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Properties of determinants: row operations, products, transposes

Cofactor expansion of a large determinant takes longer than anyone can wait. Row reduction is fast, but each row operation changes the matrix. What does each one do to the determinant, and what else follows?

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Augmented matrices and row operationsPrecalculusDeterminants by cofactor expansionLinear Algebra
Where you are 12 in Matrices
Linear algebra6
Matrices12
▴ 7 aboveWriting a system as a matrix equationInverse of a 2 by 2 matrix, and solving with itCramer's rule
Properties of determinants: row operations, products, transposes
Finding an inverse by row reducing A next to I

Why this exists

A full Determinants by cofactor expansion has n! products: 24 for n = 4, about 3.6 million for n = 10. Row reduction (Augmented matrices and row operations) reaches a triangular matrix, whose determinant is the product of its diagonal, in about n^3/3 multiplications. That works only if we know what each row operation does to the determinant. The same knowledge settles two open questions: is \det A = 0 exactly when A has no inverse, and is \det(AB) = \det A \det B for every size?

Definition

\det A is the number defined by cofactor expansion, which by the theorem of Determinants by cofactor expansion can run along any row or any column. The three row operations are a swap (R_i \leftrightarrow R_j), a scaling (R_i \to cR_i with c \neq 0), and a replacement (R_j \to R_j + kR_i, i \neq j). An elementary matrix E is I with one row operation done to it, and doing that operation to A gives EA (Finding an inverse by row reducing A next to I). A column operation is the same three moves applied to columns.

Intuition

For a 2 \times 2 matrix, ad - bc is the signed area of the parallelogram on its rows, as What the determinant means reads it on the columns. Swapping the rows flips orientation and sign. Scaling one row by c stretches one side by c, so the area scales by c. Adding a multiple of row 1 to row 2 slides the far side along a line parallel to row 1: same base, same height, same area.

\mathbf{r}_1
\mathbf{r}_2
\mathbf{r}_2 + \mathbf{r}_1
\text{both have area } 7
Figure 1Adding row 1 to row 2 slides the top side along a line parallel to the base. Base and height are unchanged, so the area is too.

With rows (4, 1) and (1, 2) the area is 8 - 1 = 7; with rows (4, 1) and (5, 3) it is 12 - 5 = 7.

Results

Day 1: row operations and invertibility

Theorem 1 (row operations). Let B come from the n \times n matrix A by one row operation.

  • A swap of two rows: \det B = -\det A.
  • Multiplying one row by c: \det B = c \det A (true for every number c, even 0).
  • Adding k times one row to a different row: \det B = \det A.

So a matrix with two equal rows, or with a row of zeros, has determinant 0. Read backward, the scaling rule says a common factor of one row comes out in front: \det \begin{bmatrix} 2 & 6 \\ 1 & 5 \end{bmatrix} = 2\det \begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix}.

Lemma (elementary matrices). A swap matrix has determinant -1, a scaling matrix c, a replacement matrix 1, and for every elementary E, \det(EA) = \det E \det A.

Theorem 2 (computing by reduction). If A reaches an echelon form U by replacements and r swaps, then

\det A = (-1)^r \, u_{11}u_{22}\cdots u_{nn}

Reducing column k takes n - k multipliers and (n - k)^2 multiply-and-subtract steps, so the total is \displaystyle 1^2 + 2^2 + \dots + (n-1)^2 = \frac{(n-1)n(2n-1)}{6}, about n^3/3: 14 for n = 4 and 2{,}470 for n = 20, against 4! = 24 and 20! \approx 2.4 \times 10^{18} products by full expansion.

Theorem 3 (invertibility). A square matrix A is invertible exactly when \det A \neq 0.

Day 2: products, transposes and columns

Theorem 4 (product rule). For n \times n matrices, \det(AB) = \det A \det B.

Theorem 5 (transpose). \det A^{T} = \det A.

Theorem 6 (column operations). Swapping two columns flips the sign, multiplying a column by c multiplies the determinant by c, and adding a multiple of one column to another leaves it unchanged. Two equal columns, or a column of zeros, give 0.

Consequences. For n \times n matrices and every number c and whole number m:

\begin{aligned} \det(A^{-1}) &= \frac{1}{\det A} \quad (A \text{ invertible}) \\ \det(cA) &= c^n \det A \\ \det(A^m) &= (\det A)^m \end{aligned}
Why it is true

Day 1: the row operations

Scaling. Let row i of B be c times row i of A. Expand both along row i. Each cofactor C_{ij} crosses out row i, the only row that differs, so A and B have the same cofactors there, and \det B = \sum_j (ca_{ij})C_{ij} = c \det A.

Swap. By induction on n. For n = 2, \det \begin{bmatrix} c & d \\ a & b \end{bmatrix} = cb - da = -(ad - bc). For n \geq 3, let B be A with rows p and q swapped, and pick a third row i. Expand both along row i: the entries and the signs (-1)^{i+j} are the same, and each minor of B is the matching minor of A with two rows swapped, an (n-1) \times (n-1) determinant, so by the induction hypothesis it is the negative. Hence \det B = -\det A. If A has two equal rows, swapping them leaves A as it was, so \det A = -\det A and \det A = 0.

Replacement. Let row j of B be row j of A plus k times row i. Expand along row j; the cofactors C_{jl} are those of A, since row j is crossed out:

\begin{aligned} \det B &= \sum_l (a_{jl} + ka_{il})C_{jl} \\ &= \det A + k\sum_l a_{il}C_{jl} \end{aligned}

The last sum is the row-j expansion of the matrix A' made from A by putting row i in place of row j. A' has two equal rows, so that sum is 0 and \det B = \det A.

The lemma. By Theorem 1, \det(EA) = f \det A with f = -1, c or 1; with A = I, \det E = f.

Theorem 2. Each swap multiplies the determinant by -1 and each replacement by 1, so \det U = (-1)^r \det A. A square echelon form is upper triangular, and a triangular determinant is the product of its diagonal (Determinants by cofactor expansion).

Theorem 3. Row reduce A to its Reduced row echelon form R, scaling only by nonzero numbers. Every step multiplies the determinant by -1, by c \neq 0, or by 1, so \det R = (\text{nonzero number}) \cdot \det A, and \det A \neq 0 exactly when \det R \neq 0. A square matrix in reduced form with a pivot in every row is I, with \det I = 1; otherwise its last row is zero and its determinant is 0. And A is invertible exactly when R = I (Finding an inverse by row reducing A next to I). \square

Day 2: products and transposes

Theorem 4. Reduce A to R: E_k \cdots E_1 A = R, so A = F_1F_2 \cdots F_k R with F_i = E_i^{-1}, each again elementary (its inverse operation undoes it). Applying the lemma once for each F_i, for any matrix X,

\begin{aligned} &\det(F_1 \cdots F_k X) \\ &\quad = \det F_1 \cdots \det F_k \det X \end{aligned}

With X = RB this gives \det(AB) = \det F_1 \cdots \det F_k \det(RB), and with X = R it gives \det A = \det F_1 \cdots \det F_k \det R. Write f = \det F_1 \cdots \det F_k. If R = I the two right sides are f\det B and f, so \det(AB) = \det A \det B. If R has a zero row, so does RB, both determinants are 0, and both sides of the product rule are 0. \square What the determinant means gave the geometric reason; this is the algebraic proof for every size.

Theorem 5. With A = F_1 \cdots F_k R as above, A^{T} = R^{T}F_k^{T} \cdots F_1^{T} (The transpose and symmetric matrices). A swap or scaling matrix equals its transpose, and a replacement matrix and its transpose are both triangular with 1s on the diagonal, so \det F^{T} = \det F for each. If R = I, R^{T} = I; otherwise R^{T} has a zero column, so \det R^{T} = 0 = \det R. Theorem 4 then gives \det A^{T} = \det R \det F_k \cdots \det F_1, which is \det A. \square

Theorem 6. A column operation on A is a row operation on A^{T}, transposed back, and transposing does not change the determinant.

Consequences. \det A \det(A^{-1}) = \det(AA^{-1}) = \det I = 1. Next, cA = (cI)A and cI is diagonal with determinant c^n (Exercises 20 and 22 of Determinants by cofactor expansion reach this and Theorem 5 by induction instead). Last, apply Theorem 4 m - 1 times.

Worked examples

Example 1

Day 1: Three operations on a known determinant

Suppose \det \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = 4. Find

\det \begin{bmatrix} g & h & i \\ 2d & 2e & 2f \\ a - 5g & b - 5h & c - 5i \end{bmatrix}
Solution

Build the new matrix a step at a time. Swapping rows 1 and 3 gives determinant -4, by the swap rule. Multiplying row 2 by 2 gives 2(-4) = -8, by the scaling rule. Finally R_3 \to R_3 - 5R_1 turns row 3 from (a, b, c) into (a - 5g, b - 5h, c - 5i), a replacement, which keeps -8. The answer is -8.

Example 2

Day 1: A 4 by 4 by row reduction

Seoyeon reduces this matrix to triangular form:

A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 2 & 5 & 1 & 3 \\ -1 & 0 & 4 & 2 \\ 3 & 6 & 1 & 5 \end{bmatrix}

Find \det A.

Solution

Use only replacements, which keep the determinant. With R_2 \to R_2 - 2R_1, R_3 \to R_3 + R_1 and R_4 \to R_4 - 3R_1, the rows become (1, 2, 0, 1), (0, 1, 1, 1), (0, 2, 4, 3), (0, 0, 1, 2). Then R_3 \to R_3 - 2R_2 gives (0, 0, 2, 1), and \displaystyle R_4 \to R_4 - \frac{1}{2}R_3 gives \displaystyle (0, 0, 0, \frac{3}{2}):

U = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & \frac{3}{2} \end{bmatrix}

No swaps, so by Theorem 2, \displaystyle \det A = 1 \cdot 1 \cdot 2 \cdot \frac{3}{2} = 3. That is 14 multiply-and-subtract steps, against 24 four-fold products for a full expansion.

Example 3

Day 1: A zero in the pivot position

Find \det B for

B = \begin{bmatrix} 0 & 2 & 1 & 3 \\ 1 & 3 & 0 & 2 \\ 2 & 6 & 1 & 1 \\ 1 & 1 & 4 & 0 \end{bmatrix}
Solution

The (1, 1) entry is 0, so swap first: R_1 \leftrightarrow R_2 makes the rows (1, 3, 0, 2), (0, 2, 1, 3), (2, 6, 1, 1), (1, 1, 4, 0), and the determinant is now -\det B. Then R_3 \to R_3 - 2R_1 gives (0, 0, 1, -3) and R_4 \to R_4 - R_1 gives (0, -2, 4, -2). Next R_4 \to R_4 + R_2 gives (0, 0, 5, 1), and R_4 \to R_4 - 5R_3 gives (0, 0, 0, 16). The echelon form has diagonal 1, 2, 1, 16, so with r = 1 swap,

\det B = (-1)^1 \cdot 1 \cdot 2 \cdot 1 \cdot 16 = -32
Example 4

Day 1: When does a family of matrices fail to be invertible?

Minjun’s test matrix has an adjustable entry k:

C = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 3 & k \\ 3 & k & 8 \end{bmatrix}

For which k is C not invertible?

Solution

By Theorem 3, exactly when \det C = 0. With R_2 \to R_2 - 2R_1 and R_3 \to R_3 - 3R_1 the rows are (1, 1, 2), (0, 1, k - 4), (0, k - 3, 2). The pivot in row 2 is 1, so R_3 \to R_3 - (k - 3)R_2 is a replacement for every k and never divides by an expression that might be 0. Row 3 becomes (0, 0, 2 - (k - 3)(k - 4)), and

\det C = 2 - (k^2 - 7k + 12) = -(k - 2)(k - 5)

So C fails to be invertible exactly for k = 2 and k = 5.

Example 5

Day 2: Products, inverses and multiples

Yerin has 4 \times 4 matrices with \det A = 2 and \det B = -5. Find \det(AB), \det(A^{T}B^{-1}), \det(3A), \det(A^3B^2) and \det(BAB^{-1}).

Solution

By the product rule, \det(AB) = 2(-5) = -10. With the transpose and inverse rules, \displaystyle \det(A^{T}B^{-1}) = 2 \cdot \frac{1}{-5} = -\frac{2}{5}. Each of the 4 rows of 3A carries a factor 3, so \det(3A) = 3^4 \cdot 2 = 162. Next \det(A^3B^2) = 2^3(-5)^2 = 200. Last, \displaystyle \det(BAB^{-1}) = (-5) \cdot 2 \cdot \frac{1}{-5} = 2: the determinants are numbers and commute even though the matrices need not.

Example 6

Day 2: Column operations

Find \det P for

P = \begin{bmatrix} 1 & 3 & -2 & 4 \\ 2 & 7 & -3 & 9 \\ -1 & -1 & 5 & 2 \\ 3 & 8 & -4 & 13 \end{bmatrix}
Solution

Column replacements can clear row 1. Replace column 2 by column 2 minus 3 times column 1, column 3 by column 3 plus 2 times column 1, and column 4 by column 4 minus 4 times column 1; by Theorem 6 none of these changes the determinant:

\det P = \det \begin{bmatrix} 1 & 0 & 0 & 0 \\ 2 & 1 & 1 & 1 \\ -1 & 2 & 3 & 6 \\ 3 & -1 & 2 & 1 \end{bmatrix}

Row 1 expansion leaves the minor with rows (1, 1, 1), (2, 3, 6), (-1, 2, 1). Subtracting its column 1 from columns 2 and 3 gives rows (1, 0, 0), (2, 1, 4), (-1, 3, 2), whose determinant is 1 \cdot (1 \cdot 2 - 4 \cdot 3) = -10. So \det P = -10.

Example 7

Day 2: The 3 by 3 Vandermonde determinant

Linear systems in use: networks, chemical equations, fitting curves fits a polynomial through points with the Vandermonde matrix V, whose rows are 1, x_i, x_i^2. For three points a, b, c, prove

\begin{aligned} \det V &= \det \begin{bmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{bmatrix} \\ &= (b - a)(c - a)(c - b) \end{aligned}
Solution

The replacements R_2 \to R_2 - R_1 and R_3 \to R_3 - R_1 keep the determinant and give rows (1, a, a^2), (0, b - a, b^2 - a^2), (0, c - a, c^2 - a^2). Since b^2 - a^2 = (b - a)(b + a), row 2 is (b - a) times (0, 1, b + a), and likewise row 3 is (c - a) times (0, 1, c + a). The scaling rule, read backward, takes both factors out; it holds for every value of the factor, 0 included:

\begin{aligned} \det V &= (b - a)(c - a) \\ &\quad \cdot \det \begin{bmatrix} 1 & a & a^2 \\ 0 & 1 & b + a \\ 0 & 1 & c + a \end{bmatrix} \end{aligned}

Expanding down column 1 leaves (c + a) - (b + a) = c - b, so \det V = (b - a)(c - a)(c - b), nonzero exactly when a, b, c are distinct. Check with a, b, c = 1, 3, 4: the formula gives 2 \cdot 3 \cdot 1 = 6, and expanding \begin{bmatrix} 1 & 1 & 1 \\ 1 & 3 & 9 \\ 1 & 4 & 16 \end{bmatrix} along row 1 gives 12 - 7 + 1 = 6. For n points it is the product of all x_j - x_i with i < j (Exercise 24 does n = 4).

Boundaries

  • Only square matrices have determinants, and the product rule needs A and B both n \times n; a 2 \times 3 times a 3 \times 2 is square, but its factors have no determinant.
  • The determinant is linear in each row separately, never in the whole matrix: \det(cA) = c^n \det A, not c\det A, and \det(A + B) is in general not \det A + \det B (for A = B = I_2 it is 4, not 2).
  • A scaling in reduction must use c \neq 0: multiplying a row by 0 cannot be undone.
  • With a letter in the matrix, dividing by an expression needs the values that make it 0 treated separately, or the division avoided, as in Example 4.
  • \det A decides invertibility, but its size does not measure how close A is to being singular: \det(0.1\,I_{10}) = 10^{-10}, yet 0.1\,I_{10} has the simple inverse 10\,I_{10}.

Where this goes next

The The invertible matrix theorem collects \det A \neq 0 with the other conditions equivalent to invertibility. Theorem 3 makes \det(A - \lambda I) = 0, the The characteristic equation and finding eigenvalues, the exact condition for A - \lambda I to have a nonzero kernel. When a matrix or transformation can be undone and Cramer's rule use these rules next.

Exercises

Day 1: the three rules

  1. Given \det \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = -3, find the determinants of \begin{bmatrix} g & h & i \\ a & b & c \\ d & e & f \end{bmatrix} and \begin{bmatrix} a & b & c \\ d & e & f \\ -2g & -2h & -2i \end{bmatrix}.
  2. For the same matrix, find the determinants of \begin{bmatrix} a + 4d & b + 4e & c + 4f \\ d & e & f \\ g & h & i \end{bmatrix} and \begin{bmatrix} 2a & 2b & 2c \\ 2d & 2e & 2f \\ 2g & 2h & 2i \end{bmatrix}.

Day 1: computing by reduction

  1. Find \det \begin{bmatrix} 1 & 3 & 2 \\ 2 & 7 & 1 \\ -1 & 0 & 5 \end{bmatrix} by row reduction.
  2. Find \det \begin{bmatrix} 1 & -1 & 2 & 0 \\ 3 & -2 & 5 & 1 \\ -2 & 4 & -3 & 2 \\ 1 & 0 & 4 & 3 \end{bmatrix}.
  3. Find \det \begin{bmatrix} 0 & 1 & 2 & -1 \\ 2 & 4 & 0 & 2 \\ 1 & 3 & 1 & 0 \\ 3 & 5 & 2 & 6 \end{bmatrix}, keeping count of the swaps.
  4. Find the determinant of the 5 \times 5 matrix with rows (1, 2, 0, 1, 3), (2, 5, 1, 2, 6), (0, 1, 3, 0, 1), (1, 2, 0, 3, 2), (-1, -2, 1, -1, -2).

Day 1: invertibility and cost

  1. For which h is \begin{bmatrix} 1 & 2 & -1 \\ 3 & h & 1 \\ 2 & 4 & h \end{bmatrix} not invertible?
  2. Is \begin{bmatrix} 2 & -1 & 3 & 0 \\ 1 & 1 & 0 & 2 \\ 3 & 0 & 3 & 2 \\ 0 & 4 & 1 & 5 \end{bmatrix} invertible? Answer with one row operation.
  3. For n = 10, how many multiply-and-subtract steps does reduction to triangular form take, by the count in Theorem 2, and how many products does a full cofactor expansion have?

Day 2: products and transposes

  1. A = \begin{bmatrix} -2 & 1 & 4 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix} and B = \begin{bmatrix} 3 & 0 & 0 \\ 2 & 1 & 0 \\ 5 & -1 & 1 \end{bmatrix}. Without multiplying any matrices, find \det(AB), \det(B^2), \det(A^{T}) and \det(A^{-1}B).
  2. For the same A and B, find \det(-A), \det(2B^{-1}) and \det(A^{T}BA^{-1}).
  3. A 4 \times 4 matrix A has \det A = 5. B is made from A by adding 3 times column 2 to column 4, then swapping columns 1 and 3. Find \det B and \det(BA^{T}).

Day 2: column operations and Vandermonde

  1. Use column operations to find \det \begin{bmatrix} 1 & 4 & 7 \\ 2 & 9 & 12 \\ -3 & -10 & -20 \end{bmatrix}.
  2. Use column operations to find \det \begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 5 & -1 & 7 \\ 0 & 1 & 4 & 2 \\ -1 & 0 & 3 & 1 \end{bmatrix}.
  3. Use the Vandermonde formula to find \det \begin{bmatrix} 1 & 2 & 4 \\ 1 & 5 & 25 \\ 1 & -1 & 1 \end{bmatrix}.
  4. For which t is \begin{bmatrix} 1 & 1 & 1 \\ 1 & t & t^2 \\ 1 & 3 & 9 \end{bmatrix} not invertible? What does that say about fitting a parabola?

Reasoning

  1. Prove that a square matrix with two equal columns has determinant 0, and that so does one in which a column is a linear combination of the other columns.
  2. Show that if A^2 = A, then \det A is 0 or 1.
  3. Show that if Q^{T}Q = I, then \det Q is 1 or -1.
  4. A matrix is skew-symmetric if A^{T} = -A. Prove that a skew-symmetric n \times n matrix with n odd has determinant 0.
  5. Show that no real 3 \times 3 matrix A has A^2 = -I, and give a real 2 \times 2 matrix that does. Why does the argument not rule out n = 2?
  6. Find 2 \times 2 matrices A and B with \det A = \det B = 0 but \det(A + B) = 1.
  7. Prove that if every row of a square matrix sums to 0, its determinant is 0.
  8. Use the column operations C_4 \to C_4 - aC_3, then C_3 \to C_3 - aC_2, then C_2 \to C_2 - aC_1 to prove that the 4 \times 4 Vandermonde matrix with rows 1, x, x^2, x^3 for x = a, b, c, d has determinant (b - a)(c - a)(d - a) times (c - b)(d - b)(d - c).
Answers
  1. -3 and 6. The rows in order 3, 1, 2 take two swaps (R_1 \leftrightarrow R_3, then R_2 \leftrightarrow R_3), so the sign flips twice. Scaling row 3 by -2 gives -2(-3) = 6.
  2. -3 and -24. R_1 \to R_1 + 4R_2 is a replacement. Each of the three rows carries a factor 2: 2^3(-3) = -24.
  3. 16. R_2 - 2R_1, R_3 + R_1, R_3 - 3R_2 give diagonal 1, 1, 16.
  4. 6. Clear column 1 with row 1, then R_3 - 2R_2, R_4 - R_2, R_4 - R_3. Diagonal 1, 1, 3, 2, no swaps.
  5. 4. R_1 \leftrightarrow R_3, clear column 1, then R_2 \leftrightarrow R_3 to bring up (0, 1, 2, -1) and clear on. Diagonal 1, 1, 2, 2, two swaps: (-1)^2 \cdot 4 = 4.
  6. 2. Clear column 1 with row 1, then R_3 - R_2 and \displaystyle R_5 - \frac{1}{2}R_3. Diagonal \displaystyle 1, 1, 2, 2, \frac{1}{2}, no swaps.
  7. h = 6 and h = -2. R_2 - 3R_1 and R_3 - 2R_1 give the triangular rows (0, h - 6, 4) and (0, 0, h + 2), so the determinant is (h - 6)(h + 2).
  8. No. Row 3 is row 1 plus row 2, so R_3 \to R_3 - R_1 - R_2 (two replacements) makes a zero row; the determinant is 0.
  9. 285 against 3{,}628{,}800. \displaystyle \frac{9 \cdot 10 \cdot 19}{6} = 285, and 10! = 3{,}628{,}800.
  10. -6, 9, -2, \displaystyle -\frac{3}{2}. Both are triangular: \det A = -2, \det B = 3. Then (-2)(3) = -6, 3^2 = 9, \det A^{T} = \det A, and \displaystyle \frac{3}{-2}.
  11. 2, \displaystyle \frac{8}{3}, 3. (-1)^3(-2) = 2; \displaystyle 2^3 \cdot \frac{1}{3} = \frac{8}{3}; \displaystyle (-2) \cdot 3 \cdot \frac{1}{-2} = 3.
  12. \det B = -5 and \det(BA^{T}) = -25. The column replacement keeps 5 and the column swap flips it; then (-5)(5) = -25.
  13. 5. Column 2 minus 4 times column 1 and column 3 minus 7 times column 1 give rows (1, 0, 0), (2, 1, -2), (-3, 2, 1); expanding along row 1, 1 \cdot 1 - (-2) \cdot 2 = 5.
  14. 6. Column 2 minus 2 times column 1, column 3 plus column 1, column 4 minus 3 times column 1 clear row 1 to (1, 0, 0, 0) and leave the minor with rows (1, 1, 1), (1, 4, 2), (2, 2, 4). Subtracting its column 1 from columns 2 and 3 gives rows (1, 0, 0), (1, 3, 1), (2, 0, 2), with determinant 3 \cdot 2 - 1 \cdot 0 = 6.
  15. 54. With a, b, c = 2, 5, -1: (5 - 2)(-1 - 2)(-1 - 5) = 3 \cdot (-3) \cdot (-6) = 54.
  16. t = 1 and t = 3. The determinant is (t - 1)(3 - 1)(3 - t). There two of the x values coincide, so no single parabola is pinned down.
  17. Swapping the two equal columns leaves the matrix unchanged but flips the sign of its determinant (Theorem 6), so \det A = -\det A = 0. If column j is \sum_{i \neq j} c_i times column i, the column replacements that subtract each c_i times column i from column j keep the determinant and produce a zero column, and expanding along it gives 0.
  18. With d = \det A: d = \det(A^2) = d^2, so d(d - 1) = 0.
  19. 1 = \det I = \det(Q^{T}Q), and by Theorems 4 and 5 this is \det Q^{T}\det Q = (\det Q)^2.
  20. \det A = \det A^{T} = \det(-A) = (-1)^n \det A = -\det A for odd n, so \det A = 0.
  21. \det(A^2) = (\det A)^2 \geq 0 for real A, but \det(-I_3) = (-1)^3 = -1. The matrix \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} squares to -I_2; for n = 2, \det(-I_2) = 1, so there is no contradiction.
  22. A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}: each has a zero row, and A + B = I_2.
  23. Add columns 2, \dots, n to column 1; these are column replacements, which keep the determinant. Each entry of the new column 1 is a row sum, 0, so the determinant is 0.
  24. Each step uses a column that has not yet changed. The row for x becomes (1, x - a, x(x - a), x^2(x - a)), so row a becomes (1, 0, 0, 0). Expanding along row 1 leaves the 3 \times 3 determinant whose rows are (x - a)(1, x, x^2) for x = b, c, d. Taking out the three row factors leaves the 3 \times 3 Vandermonde determinant of b, c, d, which by Example 7 is (c - b)(d - b)(d - c). Together, (b - a)(c - a)(d - a) times (c - b)(d - b)(d - c).
kg1128 · updated 2026-10-22 · CC-BY-SA-4.0