---
title: "Determinants and Linear Systems"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 22
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems
---

# Chapter 22 — Determinants and Linear Systems

The determinant condenses into one scalar the answer to “are these $n$ vectors a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)?” — and, geometrically, measures the volume they [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span). We characterize it by its properties (multilinear, alternating, normalized), compute it in dimensions $2$ and $3$ and by [cofactor expansion](#thm-b1-det-cofactor) in general, and put it to work on [linear systems](#def-b1-det-system), alongside the all-purpose algorithm: [Gaussian elimination](#met-b1-det-gauss).

## 22.1 The determinant

**Theorem 22.1 (Characterization).**

There is exactly one [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $\det \colon \mathcal{M}_n(K) \to K$, viewed as a function of the $n$ columns, that is:

1. *[linear](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-def) in each column* (the others fixed);
2. *alternating* : swapping two columns changes the sign (hence two equal columns give $0$ );
3. *normalized* : $\det I_n = 1$ .

For $n = 2$ and $3$:

$$
\begin{vmatrix} a & b\\ c & d\end{vmatrix} = ad - bc,
\qquad
\begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i\end{vmatrix}
= aei + bfg + cdh - ceg - bdi - afh
$$

(the $3 \times 3$ rule of Sarrus: products of descending diagonals minus ascending ones).

**Proof.** *Admitted at this level.* ∎

**Remark 22.2.**

For $n = 2$: expanding by bilinearity on the canonical columns gives the formula, which conversely satisfies the axioms — a complete proof; $n = 3$ is identical with more terms. The general case (existence via the sum over [permutations](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-objects), uniqueness by the same expansion) demands the signature of a [permutation](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-objects) and is deferred to the second year; we freely use the axioms and the consequences below.

The $n = 2$ expansion in full, since it is the template: with columns $C_1 = a\,e_1 + c\,e_2$ and $C_2 = b\,e_1 + d\,e_2$, bilinearity gives

$$
\det(C_1, C_2) = ab\det(e_1, e_1) + ad\det(e_1, e_2)
+ cb\det(e_2, e_1) + cd\det(e_2, e_2),
$$

and alternation kills the repeated pairs while flipping $\det(e_2, e_1) = -\det(e_1, e_2)$: the whole [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) collapses to $(ad - bc)\det(e_1, e_2) = ad - bc$ by normalization. Uniqueness is visible in the computation itself — the axioms left no choice at any step — and this is exactly the scaled-uniqueness fact used in the proof of the product rule below.

**Theorem 22.3 (Properties).**

For $A, B \in \mathcal{M}_n(K)$:

1. adding to a column a multiple of another does not change the determinant; multiplying a column by $\lambda$ multiplies it by $\lambda$ (so $\det(\lambda A) = \lambda^n \det A$ );
2. $\det(AB) = \det A\, \det B$ ;
3. $A$ is invertible $\iff$ $\det A \neq 0$ $\iff$ the columns form a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $K^n$ ; and then $\det(A^{-1}) = (\det  A)^{-1}$ ;
4. $\det(A^{\mathsf T}) = \det A$ — so every column rule is also a row rule;
5. the determinant of a triangular matrix is the product of its diagonal entries.

**Proof.** (1) By linearity, $\det(\dots, C_i + \lambda C_j, \dots) = \det A +
\lambda\det(\dots, C_j, \dots)$ where the second determinant has two equal columns: zero.

(2) Fix $A$ and consider $\varphi(B) = \det(AB)$ as a function of the columns of $B$: since $AB$ has columns $AB_j$, $\varphi$ is multilinear and alternating in the $B_j$. We admit, with [Theorem 22.1](#thm-b1-det-def), its uniqueness [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) in scaled form: *every* multilinear alternating [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $\varphi$ of the columns equals $\varphi(I_n) \cdot \det$. Here $\varphi(I_n) = \det A$, so $\det(AB) = \det A \cdot \det B$.

(3) If $A$ is invertible: $\det A\,\det A^{-1} = \det I = 1$, so $\det A \neq 0$ and the inverse formula holds. If $A$ is not invertible, its columns are linked ([Corollary 20.9](https://one-course.com/books/math/3/en/chapter/20-linear-maps#cor-b1-linmaps-samedim) and [Proposition 20.2](https://one-course.com/books/math/3/en/chapter/20-linear-maps#prop-b1-linmaps-basis)); expressing one column through the others and expanding by linearity leaves determinants with two equal columns: $\det A = 0$. The [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) is [Proposition 19.8](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#prop-b1-findim-twoofthree).

(4) Admitted with the general construction (it is immediate on the [permutation](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-objects) formula); we record it to use [row operations](https://one-course.com/books/math/3/en/chapter/21-matrices#met-b1-matrices-gauss).

(5) If some diagonal entry vanishes, the first $k$ columns are linked for some $k$ (rank considerations) and $\det = 0 =$ product. Otherwise, clear each column below-and-left by operations of type (1) — possible in the triangular shape — reaching the diagonal matrix, whose determinant is the product of the entries by multilinearity from $I_n$. ∎

**Example 22.4 (The rules, checked on numbers).**

Take $A = \begin{pmatrix} 1 & 2\\ 3 & 4\end{pmatrix}$ ($\det A =
-2$) and $B = \begin{pmatrix} 0 & 1\\ 1 & 1\end{pmatrix}$ ($\det
B = -1$). Then

$$
AB = \begin{pmatrix} 2 & 3\\ 4 & 7\end{pmatrix},
\quad \det(AB) = 14 - 12 = 2 = (-2)(-1) ;
\qquad
\det(A^{\mathsf T}) = \begin{vmatrix} 1 & 3\\ 2 & 4
\end{vmatrix} = -2 = \det A .
$$

Multiplicativity and transpose-invariance confirmed — while the *false* additivity fails on the same pair:

$$
\det(A + B) = \begin{vmatrix} 1 & 3\\ 4 & 5\end{vmatrix} = -7
\neq \det A + \det B = -3 .
$$

Thirty seconds of arithmetic of this kind, after any determinant identity is invoked, is the cheapest error insurance available.

**Example 22.5 (Determinants as areas).**

The parallelogram spanned by $u = (2, 0)$ and $v = (1, 3)$ has base $2$ and height $3$: area $6$. And

$$
\begin{vmatrix} 2 & 1\\ 0 & 3\end{vmatrix} = 6 :
$$

the $2\times2$ determinant *is* the signed area of the parallelogram of its columns. The axioms retell the geometry: adding a multiple of one column to the other is a *shear*, sliding the parallelogram parallel to a side without changing base or height (operation (1) of [Theorem 22.3](#thm-b1-det-props)); scaling a column scales the area; swapping the columns flips the orientation, whence the sign, $\det(v, u) = -6$. In $\R^3$ the same reading gives signed volumes, and $\abs{\det}$ becomes the universal volume-scaling factor of [linear maps](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-def) — the fact behind the change-of-variables formula for multiple [integrals](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) in the Year 2 volume.

**Theorem 22.6 (Cofactor expansion).**

Let $A \in \mathcal{M}_n(K)$ and $\Delta_{ij}$ the determinant of $A$ with row $i$ and column $j$ deleted. Then, for any fixed column $j$ (or row, by transposition):

$$
\det A = \sum_{i=1}^{n} (-1)^{i+j}\, a_{ij}\, \Delta_{ij} .
$$

**Proof.** *Admitted at this level.* ∎

**Example 22.7.**

Expanding along the first column:

$$
\begin{vmatrix}
2 & 1 & 0\\
1 & 2 & 1\\
0 & 1 & 2
\end{vmatrix}
= 2\begin{vmatrix} 2 & 1\\ 1 & 2\end{vmatrix}
- 1\begin{vmatrix} 1 & 0\\ 1 & 2\end{vmatrix}
= 2 \times 3 - 2 = 4 .
$$

Strategy: create zeros first (row/column operations), then expand along the emptiest line.

**Example 22.8 (The cofactor inverse, once by hand).**

For $A = \begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 1 & 0 &
1\end{pmatrix}$: $\det A = 1(1) - 1(-1) + 0 = 2$. The nine cofactors $(-1)^{i+j}\Delta_{ij}$ assemble into

$$
\operatorname{Com}(A) = \begin{pmatrix}
1 & 1 & -1\\
-1 & 1 & 1\\
1 & -1 & 1
\end{pmatrix},
\qquad
A^{-1} = \frac{1}{\det A}\operatorname{Com}(A)^{\mathsf T}
= \frac12\begin{pmatrix}
1 & -1 & 1\\
1 & 1 & -1\\
-1 & 1 & 1
\end{pmatrix},
$$

the formula quoted in [Exercise 22.8](#exo-b1-det-8). Check one row-column pair: (row $1$ of $A$)(column $1$ of $A^{-1}$) $= \frac12(1 + 1
+ 0) = 1$, and against column $2$: $\frac12(-1 + 1 + 0) = 0$. Nine $2\times2$ determinants for one $3\times3$ inverse: already at this size, row reduction ([Exercise 22.3](#exo-b1-det-3)) is cheaper — the cofactor formula’s value is theoretical (integrality in [Exercise 22.8](#exo-b1-det-8), differentiability of the inverse in later volumes), not computational.

**Example 22.9 (The block-triangular rule, in size 444).**

Claim: $\det\begin{pmatrix} M & N\\ 0 & P\end{pmatrix} = \det
M\,\det P$ for $2\times2$ blocks. Clear the $N$-block by column operations: adding to columns $3, 4$ suitable combinations of columns $1, 2$ removes $N$ *when $M$ is invertible* (solve $M\Lambda = -N$ for the combination coefficients $\Lambda$), leaving $\det\begin{pmatrix} M & 0\\ 0 & P\end{pmatrix}$; then [cofactor expansion](#thm-b1-det-cofactor) along the first column, twice, gives $\det
M\det P$ for this block-diagonal shape. If $M$ is not invertible, its columns are linked, so the first two columns of the big matrix are linked (their bottom halves are zero): both sides vanish. The rule extends to any block sizes by the same two-case argument — and it is the engine of [Exercise 22.10](#exo-b1-det-10).

**Example 22.10 (A 4×44 \times 44×4 determinant, strategically).**

$$
\Delta = \begin{vmatrix}
1 & 2 & 3 & 4\\
2 & 3 & 4 & 1\\
3 & 4 & 1 & 2\\
4 & 1 & 2 & 3
\end{vmatrix}.
$$

Every row sums to $10$: the operation $C_1 \leftarrow C_1 + C_2 +
C_3 + C_4$ makes the first column constant, and factoring $10$ leaves ones. Then $L_i \leftarrow L_i - L_1$ ($i \geq 2$) clears the first column:

$$
\Delta = 10\begin{vmatrix}
1 & 2 & 3 & 4\\
0 & 1 & 1 & -3\\
0 & 2 & -2 & -2\\
0 & -1 & -1 & -1
\end{vmatrix}
= 10\begin{vmatrix}
1 & 1 & -3\\
2 & -2 & -2\\
-1 & -1 & -1
\end{vmatrix}
= 10 \times 16 = 160,
$$

the last $3\times3$ determinant expanding along its first row: $1(2 - 2) - 1(-2 - 2) + (-3)(-2 - 2) = 0 + 4 + 12 = 16$. Moral: one well-chosen operation (spotting the constant row sum) beats sixteen cofactors.

**Method 22.11 (Choosing a determinant strategy).**

Scan the matrix before computing anything.

1. *Constant row or column sums* : add everything into one line, factor the common value ( [Example 22.10](#ex-b1-det-fourbyfour) , [Exercise 22.7](#exo-b1-det-7) ).
2. *Repetitive structure* : subtract neighboring rows or columns to create zeros; staircase patterns collapse toward triangular form, whose determinant is read on the diagonal.
3. *Isolated zeros* : expand along the emptiest line ( [Example 22.7](#ex-b1-det-cofactor) ); recursive families (tridiagonal, [Exercise 22.6](#exo-b1-det-6) ) yield recurrences this way.
4. *A parameter* : the determinant is a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) in it; find its roots by spotting the degenerate values (equal rows, proportional columns), then pin the [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) down by degree and leading coefficient. For the matrix of [Exercise 22.7](#exo-b1-det-7) : $m = 1$ gives three equal rows (rank $1$ , a double root), $m = -2$ makes the rows sum to zero (one more root); the determinant has degree $3$ in $m$ with leading term $-m^3$ (the anti-diagonal product $m\cdot m\cdot m$ , whose Sarrus sign is $-1$ ), so it must be $-(m+2)(m-1)^2$ — no expansion needed, and the two methods check each other.

**Example 22.12 (Vandermonde determinant).**

For scalars $x_1, \dots, x_n$:

$$
V(x_1, \dots, x_n) =
\begin{vmatrix}
1 & x_1 & x_1^2 & \cdots & x_1^{n-1}\\
1 & x_2 & x_2^2 & \cdots & x_2^{n-1}\\
\vdots & & & & \vdots\\
1 & x_n & x_n^2 & \cdots & x_n^{n-1}
\end{vmatrix}
= \prod_{1 \leq i < j \leq n} (x_j - x_i) .
$$

Proof sketch (detailed in [Exercise 22.5](#exo-b1-det-5)): column operations $C_k \leftarrow C_k - x_1 C_{k-1}$ from the right clear the first row, and factoring each remaining row reduces to $V(x_2, \dots, x_n)$. Nonzero iff the $x_i$ are pairwise distinct — the determinant behind [Lagrange interpolation](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-lagrange) ([Example 20.10](https://one-course.com/books/math/3/en/chapter/20-linear-maps#ex-b1-linmaps-interpolation)).

## 22.2 Linear systems

**Definition 22.13.**

A [linear](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-def) system of $n$ equations in $p$ unknowns is $AX = B$ with $A \in \mathcal{M}_{n,p}(K)$, $B \in
K^n$; it is *homogeneous* when $B = 0$. Its solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets), when nonempty, is $X_0 + \ker A$: a particular solution plus the general homogeneous solution — an affine [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of dimension $p -
\operatorname{rk} A$ (rank–nullity).

**Example 22.14 (The affine structure, made visible).**

Solve

$$
\begin{cases}
x + y + z = 3\\
x - y + 2z = 2 .
\end{cases}
$$

Subtracting the equations: $2y - z = 1$, so $z = 2y - 1$ and $x =
3 - y - z = 4 - 3y$. The solutions form the line

$$
(x, y, z) = (4,\ 0,\ -1) + y\,(-3,\ 1,\ 2)
\qquad (y \in \R):
$$

the particular solution $X_0 = (4, 0, -1)$ (the choice $y = 0$) plus the kernel line $\ker A = \operatorname{Vect}(-3, 1, 2)$ of the associated homogeneous system — check: $(-3) + 1 + 2 = 0$ and $(-3) - 1 + 4 = 0$. Geometrically, two non-parallel planes of $\R^3$ intersect along a line, and the dimension count $p -
\operatorname{rk} A = 3 - 2 = 1$ knew it before we solved anything. Changing the particular solution (say $y = 1$: $X_0' =
(1, 1, 1)$) changes the description, not the line: an affine [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) has many origins and one direction.

**Theorem 22.15 (Square Cramer systems).**

If $A \in GL_n(K)$, the system $AX = B$ has the unique solution $X =
A^{-1}B$, whose [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) are

$$
x_j = \frac{\det A_j}{\det A},
\qquad A_j = A \text{ with column } j \text{ replaced by } B .
$$

**Proof.** Uniqueness and existence are the invertibility. For the formula: write $B = \sum_k x_k C_k$ (columns of $A$); then, by multilinearity and alternation,

$$
\det A_j = \det\Bigl(C_1, \dots, \sum_k x_k C_k, \dots, C_n\Bigr)
= \sum_k x_k \det(C_1, \dots, C_k, \dots, C_n)
= x_j \det A ,
$$

every term except $k = j$ having a repeated column. ∎

**Example 22.16 (Cramer with a parameter, in full).**

For $m \in \R$, solve

$$
\begin{cases}
x + m y = 1\\
m x + y = 2 .
\end{cases}
$$

The determinant is $1 - m^2$. *Generic case* $m \neq \pm1$: Cramer gives

$$
x = \frac{\begin{vmatrix} 1 & m\\ 2 & 1\end{vmatrix}}{1 - m^2}
= \frac{1 - 2m}{1 - m^2},
\qquad
y = \frac{\begin{vmatrix} 1 & 1\\ m & 2\end{vmatrix}}{1 - m^2}
= \frac{2 - m}{1 - m^2},
$$

one clean solution for each admissible $m$ (check at $m = 0$: $(1, 2)$, obviously right). *Degenerate cases*: at $m = 1$ the equations read $x + y = 1$ and $x + y = 2$: incompatible; at $m = -1$ they read $x - y = 1$ and $-x + y = 2$, i.e. $x - y =
1$ and $x - y = -2$: incompatible again. The vanishing of the determinant announces that *something* degenerates, but never says what — empty or infinite must be decided by looking at the right-hand side. Note also how the formulas signal their own limits: as $m \to 1^{-}$, $x = \frac{1 - 2m}{1 - m^2} \to
-\infty$; the solution point runs away as the two lines become parallel.

**Method 22.17 (Gaussian elimination on systems).**

Row-reduce the augmented matrix $(A \mid B)$ to echelon form.

1. If a pivot appears in the last column ( $0 = 1$ line): no solution.
2. Otherwise, the unknowns split into *pivot unknowns* and *[free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) unknowns* (parameters); back-substitution expresses the former in the latter: the solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) is an affine [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of dimension $=$ number of [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) unknowns.

Cramer’s formulas are for theory and small systems; elimination is the practical algorithm.

**Example 22.18 (A discussion with parameter).**

For $m \in \R$, consider

$$
\begin{cases}
x + y + mz = 1\\
x + my + z = 1\\
mx + y + z = 1 .
\end{cases}
$$

The matrix has determinant $-(m+2)(m-1)^2$ (computed in [Exercise 22.7](#exo-b1-det-7) by adding all columns to the first). For $m
\neq 1, -2$: unique solution $x = y = z = \frac{1}{m+2}$ (by symmetry). For $m =
1$: one equation repeated thrice, a plane of solutions. For $m =
-2$: adding the three equations gives $0 = 3$, no solution.

**Remark 22.19 (Common pitfalls).**

*The determinant is not [linear](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-def) in the matrix*: $\det(A + B)
\neq \det A + \det B$ (already $\det(I_2 + I_2) = 4 \neq 2$); it is [linear](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-def) in each *column* separately, which is a different thing entirely. *Scaling*: $\det(\lambda A) = \lambda^n\det A$, not $\lambda\det A$ — each of the $n$ columns is scaled. *[Row operations](https://one-course.com/books/math/3/en/chapter/21-matrices#met-b1-matrices-gauss) are not all [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)*: $L_i \leftarrow L_i +
\lambda L_j$ preserves the determinant, but a swap changes the sign and $L_i \leftarrow \lambda L_i$ multiplies it by $\lambda$ — bookkeeping errors here are the classic source of wrong signs in elimination-based computations. *A zero determinant is the beginning, not the end*: it says “rank $< n$” but not which rank; only further work (echelon form, or the minors of [Exercise 22.12](#exo-b1-det-12)) locates it — cf. the case $m = 1$ vs $m = -2$ in [Example 22.18](#ex-b1-det-parameter). *Cramer needs invertibility*: when $\det A = 0$ the formulas $x_j = \det A_j/\det A$ are meaningless, and the system may perfectly well have (infinitely many) solutions. *Only square matrices have determinants*: for a rectangular system, elimination is the only tool.

**Remark 22.20 (Where determinants go).**

Three lives await this scalar. *Geometric*: $\abs{\det}$ is the area or volume scaling factor of the associated [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) — made precise for the plane in [Chapter 23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid) and, as the Jacobian of a change of variables, in the multiple [integrals](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) of the Year 2 volume. *Algebraic*: $\det(A - \lambda I)$, the characteristic [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def), [opens](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-open) eigenvalue theory in Year 2 — the identity $A^2 - (\operatorname{tr} A)A + (\det A)I = 0$ of the [Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices) weekend problem is its first shadow. *Analytic*: determinants of special matrices (Vandermonde, Cauchy, Gram) decide when interpolation, decomposition and [projection](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-projection) problems are well posed; the weekend problem below evaluates the first two families completely.

**Remark 22.21 (Perspectives inside Book 3).**

This chapter closes the linear-algebra spine of the volume, and its two remaining chapters cash the dividends. In [Chapter 23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid): the Gram matrix $\bigl(\langle v_i, v_j\rangle\bigr)$ tests freeness by a determinant ([Exercise 23.11](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#exo-b1-euclid-11)), and the plane isometries split into rotations and reflections along the sign of their determinant — the classification of the weekend problem there runs on it. In [Chapter 25](https://one-course.com/books/math/3/en/chapter/25-functions-of-two-variables#ch-b1-multivar): the Monge quantity $rt -
s^2$ is the determinant of the symmetric matrix of second [derivatives](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def), and the normal equations of least squares are a Cramer system whose matrix is a Gram (hence moment) matrix — invertible precisely by the Vandermonde-flavored criteria established here. When those chapters assert “invertible” or “positive”, the receipts are in this one.

## 22.3 Exercises

**Exercise 22.1 ★.**

Compute:

$$
\begin{vmatrix} 3 & 1\\ 5 & 2 \end{vmatrix},
\qquad
\begin{vmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\end{vmatrix},
\qquad
\begin{vmatrix} 1 & 1 & 1\\ 1 & 2 & 4\\ 1 & 3 & 9\end{vmatrix}.
$$

**Solution of Exercise 22.1.**

$3 \times 2 - 1 \times 5 = 1$.

Second: $L_2 \leftarrow L_2 - L_1$, $L_3 \leftarrow L_3 - L_2$ (on the original rows) give rows $(1,2,3), (3,3,3), (3,3,3)$: two equal rows, determinant $0$. (Sarrus confirms: $45 + 84 + 96 - 105 - 48 -
72 = 0$.)

Third: it is Vandermonde with $x = 1, 2, 3$ ([Example 22.12](#ex-b1-det-vandermonde)): $(2-1)(3-1)(3-2) = 2$.

**Exercise 22.2 ★.**

For which $\lambda \in \R$ is the family $\bigl((1, 1, \lambda),
(1, \lambda, 1), (\lambda, 1, 1)\bigr)$ a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $\R^3$?

**Solution of Exercise 22.2.**

The determinant equals (add all columns to the first, factor) $(\lambda + 2)$ times

$$
\begin{vmatrix}
1 & 1 & \lambda\\ 1 & \lambda & 1\\ 1 & 1 & 1
\end{vmatrix}
= -(\lambda - 1)^2
$$

(clear with $L_1 \leftarrow L_1 - L_3$, $L_2 \leftarrow L_2 - L_3$ and expand), giving $\det = -(\lambda+2)(\lambda-1)^2$. [Basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $\iff \det \neq 0 \iff \lambda \notin \{1, -2\}$.

**Exercise 22.3 ★.**

Solve by Cramer’s rule:

$$
\begin{cases}
2x + y = 5\\
3x - 2y = 4 ,
\end{cases}
\qquad\text{then}\qquad
\begin{cases}
x + y + z = 6\\
x - y + z = 2\\
2x + y - z = 1 .
\end{cases}
$$

**Solution of Exercise 22.3.**

First system: $\det = -7$; $x = \frac{1}{-7}\begin{vmatrix} 5 & 1\\ 4 & -2\end{vmatrix}
= \frac{-14}{-7} = 2$, $y = \frac{1}{-7}\begin{vmatrix} 2 & 5\\ 3 & 4\end{vmatrix}
= \frac{-7}{-7} = 1$. Check: $2(2) + 1 = 5$; $3(2) - 2 = 4$.

Second system: after $L_2 - L_1$ and $L_3 - 2L_1$, the rows become $(1,1,1)$, $(0,-2,0)$, $(0,-1,-3)$, so

$$
\det A = \begin{vmatrix} 1&1&1\\ 1&-1&1\\ 2&1&-1\end{vmatrix}
= 1 \times \begin{vmatrix} -2 & 0\\ -1 & -3\end{vmatrix} = 6 .
$$

Cramer, replacing columns by $(6,2,1)^{\mathsf T}$:

$$
x = \frac{6}{6} = 1, \qquad
y = \frac{12}{6} = 2, \qquad
z = \frac{18}{6} = 3
$$

(the numerators computed the same way). Check: $1 + 2 + 3 = 6$; $1 -
2 + 3 = 2$; $2 + 2 - 3 = 1$.

**Exercise 22.4 ★.**

Solve by [Gaussian elimination](#met-b1-det-gauss), describing the solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets):

$$
\begin{cases}
x + 2y - z + t = 1\\
2x + 4y + z - t = 5\\
x + 2y + 2z - 2t = 4 .
\end{cases}
$$

**Solution of Exercise 22.4.**

Reduce the augmented matrix: $L_2 \leftarrow L_2 - 2L_1$, $L_3
\leftarrow L_3 - L_1$:

$$
\begin{pmatrix}
1 & 2 & -1 & 1 & 1\\
0 & 0 & 3 & -3 & 3\\
0 & 0 & 3 & -3 & 3
\end{pmatrix}
\to
\begin{pmatrix}
1 & 2 & -1 & 1 & 1\\
0 & 0 & 1 & -1 & 1\\
0 & 0 & 0 & 0 & 0
\end{pmatrix}.
$$

Pivot unknowns $x, z$; [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) unknowns $y, t$. Back-substitution: $z =
1 + t$, $x = 1 - 2y + z - t = 2 - 2y$. Solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets):

$$
\{(2 - 2y,\; y,\; 1 + t,\; t) : y, t \in \R\}
= (2, 0, 1, 0) + \operatorname{Vect}\bigl((-2,1,0,0),\,
(0,0,1,1)\bigr),
$$

an affine plane (dimension $2 = 4 - \operatorname{rk} 2$) of $\R^4$.

**Exercise 22.5 ★★.**

Prove the Vandermonde formula of [Example 22.12](#ex-b1-det-vandermonde) by induction on $n$, with the column operations $C_k \leftarrow C_k -
x_1 C_{k-1}$ performed from $k = n$ down to $k = 2$.

**Solution of Exercise 22.5.**

Induction; $n = 1$ is empty product $= 1$. For the step, perform $C_k \leftarrow C_k - x_1 C_{k-1}$ for $k = n, n-1, \dots, 2$ (in this order, so each operation uses a not-yet-modified column). The first row becomes $(1, 0, \dots, 0)$; in row $i \geq 2$, the $k$-th entry becomes $x_i^{k-1} - x_1 x_i^{k-2} = x_i^{k-2}(x_i - x_1)$. Expanding along the first row and factoring $(x_i - x_1)$ out of each row $i$:

$$
V(x_1, \dots, x_n)
= \prod_{i=2}^{n} (x_i - x_1)\cdot V(x_2, \dots, x_n),
$$

and the induction hypothesis completes the product $\prod_{i<j}(x_j
- x_i)$.

**Exercise 22.6 ★★.**

(Tridiagonal) Let $D_n$ be the $n \times n$ determinant with $2$ on the diagonal, $1$ on the two adjacent diagonals, $0$ elsewhere. Expanding along the first row, prove $D_n = 2D_{n-1} - D_{n-2}$ and compute $D_n$ ($D_1 = 2$, $D_2 = 3$).

**Solution of Exercise 22.6.**

Expanding $D_n$ along the first row: $D_n = 2 D_{n-1} -
1\cdot\begin{vmatrix} 1 & \ast\\ 0 & D_{n-2}\text{-block}
\end{vmatrix}$; the second determinant, expanded along its first column, is $D_{n-2}$. So $D_n = 2D_{n-1} - D_{n-2}$, i.e. $D_n -
D_{n-1} = D_{n-1} - D_{n-2}$: the differences are constant, equal to $D_2 - D_1 = 1$. Hence $D_n = D_1 + (n - 1) = n + 1$. (Check: $D_2 = 3$, and the $3\times3$ case is [Example 22.7](#ex-b1-det-cofactor): $D_3 = 4$.)

**Exercise 22.7 ★★.**

Complete [Example 22.18](#ex-b1-det-parameter): compute the determinant $\begin{vmatrix} 1 & 1 & m\\ 1 & m & 1\\ m & 1 & 1\end{vmatrix}$ by the operation $C_1 \leftarrow C_1 + C_2 + C_3$, and carry out the full discussion of the system.

**Solution of Exercise 22.7.**

$C_1 \leftarrow C_1 + C_2 + C_3$ makes the first column constant $(m+2)$; factor it:

$$
\det = (m+2)\begin{vmatrix}
1 & 1 & m\\ 1 & m & 1\\ 1 & 1 & 1
\end{vmatrix}
\overset{L_1 - L_3,\ L_2 - L_3}{=}
(m+2)\begin{vmatrix}
0 & 0 & m-1\\ 0 & m-1 & 0\\ 1 & 1 & 1
\end{vmatrix}
= (m+2)\cdot\bigl(-(m-1)^2\bigr)
$$

(expand along the first column: the single entry $1$ carries sign $+$, and the remaining $2 \times 2$ determinant is $0 \cdot 0 -
(m-1)(m-1) = -(m-1)^2$).

Discussion. $m \notin \{1, -2\}$: Cramer system; by the symmetry of the equations, $x = y = z$, and each equation gives $(m + 2)x = 1$: unique solution $\bigl(\frac{1}{m+2}, \frac{1}{m+2},
\frac{1}{m+2}\bigr)$. $m = 1$: the three equations all read $x + y
+ z = 1$: solutions form the affine plane $x + y + z = 1$. $m = -2$: summing the three equations gives $0 = 3$: empty solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets).

**Exercise 22.8 ★★.**

Let $A \in \mathcal{M}_n(\R)$ with *integer* entries. Prove that $A$ has an inverse with integer entries if and only if $\det A
= \pm 1$. *(For the direct sense, take determinants; for the converse, admit — or prove for $n \leq 3$ via cofactors — that $A^{-1} = \frac{1}{\det A}\,\operatorname{Com}(A)^{\mathsf T}$ with integer cofactor matrix.)*

**Solution of Exercise 22.8.**

($\Rightarrow$) If $A^{-1}$ has integer entries: $\det A \cdot \det
A^{-1} = 1$ with both determinants integers (sums of products of entries): two integers with product $1$ are both $\pm1$.

($\Leftarrow$) The cofactor formula $A^{-1} = \frac{1}{\det
A}\operatorname{Com}(A)^{\mathsf T}$ (checked for $n \leq 3$ by direct expansion, admitted in general) has $\operatorname{Com}(A)$ with integer entries (each cofactor is an integer determinant); dividing by $\det A = \pm 1$ keeps integers.

**Exercise 22.9 ★★★.**

Compute the $n \times n$ determinant of the matrix $aI + bJ$ ([Exercise 21.9](https://one-course.com/books/math/3/en/chapter/21-matrices#exo-b1-matrices-9)), i.e. with $a + b$ on the diagonal and $b$ elsewhere. *(Add all columns to the first, factor, then clear.)* Recover the invertibility condition $a \neq 0$, $a + nb
\neq 0$.

**Solution of Exercise 22.9.**

Add all columns to the first: each entry of the new first column is $a + nb$; factor it out, so that the first column is all ones. Then the [row operations](https://one-course.com/books/math/3/en/chapter/21-matrices#met-b1-matrices-gauss) $L_i \leftarrow L_i - L_1$ ($i \geq 2$) clear every entry below the top-left $1$ and leave $a$ on the diagonal and $0$ elsewhere in those rows: the matrix is upper triangular with diagonal $(1, a, \dots, a)$. Hence

$$
\det(aI + bJ) = (a + nb)\, a^{\,n-1} ,
$$

nonzero iff $a \neq 0$ and $a + nb \neq 0$: the condition of [Exercise 21.9](https://one-course.com/books/math/3/en/chapter/21-matrices#exo-b1-matrices-9).

**Exercise 22.10 ★★★.**

Let $A, B \in \mathcal{M}_n(\R)$. Prove that

$$
\det\begin{pmatrix} A & B\\ B & A \end{pmatrix}
= \det(A + B)\,\det(A - B),
$$

by block column and [row operations](https://one-course.com/books/math/3/en/chapter/21-matrices#met-b1-matrices-gauss) ($C_1 \leftarrow C_1 + C_2$, then $L_2 \leftarrow L_2 - L_1$, in block form), assuming the natural block-triangular rule $\det\begin{pmatrix} M & N\\ 0 &
P\end{pmatrix} = \det M \det P$ — proved for $2 \times 2$ blocks in [Example 22.9](#ex-b1-det-blocktriangular).

**Solution of Exercise 22.10.**

Block operations (each a composition of the corresponding $n$ scalar operations, allowed by [Theorem 22.3](#thm-b1-det-props) (1)):

$$
\begin{vmatrix} A & B\\ B & A\end{vmatrix}
\overset{C_1 \leftarrow C_1 + C_2}{=}
\begin{vmatrix} A + B & B\\ A + B & A\end{vmatrix}
\overset{L_2 \leftarrow L_2 - L_1}{=}
\begin{vmatrix} A + B & B\\ 0 & A - B\end{vmatrix}
= \det(A+B)\,\det(A-B),
$$

using the block-triangular rule for the last step.

**Exercise 22.11 ★★.**

(Circulant of order $3$) Let $a, b, c \in \C$ and

$$
\Delta = \begin{vmatrix}
a & b & c\\
c & a & b\\
b & c & a
\end{vmatrix}.
$$

Prove that $\Delta = (a + b + c)(a^2 + b^2 + c^2 - ab - bc -
ca)$, and factor completely over $\C$ using $j =
\eu^{2\iu\pi/3}$:

$$
\Delta = (a + b + c)(a + jb + j^2c)(a + j^2b + jc) .
$$

*(Start with $C_1 \leftarrow C_1 + C_2 + C_3$; for the complex form, note that the column $(1, j, j^2)^{\mathsf T}$ almost behaves like an eigenvector.)*

**Solution of Exercise 22.11.**

$C_1 \leftarrow C_1 + C_2 + C_3$ makes the first column constant $(a + b + c)$; factor it out, then $L_2 \leftarrow L_2 - L_1$, $L_3 \leftarrow L_3 - L_1$:

$$
\Delta = (a+b+c)\begin{vmatrix}
1 & b & c\\
0 & a - b & b - c\\
0 & c - b & a - c
\end{vmatrix}
= (a+b+c)\bigl[(a-b)(a-c) + (b-c)^2\bigr],
$$

and expanding, $(a-b)(a-c) + (b-c)^2 = a^2 + b^2 + c^2 - ab - bc
- ca$. Over $\C$, with $j^3 = 1$ and $1 + j + j^2 = 0$:

$$
\begin{align*}
(a + jb + j^2c)(a + j^2b + jc)
&= a^2 + b^2 + c^2 + (j + j^2)(ab + bc + ca)\\
&= a^2 + b^2 + c^2 - ab - bc - ca ,
\end{align*}
$$

whence the complete factorization. (Structurally: the column $(1, j, j^2)^{\mathsf T}$ satisfies $M\,(1, j, j^2)^{\mathsf T}
= (a + jb + j^2c)(1, j, j^2)^{\mathsf T}$, and similarly for $j^2$ and $1$: the three factors are the three “eigenvalues” of the circulant, a story systematized in the Year 2 volume.)

**Exercise 22.12 ★★★.**

(Rank and minors) Let $A \in \mathcal{M}_{n,p}(K)$. Prove that $\operatorname{rk} A$ equals the largest size $r$ of an invertible $r \times r$ submatrix of $A$ (a submatrix keeps the entries at the crossings of $r$ chosen rows and $r$ chosen columns). *(If $\operatorname{rk} A = r$, select $r$ [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) columns, then $r$ [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) rows of the resulting $n \times r$ block; conversely, an invertible submatrix forces the corresponding columns of $A$ to be [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free).)*

**Solution of Exercise 22.12.**

Write $r = \operatorname{rk} A$.

*An invertible $r \times r$ submatrix exists.* Choose $r$ [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) columns of $A$ and let $B \in \mathcal{M}_{n,r}$ be the matrix they form: $\operatorname{rk} B = r$. Since row rank equals column rank ([Theorem 21.13](https://one-course.com/books/math/3/en/chapter/21-matrices#thm-b1-matrices-rank)), $B$ has $r$ [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) rows; keeping those rows yields an $r \times r$ submatrix of $A$ of rank $r$, i.e. invertible.

*No larger one does.* Let $S$ be an invertible $s \times s$ submatrix, taken from columns $j_1, \dots, j_s$ and rows $i_1,
\dots, i_s$ of $A$. If a combination $\sum_k \lambda_k
C_{j_k} = 0$ of the corresponding *full* columns vanishes, then reading only the rows $i_1, \dots, i_s$ gives $\sum_k
\lambda_k S_k = 0$ on the columns of $S$, so all $\lambda_k = 0$ ($S$ invertible): the columns $C_{j_1}, \dots, C_{j_s}$ of $A$ are [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), and $s \leq \operatorname{rk} A = r$.

Hence $\operatorname{rk} A$ is exactly the largest size of an invertible submatrix.

## 22.4 Problem: Cauchy’s double alternant

**Problem 22.1.**

Two determinants rule the applications of this chapter: the [Vandermonde determinant](#ex-b1-det-vandermonde), evaluated in [Exercise 22.5](#exo-b1-det-5), and the *[Cauchy determinant](#pb-b1-det-1)* $\det\bigl(\frac{1}{a_i +
b_j}\bigr)$, evaluated here. Around them this problem collects the alternant toolbox: [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) column tricks, interpolation by Cramer, the [Hilbert matrix](#pb-b1-det-1), the discriminant of a cubic, and the method of alternating [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def). Throughout, $V(x_1,
\dots, x_n) = \prod_{i < j}(x_j - x_i)$ denotes the Vandermonde value.

**Part I — The Vandermonde toolbox.**

1. Compute $V(1, 2, 3, 4)$ , and recall why interpolation at $n$ pairwise distinct nodes is a Cramer system.
2. ([Polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) alternant) Let $P_0, \dots, P_{n-1}$ be *[monic](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def)* with $\deg P_k = k$. Prove $$\det\bigl(P_{j-1}(x_i)\bigr)_{1 \leq i, j \leq n}  = V(x_1, \dots, x_n) :$$ column operations replace each power column by any [monic](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) staircase, for free.
3. Apply question 2 to the binomial [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) $B_k =  \frac{X(X-1)\cdots(X-k+1)}{k!}$: prove that for *integers* $m_1 < m_2 < \dots < m_n$, $$\frac{V(m_1, \dots, m_n)}{0!\,1!\,2!\cdots(n-1)!}  \in \N :$$ the product of all pairwise differences of $n$ integers is divisible by the superfactorial $0!\,1!\cdots(n-1)!$.
4. Prove $\det\bigl(x_i^{\,j}\bigr)_{1 \leq i, j \leq n} =  x_1 x_2 \cdots x_n\, V(x_1, \dots, x_n)$ (powers now starting at $1$ ).
5. (Moment matrix) Let $S = \bigl(p_{i+j-2}\bigr)_{1 \leq  i, j \leq n}$ where $p_k = x_1^k + \dots + x_n^k$. Prove that $S = W^{\mathsf T} W$ for the matrix $W =  (x_i^{\,j-1})_{ij}$, deduce $$\det S = V(x_1, \dots, x_n)^2 ,$$ and conclude: $n$ *real* numbers are pairwise distinct if and only if their moment matrix is invertible, and $\det S \geq 0$ always.

**Part II — Interpolation revisited.** Nodes $x_1 < \dots < x_n$, values $y_1, \dots, y_n$.

6. Write the conditions “ $P = c_0 + c_1X + \dots +  c_{n-1}X^{n-1}$ interpolates” as a [linear system](#def-b1-det-system) in the $c_k$ with matrix $W$ , and recover from $\det W = V \neq  0$ the existence and uniqueness of the interpolant (compare the two earlier proofs, [Theorem 8.23](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-lagrange) and [Example 20.10](https://one-course.com/books/math/3/en/chapter/20-linear-maps#ex-b1-linmaps-interpolation) ).
7. By Cramer’s rule and [cofactor expansion](#thm-b1-det-cofactor) of the relevant determinant along its last column, prove that the leading coefficient of the interpolant is $$c_{n-1} = \sum_{i=1}^{n}  \frac{y_i}{\prod_{j \neq i}(x_i - x_j)} .$$
8. (Confluent Vandermonde) Compute $$\begin{vmatrix}  1 & x_1 & x_1^2\\  0 & 1 & 2x_1\\  1 & x_2 & x_2^2  \end{vmatrix}  = (x_2 - x_1)^2 ,$$ and interpret: the data $\bigl(P(x_1), P'(x_1),  P(x_2)\bigr)$ determine a unique $P \in \R_2[X]$ when $x_1 \neq x_2$ (Hermite interpolation).
9. Find the unique $P \in \R_2[X]$ with $P(0) = 1$ , $P'(0)  = 0$ , $P(1) = 2$ , and check your answer against question 8.

**Part III — The [Cauchy determinant](#pb-b1-det-1).** Let $a_1,
\dots, a_n$ and $b_1, \dots, b_n$ be scalars with $a_i + b_j
\neq 0$ for all $i, j$, and

$$
C_n = \det\Bigl(\frac{1}{a_i + b_j}\Bigr)_{1 \leq i, j \leq n} .
$$

10. Compute $C_2$ by hand and put it in the form “products of differences over products of sums”.
11. For $n \geq 2$, perform $L_i \leftarrow L_i - L_n$ ($i <  n$) and factor rows and columns to prove $$C_n = \frac{\prod_{i<n}(a_n - a_i)}{\prod_{j}(a_n +  b_j)}\;\det M,$$ where $M$ agrees with the Cauchy matrix on rows $i < n$ and has last row $(1, 1, \dots, 1)$.
12. Perform $C_j \leftarrow C_j - C_n$ ($j < n$) on $M$, factor again, and conclude by induction *Cauchy’s double alternant*: $$C_n = \frac{\prod_{1 \leq i < j \leq n}(a_j - a_i)(b_j -  b_i)}{\prod_{i, j}(a_i + b_j)} .$$
13. Deduce the invertibility criterion (the $a_i$ pairwise distinct and the $b_j$ pairwise distinct). For the *[Hilbert matrix](#pb-b1-det-1)* $H_n = \bigl(\frac{1}{i + j -  1}\bigr)$ : compute $\det H_2$ and $\det H_3$ from the formula, and verify that $H_2^{-1}$ has integer entries.
14. Show that for pairwise distinct $b_j$ and any right-hand side, the system $\sum_j \frac{c_j}{a_i + b_j} = y_i$ ( $i = 1, \dots, n$ ) has a unique solution, and relate this to the existence and uniqueness of partial fraction decompositions with simple [poles](https://one-course.com/books/math/3/en/chapter/9-rational-fractions#def-b1-fractions-field) ( [Theorem 9.5](https://one-course.com/books/math/3/en/chapter/9-rational-fractions#thm-b1-fractions-complex) ).

**Part IV — The discriminant of a cubic.** Let $\lambda_1, \lambda_2, \lambda_3$ be the roots (in $\C$) of $X^3
+ pX + q$, and $p_k = \lambda_1^k + \lambda_2^k + \lambda_3^k$.

15. Using $\lambda^3 = -p\lambda - q$ at each root and Vieta ( $p_1 = 0$ ), compute $p_2 = -2p$ , $p_3 = -3q$ , and $p_4 = 2p^2$ .
16. With question 5 (over $\C$, keeping $\det S = V^2$), compute $$\operatorname{disc} = V(\lambda_1, \lambda_2,  \lambda_3)^2 = \begin{vmatrix}  3 & 0 & -2p\\  0 & -2p & -3q\\  -2p & -3q & 2p^2  \end{vmatrix}  = -4p^3 - 27q^2 .$$
17. Deduce: $X^3 + pX + q$ has a repeated root if and only if $4p^3 + 27q^2 = 0$ ; check on $X^3 - 3X + 2 = (X -  1)^2(X + 2)$ .
18. Suppose $p, q$ real. Prove that the cubic has three distinct real roots if and only if $\operatorname{disc}  > 0$ , and one real plus two nonreal conjugate roots if and only if $\operatorname{disc} < 0$ . *(If $\lambda_3 = \conj{\lambda_2} \neq \lambda_2$ and $\lambda_1 \in \R$, show $V$ is purely imaginary.)*

**Part V — Dividends, and the alternating method.**

19. For $0 < a_1 < a_2 < \dots < a_n$ , show $\det\bigl(\frac{1}{a_i + a_j}\bigr) > 0$ .
20. Compute $\det\bigl(\binom{m_i}{j-1}\bigr)_{1 \leq i, j  \leq 3}$ for $(m_1, m_2, m_3) = (2, 4, 7)$ , first by questions 2–3, then by direct expansion.
21. Let $\lambda_1, \dots, \lambda_n$ be pairwise distinct and nonzero. Using an invertible Vandermonde matrix, prove again that the geometric sequences $\bigl((\lambda_i^{\,k})_{k \geq 0}\bigr)_{1 \leq i \leq  n}$ form a [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of the space of sequences.
22. Compute $\det\bigl(\frac{1}{i + j}\bigr)_{1 \leq i, j  \leq 3}$ from the double alternant.
23. (Alternating [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) ) Call a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) $F$ in $x_1,  \dots, x_n$ *alternating* when swapping any two variables changes its sign. Show that an alternating $F$ vanishes whenever $x_i = x_j$ ( $i \neq j$ ), and deduce — one variable at a time, by the factor theorem — that $F$ is divisible by $\prod_{i<j}(x_j - x_i)$ .
24. Use question 23 to reprove the Vandermonde formula with no induction: the determinant $\det(x_i^{\,j-1})$ is an alternating [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of total degree $\binom n2$ , hence a *constant* multiple of $\prod_{i<j}(x_j -  x_i)$ ; identify the constant by comparing one monomial.
25. Synthesis, in four sentences: what single property of the determinant (which axiom) generates all the factorizations of this problem; why the moment-matrix identity of question 5 turns a [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) about *complex* distinctness into a computable *real* sign test; which two classical matrices were completely evaluated here and which [linear](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-def) problems they govern; and how the alternating method of questions 23–24 explains, in one stroke, why $\prod_{i<j}(x_j -  x_i)$ keeps appearing. Name the theorem of Part III.

**Solution of Problem 22.1.**

**1.** $V(1,2,3,4) = (2-1)(3-1)(4-1)(3-2)(4-2)(4-3) = 1
\cdot 2\cdot 3\cdot 1\cdot 2\cdot 1 = 12$. Interpolation at distinct nodes asks for the coefficients of $P$ solving $W c =
y$ with $W = (x_i^{\,j-1})$, and $\det W = V \neq 0$: a Cramer system.

**2.** Work through the columns from left to right. $C_1$ is the constant column $P_0(x_i) = 1$ ($P_0$ [monic](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of degree $0$). Assume columns $1, \dots, j-1$ have already been reduced to the pure powers $1, x_i, \dots, x_i^{\,j-2}$. Since $P_{j-1}
= X^{j-1} + \sum_{k < j-1}\alpha_k X^k$, subtracting from $C_j$ the combination $\sum_k \alpha_k\,(\text{column of } x_i^k)$ — an operation that does not change the determinant — leaves the pure power column $x_i^{\,j-1}$. After the last column the matrix is the Vandermonde matrix: $\det = V(x_1, \dots, x_n)$.

**3.** The [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) $(j-1)!\,B_{j-1}$ are [monic](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of degree $j - 1$, so question 2 gives

$$
\det\bigl(B_{j-1}(m_i)\bigr)
= \frac{V(m_1, \dots, m_n)}{0!\,1!\cdots(n-1)!} .
$$

The left side is the determinant of a matrix with *integer* entries ($B_k$ is integer-valued on $\Z$: questions 16–17 of the weekend problem [Problem 18.1](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#pb-b1-vspaces-1)), hence an integer; and it is positive since $V(m_1, \dots, m_n) > 0$ for $m_1 <
\dots < m_n$. So the superfactorial [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) the product of all pairwise differences.

**4.** Factor $x_i$ out of each row $i$: $\det(x_i^{\,j})_{j = 1..n} = x_1\cdots x_n\,
\det(x_i^{\,j-1}) = x_1\cdots x_n\,V$.

**5.** $(W^{\mathsf T}W)_{ij} = \sum_k x_k^{\,i-1}
x_k^{\,j-1} = p_{i+j-2}$: $S = W^{\mathsf T}W$. Hence $\det S =
\det(W^{\mathsf T})\det W = V^2$ ([Theorem 22.3](#thm-b1-det-props) (2),(4)). For real $x_i$: $\det S = V^2
\geq 0$, and $S$ is invertible iff $V \neq 0$ iff the $x_i$ are pairwise distinct — a sign-definite test computable from the power sums alone.

**6.** The interpolation conditions $\sum_{k}
c_k\,x_i^{\,k} = y_i$ form the system $Wc = y$; $\det W = V \neq
0$ gives existence and uniqueness at once. This is the third proof in the book: explicit formula in [Theorem 8.23](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-lagrange), kernel argument in [Example 20.10](https://one-course.com/books/math/3/en/chapter/20-linear-maps#ex-b1-linmaps-interpolation), Cramer here.

**7.** Cramer: $c_{n-1} = \det W'/\det W$ where $W'$ is $W$ with its last column replaced by $y$. Expanding $\det W'$ along that column:

$$
\det W' = \sum_{i=1}^n (-1)^{i+n} y_i\,V(x_1, \dots, \widehat{x_i},
\dots, x_n) .
$$

Now $V = V(\setminus i)\cdot\prod_{j<i}(x_i - x_j)\prod_{j>i}
(x_j - x_i)$, and converting the second product costs $(-1)^{n-i}$:

$$
(-1)^{i+n}\,\frac{V(\setminus i)}{V}
= \frac{(-1)^{i+n}(-1)^{n-i}}{\prod_{j\neq i}(x_i - x_j)}
= \frac{1}{\prod_{j\neq i}(x_i - x_j)} ,
$$

whence $c_{n-1} = \sum_i y_i/\prod_{j \neq i}(x_i - x_j)$ — the divided-difference formula again.

**8.** $L_3 \leftarrow L_3 - L_1$ gives the rows $(1, x_1,
x_1^2)$, $(0, 1, 2x_1)$, $(0,\ x_2 - x_1,\ (x_2-x_1)(x_2+x_1))$; expanding along the first column and factoring $(x_2 - x_1)$:

$$
(x_2 - x_1)\begin{vmatrix} 1 & 2x_1\\ 1 & x_2 + x_1
\end{vmatrix} = (x_2 - x_1)(x_2 - x_1) = (x_2 - x_1)^2 .
$$

Nonzero for $x_1 \neq x_2$: the [linear system](#def-b1-det-system) expressing $P(x_1) = u$, $P'(x_1) = v$, $P(x_2) = w$ on the coefficients of $P \in \R_2[X]$ is Cramer — Hermite interpolation with a doubled node is well posed.

**9.** $P = a + bX + cX^2$ with $a = P(0) = 1$, $b = P'(0)
= 0$, $a + b + c = P(1) = 2$: $c = 1$, so $P = 1 + X^2$, unique. Consistency: here $x_1 = 0$, $x_2 = 1$ and the determinant of question 8 is $(1 - 0)^2 = 1 \neq 0$.

**10.** Direct computation:

$$
C_2 = \frac{1}{(a_1+b_1)(a_2+b_2)} - \frac{1}{(a_1+b_2)(a_2+b_1)}
= \frac{(a_1+b_2)(a_2+b_1) - (a_1+b_1)(a_2+b_2)}
{\prod_{i,j}(a_i+b_j)} ,
$$

and the numerator expands to $a_1b_1 + a_2b_2 - a_1b_2 - a_2b_1
= (a_2 - a_1)(b_2 - b_1)$: differences over sums.

**11.** For $i < n$, the new entry of row $i$ is

$$
\frac{1}{a_i + b_j} - \frac{1}{a_n + b_j}
= \frac{a_n - a_i}{(a_i + b_j)(a_n + b_j)} .
$$

Factor $(a_n - a_i)$ out of each row $i < n$, then $\frac1{a_n + b_j}$ out of each column $j$: what remains has entries $\frac1{a_i + b_j}$ in rows $i < n$ and constant $1$ in row $n$ — the matrix $M$, with the announced prefactor.

**12.** On $M$, for $j < n$ the operation $C_j \leftarrow
C_j - C_n$ turns row $n$ into $(0, \dots, 0, 1)$ and, in row $i
< n$,

$$
\frac{1}{a_i + b_j} - \frac{1}{a_i + b_n}
= \frac{b_n - b_j}{(a_i + b_j)(a_i + b_n)} .
$$

Factor $(b_n - b_j)$ from each column $j < n$ and $\frac1{a_i +
b_n}$ from each row $i < n$, then expand along the last row (sign $(-1)^{n+n} = +1$): the remaining determinant is $C_{n-1}$. Collecting the factors of questions 11–12:

$$
C_n = \frac{\prod_{i<n}(a_n - a_i)\,\prod_{j<n}(b_n - b_j)}
{\prod_{j}(a_n + b_j)\,\prod_{i<n}(a_i + b_n)}\;C_{n-1},
$$

and the induction (base $C_1 = \frac1{a_1+b_1}$) assembles exactly Cauchy’s double alternant: the factors $(a_j - a_i)(b_j
- b_i)$ for all pairs, over all sums $(a_i + b_j)$.

**13.** The formula vanishes iff some $a_j = a_i$ or $b_j =
b_i$: the Cauchy matrix is invertible iff both families are pairwise distinct. Hilbert: $a_i = i$, $b_j = j - 1$. For $n =
2$: numerator $(2-1)(1-0) = 1$, denominator $1\cdot2\cdot2\cdot3
= 12$: $\det H_2 = \frac1{12}$. For $n = 3$: numerator $\bigl[(1)(2)(1)\bigr]^2 = 4$, denominator $(1\cdot2\cdot3)
(2\cdot3\cdot4)(3\cdot4\cdot5) = 6\cdot24\cdot60 = 8640$: $\det H_3 = \frac{4}{8640} = \frac1{2160}$. Inverse for $n = 2$:

$$
H_2^{-1} = 12\begin{pmatrix} \frac13 & -\frac12\\[2pt]
-\frac12 & 1\end{pmatrix}
= \begin{pmatrix} 4 & -6\\ -6 & 12 \end{pmatrix},
$$

all integers (a phenomenon true for every $n$).

**14.** The matrix of the system is the Cauchy matrix, invertible by question 13 when the $b_j$ (and the $a_i$) are pairwise distinct: unique solution. Interpretation: a rational function $R = \sum_j \frac{c_j}{X + b_j}$ with simple [poles](https://one-course.com/books/math/3/en/chapter/9-rational-fractions#def-b1-fractions-field) is determined by $n$ of its values $R(a_1), \dots, R(a_n)$, and conversely any such data sheet is realized exactly once — the sampling counterpart of the partial-fraction existence and uniqueness theorem ([Theorem 9.5](https://one-course.com/books/math/3/en/chapter/9-rational-fractions#thm-b1-fractions-complex)).

**15.** Vieta for $X^3 + pX + q$: $\lambda_1 + \lambda_2 +
\lambda_3 = 0$, $\sum_{i<j}\lambda_i\lambda_j = p$, so $p_1 = 0$ and $p_2 = p_1^2 - 2p = -2p$. Each root satisfies $\lambda^3 =
-p\lambda - q$; summing: $p_3 = -p\,p_1 - 3q = -3q$. Multiplying by $\lambda$ and summing: $p_4 = -p\,p_2 - q\,p_1 = 2p^2$.

**16.** By question 5 (the identity $S = W^{\mathsf T}W$ and $\det S = V^2$ are valid over $\C$),

$$
V^2 = \begin{vmatrix}
3 & 0 & -2p\\
0 & -2p & -3q\\
-2p & -3q & 2p^2
\end{vmatrix}
= 3\bigl(-4p^3 - 9q^2\bigr) + (-2p)\bigl(0 - 4p^2\bigr)
= -4p^3 - 27q^2 ,
$$

expanding along the first row.

**17.** A repeated root means two equal $\lambda_i$, i.e. $V = 0$, i.e. $\operatorname{disc} = -4p^3 - 27q^2 = 0$. For $X^3 - 3X + 2$: $4(-3)^3 + 27\cdot4 = -108 + 108 = 0$, matching the double root $1$ of $(X-1)^2(X+2)$.

**18.** The nonreal roots of a real cubic come in conjugate pairs, so exactly two cases occur when $\operatorname{disc} \neq
0$. Three distinct real roots: $V$ is real and nonzero, so $\operatorname{disc} = V^2 > 0$. One real root $\lambda_1$ and $\lambda_3 = \conj{\lambda_2} \notin \R$: then

$$
(\lambda_2 - \lambda_1)(\lambda_3 - \lambda_1) =
\abs{\lambda_2 - \lambda_1}^2 > 0,
\qquad
\lambda_3 - \lambda_2 = -2\iu\,\operatorname{Im}\lambda_2 \neq 0,
$$

so $V$ is a nonzero purely imaginary number and $\operatorname{disc} = V^2 < 0$. The two signs characterize the two cases.

**19.** Take $b_i = a_i$ in the double alternant: the numerator is $\prod_{i<j}(a_j - a_i)^2 > 0$ and the denominator $\prod_{i,j}(a_i + a_j) > 0$ (all entries positive): the determinant is positive. (In later language: the kernel $\frac1{x+y}$ is positive definite.)

**20.** By questions 2–3, the determinant equals $V(2,4,7)/(0!\,1!\,2!) = \frac{(4-2)(7-2)(7-4)}{2} =
\frac{30}{2} = 15$. Directly, the matrix is

$$
\begin{pmatrix}
1 & 2 & 1\\
1 & 4 & 6\\
1 & 7 & 21
\end{pmatrix},
\qquad
\det = (84 - 42) - 2(21 - 6) + (7 - 4) = 42 - 30 + 3 = 15 .
$$

**21.** Suppose $\sum_i c_i\,(\lambda_i^{\,k})_{k} = 0$ as a sequence. Reading off $k = 0, 1, \dots, n-1$ gives $W^{\mathsf
T}c = 0$ with $W = (\lambda_i^{\,j-1})$ invertible ($\det = V
\neq 0$, distinct $\lambda_i$): $c = 0$. The geometric sequences are [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free).

**22.** $a = b = (1, 2, 3)$: numerator $\bigl[(2-1)(3-1)
(3-2)\bigr]^2 = 4$; denominator $\prod_{i,j}(i + j) =
(2\cdot3\cdot4)(3\cdot4\cdot5)(4\cdot5\cdot6) = 24\cdot60\cdot120
= 172800$. Hence $\det\bigl(\frac1{i+j}\bigr) = \frac{4}{172800}
= \frac1{43200}$.

**23.** If $x_i = x_j$, the swap of the two variables fixes the point but must change the sign of $F$: $F = -F$, so $F = 0$ there. [Divisibility](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides): view $F$ as a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) in the single variable $x_n$ with coefficients in the other variables; it vanishes at the $n - 1$ “values” $x_1, \dots, x_{n-1}$, so repeated factoring ([Theorem 8.7](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-factor)) gives $F =
\prod_{i<n}(x_n - x_i)\cdot G$ with $G$ [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def). The prefactor is invariant under swaps of two indices $i, j < n$, so $G$ is alternating in $x_1, \dots, x_{n-1}$, and induction completes: $\prod_{i<j}(x_j - x_i)$ [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) $F$.

**24.** $D = \det(x_i^{\,j-1})$ is a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) in the $x_i$; swapping two variables swaps two rows, so $D$ is alternating, and by question 23, $D = c\,\prod_{i<j}(x_j - x_i)$ for some [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) $c$. Total degrees: $D$ has degree $\leq 0 +
1 + \dots + (n-1) = \binom n2$, the product has degree exactly $\binom n2$: $c$ is a constant. The monomial $x_2\,x_3^2\cdots
x_n^{\,n-1}$ has coefficient $1$ in $D$ (diagonal product) and $1$ in the product (choose the larger-index variable in every factor): $c = 1$, and Vandermonde’s formula falls out with no induction.

**25.** (i) Alternation — the axiom “two equal columns kill the determinant” — is the engine: it produced every factor $(x_j - x_i)$, $(a_j - a_i)$, $(b_j - b_i)$ in the problem. (ii) The identity $\det S = V^2$ replaces the individually complex, unreachable roots by their power sums, which are real [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) in the coefficients, so distinctness becomes the sign of a computable real number. (iii) The [Vandermonde determinant](#ex-b1-det-vandermonde) governs [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) interpolation, the [Cauchy determinant](#pb-b1-det-1) governs partial fractions and sampled rational functions (with the [Hilbert matrix](#pb-b1-det-1) as its most famous special case). (iv) Any alternating [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) is divisible by $\prod_{i<j}(x_j - x_i)$, and a degree count then pins such a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) down up to a constant — which is why this product keeps reappearing wherever a determinant vanishes on coincidences. The theorem of Part III is *Cauchy’s double alternant*.