Матрицы, системы, определительMatrices, systems, and determinants

Opening: a system whose answer depends on a letter

In school examples, a system is usually solved by substitution, and you immediately look for a pair of numbers without asking separately “how many solutions are there?” Put a letter in the coefficients: $ax + y = 1$, $x + ay = 1$. From the first equation, $y = 1-ax$; substituting into the second gives $(1-a^2)x = 1-a$. In the general case, you want to divide by $1-a^2$, but at $a = \pm1$ this coefficient is zero — and those two values must be analyzed separately. At $a = 1$, both equations become $x + y = 1$, giving infinitely many solutions. At $a = -1$, they become $-x + y = 1$ and $x - y = 1$; adding them gives $0 = 2$, so there are no solutions. For every other $a$, the solution is unique. Three outcomes instead of one — that is the content of the day, with three tools: Gaussian elimination (universal but slow), rank (how much genuine, non-repeated information the system contains), and the determinant (one number that answers “is the solution unique?” before the system itself is solved).

Matrix and system

A matrix of size $m \times n$ is a table of numbers with $m$ rows and $n$ columns. Its purpose is to record a system compactly: $2x + y = 5$, $x - 3y = -1$ becomes $A\mathbf{x} = \mathbf{b}$ with rows $(2,1)$ and $(1,-3)$, column $\mathbf{x} = (x, y)^{T}$, and right-hand side $\mathbf{b} = (5, -1)^{T}$. This works because the $i$th coordinate of $A\mathbf{x}$ is the dot product of the matrix’s $i$th row with $\mathbf{x}$ — yesterday’s operation applied row by row. Matrix multiplication follows the same rule: $(AB)_{ij}$ is the dot product of row $i$ of $A$ and column $j$ of $B$. Therefore $A$ must have as many columns as $B$ has rows, and generally $AB \ne BA$, even where both products exist. Transposition $A^{T}$ flips the table across its main diagonal.

Gaussian elimination

The idea is to turn a system into row-echelon form, where each successive equation contains fewer unknowns, and then work upward from the bottom. We operate not on equations but on the augmented matrix $[A \mid \mathbf{b}]$ — the system matrix with the right-hand-side column appended: there is no need to rewrite $x$, $y$, and $z$ at every step, because all information is in the numbers. Exactly three operations are allowed, and none changes the solution set: swap two rows; multiply a row by a nonzero number; add to one row another row multiplied by a number.

Example. System $x + y + z = 6$, $2x + 3y - z = 5$, $x - y + 2z = 5$. Subtract twice the first row from the second and the first from the third, then add twice the second to the third:

The last row reads $-5z = -15$, so $z = 3$. Back-substitution: $y - 9 = -7$, hence $y = 2$; then $x + 2 + 3 = 6$, hence $x = 1$. Check by substituting into the second equation: $2 + 6 - 3 = 5$ ✓.

The third case is often skipped at school, but it appears constantly on exams.

Rank and the Rouché–Capelli theorem

The row $0 = 0$ does not appear by accident: it means one equation added no new information and was generated by the others.

The row $0 = c$ with $c \ne 0$ is literally $\rank A < \rank[A \mid \mathbf{b}]$: the appended column added an independent row, which $A$ did not have. Take $x + 2y + 3z = 1$, $2x + 4y + 6z = \beta$, $x + 3y + 5z = 3$. The second row of $A$ is twice the first, so $\rank A = 2$ for every $\beta$. At $\beta = 2$, the right side also doubles and the second equation is entirely redundant; the ranks match, there are $3 - 2 = 1$ free variables, and the solutions form the line $(-3 + t,\ 2 - 2t,\ t)$. At $\beta = 3$, the second equation contradicts the first, the augmented rank becomes 3, and there are no solutions.

Determinant

For a square matrix, one number answers the question of whether the solution is unique. For $2\times2$ it is $ad - bc$. For $3\times3$, expand along a row: multiply each entry by the determinant of the matrix left after crossing out its row and column, and alternate the signs. For the matrix with rows $(2,1,0)$, $(1,3,1)$, $(0,2,4)$:

You may expand along any row or column; the sign at position $(i,j)$ is $(-1)^{i+j}$, a checkerboard pattern beginning with plus in the upper-left corner. Four properties save time:

• $\det A^{T} = \det A$, so columns can be treated just like rows;
• adding a multiple of one row to another does not change the determinant; swapping two rows changes its sign, and multiplying a row by $\lambda$ multiplies the entire determinant by $\lambda$;
• if rows are linearly dependent (one is proportional to another or equals a sum of the others), the determinant is zero;
• $\det(AB) = \det A \cdot \det B$.

The second property means you can calculate a determinant with the same Gaussian elimination: reduce the matrix to triangular form using additions only, and it equals the product of the remaining diagonal entries. In the Gaussian-elimination example, the diagonal entries are $1$, $1$, and $-5$; the determinant is $-5$, not zero, which agrees with the unique solution.

Vandermonde. One structure is worth recognizing at a glance: if each row contains powers of the same node from zero through $(n-1)$, so the rows look like $(1,\ x_i,\ x_i^2,\ \ldots)$, this is a Vandermonde matrix, and its determinant is $\prod_{i<j}(x_j - x_i)$ — the product of all node differences. Since $\det A^{T} = \det A$, the same is true when this pattern appears in columns rather than rows. The consequence matters more than the formula: it is zero if and only if two nodes coincide.

Geometrically, $|\det A|$ is how much the transformation $\mathbf{x} \mapsto A\mathbf{x}$ scales area or volume; zero means space has been flattened into a line or a point, information has been lost, and the original vector can no longer be recovered uniquely. When $\det A \ne 0$, an inverse matrix $A^{-1}$ exists with $AA^{-1} = I$, and the solution is $\mathbf{x} = A^{-1}\mathbf{b}$. For $2\times2$, it is $\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}$. There is also a free test: if in every row the diagonal entry’s absolute value is strictly greater than the sum of the absolute values of the other entries (strict diagonal dominance), the matrix is nonsingular.

Parameterized system: the day’s main storyline

The fourth point is where marks are lost. Determinant roots cannot be divided in advance into “bad” and “good”: one may give an inconsistent system and another infinitely many solutions, and only substitution distinguishes them. That is exactly what the model problem below demonstrates.

Gram matrix

This is the thread that puts today immediately before the YMLS exam. In this type of problem, several vectors are given, their lengths and some pairwise dot products are known, the vectors are linearly dependent, and a missing product must be found. There are no coordinates and nowhere to introduce them; the direct route of finding decomposition coefficients is long.

Why this is true is already visible for two vectors: $\det G = \|u\|^2\|v\|^2 - \langle u,v\rangle^2 = \|u\|^2\|v\|^2(1 - \cos^2\varphi) = (\|u\|\,\|v\|\sin\varphi)^2$, exactly the squared area of the parallelogram. It is nonnegative and zero precisely when $\sin\varphi = 0$, meaning the vectors are collinear — the same equality case from Cauchy–Schwarz. In dimension $k$, the reasoning is the same: dependent vectors lie in a lower-dimensional subspace, so their “volume” is zero. The calculation is therefore: call the unknown product a letter and put it in the two symmetric cells of the table; $\det G$ becomes a polynomial in that letter; dependence gives $\det G = 0$; for three vectors, this is a quadratic equation.

Remember one special fact: in $\mathbb{R}^n$, any $n+1$ vectors are linearly dependent. Therefore, in the planar case $\det G = 0$ for three vectors automatically; there is no need to prove it — use it.

Rank as an argument about size

The final storyline of the day uses rank not to calculate something but to prove something. Three facts are needed.

• $\rank(AB) \le \min(\rank A, \rank B)$: a product cannot contain more information than either factor.
• $\rank(A^{T}A) = \rank A$ for every real matrix. From $A^{T}A\mathbf{x} = 0$ we get $\mathbf{x}^{T}A^{T}A\mathbf{x} = \|A\mathbf{x}\|^2 = 0$, hence $A\mathbf{x} = 0$; the reverse implication is obvious. $A$ and $A^{T}A$ have the same solution set for the homogeneous system, and therefore the same number of free variables and the same rank. The same argument gives $\rank(AA^{T}) = \rank A^{T} = \rank A$.
• If $J$ is a square matrix of ones and $D$ is diagonal with positive entries, then $J + D$ is nonsingular: for any nonzero $\mathbf{x}$, $\mathbf{x}^{T}(J + D)\mathbf{x} = \left(\sum_i x_i\right)^2 + \sum_i d_i x_i^2$ is strictly positive, so $(J + D)\mathbf{x} = 0$ is impossible.

Together these give a technique: encode the configuration in the statement with a zero–one matrix and see what $AA^{T}$ and $A^{T}A$ mean. If both have the form “ones plus a positive diagonal,” both are nonsingular and their ranks equal their sizes — so two different sizes must be equal because both equal $\rank A$. This turns a combinatorial problem into an equality between two numbers. That is the content of s1001.

What should stay in your head

1. A matrix encodes a system; multiplication by a vector is a row-wise collection of dot products.
2. Gaussian elimination uses the augmented matrix and three operations; read the outcome from the last row: echelon form, $0 = c$, or $0 = 0$.
3. $\rank A$ is the number of nonzero rows in echelon form, and $\rank A^{T} = \rank A$; Rouché–Capelli says consistency is $\rank A = \rank[A\mid\mathbf{b}]$, uniqueness is equality of this rank with the number of unknowns, and the number of free variables is $n - \rank A$.
4. Determinant: $ad - bc$, expansion along a row with alternating signs, and triangular Gaussian form as a second method; zero means dependent rows, while $\det A \ne 0$ means a unique solution, but only for a square system.
5. Recognize Vandermonde by powers of the same nodes in rows (or equivalently columns), and calculate it as the product of differences $\prod_{i<j}(x_j - x_i)$.
6. Parameter: calculate the determinant, factor it, treat all other values as one case, and analyze each root separately by hand.
7. Gram matrix: diagonal of squared lengths, $\det G = 0 \iff$ vectors are dependent, and $\det G$ is squared volume; use the problem’s condition to distinguish the two roots.
8. $\rank(AB) \le \min(\rank A, \rank B)$ and $\rank(A^{T}A) = \rank A$ are proof tools, not calculation tools.