.

$A_k$ is invertible if and only if $\det(A_k) \neq 0$, that is $k \neq 2$ and $k \neq -1$. At those two values the columns are dependent and the matrix is singular; at $k = 2$ you can even see it, since rows $1$ and $3$ both become $(1, 1, 2)$. State the answer as a condition on $k$, every real $k$ except $2$ and $-1$. Listing the two bad values without saying what they mean loses the conclusion mark, because the question asked where $A_k$ IS invertible.

$\det(A_k^{T}A_k) = \det(A_k^{T})\,\det(A_k) = \det(A_k)^{2}$, so the condition is $\det(A_k)^{2} = 16$, that is $\det(A_k) = 4$ OR $\det(A_k) = -4$. First branch: $(2 - k)(k + 1) = 4$ becomes $k^{2} - k + 2 = 0$, whose discriminant $1 - 8 = -7$ is negative, so no real $k$. Second branch: $(2 - k)(k + 1) = -4$ becomes $k^{2} - k - 6 = 0$, that is $(k - 3)(k + 2) = 0$, so $k = 3$ or $k = -2$. The classic loss is to keep only $\det(A_k) = 4$, find no solution and conclude that no $k$ works, the exact opposite of the truth. The figure of the solution shows both branches: the parabola never reaches height $4$ and crosses height $-4$ twice.

For $k = 0$, $\det(A_0) = 2 \neq 0$, so $A_0^{-1}$ exists, and $\det(A_0^{-1}) = \dfrac{1}{2}$. Then $\det(2A_0^{-1}A_0^{T}) = 2^{3} \times \dfrac{1}{2} \times 2 = 8$: the inverse and the transpose cancel in the determinant, and only the $2^{3}$ of the scalar is left. The sentence that says $k = 0$ is not a forbidden value comes BEFORE any use of $A_0^{-1}$; without it, the inverse is used before it is known to exist.

For $k = 2$, $\det(A_2) = 0$. For any $3 \times 3$ matrix $B$, $\det(A_2B) = \det(A_2)\,\det B = 0 \times \det B = 0$, so $A_2B$ is singular however well behaved $B$ is, and the same argument covers $BA_2$ and $A_2^{5}$. In words: once one factor has flattened space onto a plane, nothing applied before or after it can restore the lost dimension, and the determinant, being a product, records this as a factor of zero.

Row $1$, $(1, 1, 0)$, has a zero, so its third term vanishes: $\det A = 1 \cdot C_{11} + 1 \cdot C_{12} + 0$. With $C_{11} = +\det\begin{pmatrix} 0 & -1 \\ 2 & -1 \end{pmatrix} = 0 + 2 = 2$ and $C_{12} = -\det\begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix} = -(-1 + 1) = 0$, we get $\det A = 2$. Choosing the row with a zero saves one $2 \times 2$ determinant; here it also produces two cofactors needed in b), so nothing is computed twice. The determinant is not zero, so $A$ is invertible and the adjugate formula applies.

The signs follow the checkerboard $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$, and each minor deletes the row and column of its entry. Row $1$: $C_{11} = 2$, $C_{12} = 0$, $C_{13} = +\det\begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} = 2$. Row $2$: $C_{21} = -\det\begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix} = 1$, $C_{22} = +\det\begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} = -1$, $C_{23} = -\det\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = -1$. Row $3$: $C_{31} = +\det\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = -1$, $C_{32} = -\det\begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} = 1$, $C_{33} = +\det\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = -1$. The matrix of cofactors is $\begin{pmatrix} 2 & 0 & 2 \\ 1 & -1 & -1 \\ -1 & 1 & -1 \end{pmatrix}$. The minus signs of the checkerboard are where the marks go: $C_{21}$ is MINUS a determinant that equals $-1$.

Transpose: row $i$ of $\text{adj}(A)$ is column $i$ of the cofactor matrix, $\text{adj}(A) = \begin{pmatrix} 2 & 1 & -1 \\ 0 & -1 & 1 \\ 2 & -1 & -1 \end{pmatrix}$. Row $1$ of $A$, $(1, 1, 0)$, against the three columns of $\text{adj}(A)$: $2 + 0 + 0 = 2$, $1 - 1 + 0 = 0$, $-1 + 1 + 0 = 0$. That is $(2, 0, 0)$, the first row of $2I$. The zeros are not luck: a row of $A$ against the cofactors of ANOTHER row is the expansion of a matrix with two equal rows, hence zero. This check costs three dot products and catches almost every sign error.

$A^{-1} = \dfrac{1}{2}\,\text{adj}(A) = \begin{pmatrix} 1 & \frac{1}{2} & -\frac{1}{2} \\ 0 & -\frac{1}{2} & \frac{1}{2} \\ 1 & -\frac{1}{2} & -\frac{1}{2} \end{pmatrix}$. Every entry has denominator $2 = \det A$ or a divisor of it, which is the adjugate formula made visible: the only division in the whole computation is by the determinant. On a copy, leave the answer as $\frac{1}{2}$ times an integer matrix if the question allows it; it is shorter and easier to check.

With the matrix of cofactors $C$ instead of its transpose, entry $(1, 1)$ of $A \cdot \dfrac{1}{2}C$ is row $1$ of $A$ against column $1$ of $C$: $\dfrac{1}{2}(1 \cdot 2 + 1 \cdot 1 + 0 \cdot (-1)) = \dfrac{3}{2}$, not $1$. One wrong entry is enough to reject a proposed inverse. The transpose is not a convention: the product $A\,\text{adj}(A)$ pairs row $i$ of $A$ with the cofactors of row $i$, which is the expansion of $\det A$ along row $i$, and that pairing only happens when those cofactors sit in COLUMN $i$ of the adjugate.

Expand along row $3$, $(1, 0, 1)$, which has a zero. The signs in row $3$ are $+, -, +$: $\det M = +1 \cdot \det\begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix} + 0 + 1 \cdot \det\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} = (2 + 3) + (6 - 1) = 10$. Since $\det M \neq 0$, $M$ is invertible and Cramer's rule applies: say it in one sentence, it is the hypothesis of everything that follows.

The entry in row $2$, column $3$ of $M^{-1}$ is the cofactor of position $(3, 2)$ of $M$, divided by $\det M$. Delete row $3$ and column $2$: the minor is $\det\begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix} = 4 + 1 = 5$, and the sign is $(-1)^{3+2} = -1$, so $C_{32} = -5$ and $(M^{-1})_{23} = -\dfrac{5}{10} = -\dfrac{1}{2}$. The trap is to take $C_{23}$ instead: deleting row $2$ and column $3$ gives $\det\begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix} = -1$, so $C_{23} = 1$ and the wrong answer $\dfrac{1}{10}$. The two differ here, and in general they differ whenever $M$ is not symmetric. The swap comes from the transpose in $\text{adj}(M)$, nothing else.

Replace column $2$ by $\mathbf{b}$: $\det\begin{pmatrix} 2 & -3 & -1 \\ 1 & 1 & 2 \\ 1 & 4 & 1 \end{pmatrix}$, expanded along row $1$, is $2(1 - 8) + 3(1 - 2) - 1(4 - 1) = -14 - 3 - 3 = -20$, so $y = \dfrac{-20}{10} = -2$. Replace column $3$ by $\mathbf{b}$: $\det\begin{pmatrix} 2 & 1 & -3 \\ 1 & 3 & 1 \\ 1 & 0 & 4 \end{pmatrix}$, expanded along row $3$, is $1 \cdot (1 + 9) + 4 \cdot (6 - 1) = 30$, so $z = 3$. The third equation, $x + z = 4$, then gives $x = 1$, and all three equations check: $2 - 2 - 3 = -3$, $1 - 6 + 6 = 1$, $1 + 3 = 4$. The column that is replaced is the column of the unknown you want, and $\mathbf{b}$ goes in as a COLUMN, never as a row.