Linear Systems & Matrices

A system of two linear equations asks for the point $(x,y)$ that satisfies both equations at once. There are three standard methods:

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• Graphing: plot both lines; the solution is where they cross.
• Substitution: solve one equation for a variable, then substitute into the other.
• Elimination: add or subtract multiples of the equations to cancel one variable.

Slope is rise over run.

Compare the two lines:

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• Different slopes $\Rightarrow$ lines cross once $\Rightarrow$ one solution (consistent, independent).
• Same slope, different intercepts $\Rightarrow$ parallel lines $\Rightarrow$ no solution (inconsistent).
• Same slope and same intercept $\Rightarrow$ same line $\Rightarrow$ infinitely many solutions (dependent).

For a system of linear inequalities, the solution set is the region where all shaded half-planes overlap.

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• Graph each boundary line: use a solid line for $\le$ or $\ge$, a dashed line for $<$ or $>$.
• Shade the side that makes each inequality true (use a test point such as $(0,0)$ if it is not on the line).
• The overlap of all shadings is the solution region.

A system in $x,y,z$ needs three equations. To solve:

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• Pick a variable to eliminate. Combine two equations to remove it, then combine a different pair to remove the same variable. This leaves a $2\times2$ system.
• Solve that $2\times2$ system for the remaining two variables.
• Back-substitute to find the third variable.
• Always check the solution in all three original equations.

A matrix is a rectangular array of numbers. Its order (dimension) is $\text{rows} \times \text{columns}$; the entry in row $i$, column $j$ is $a_{ij}$.

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• Equal matrices: same order and every matching entry is equal.
• Addition/subtraction: only for matrices of the same order; add or subtract entry by entry.
• Scalar multiplication: multiply every entry by the scalar $k$.

The product $AB$ is defined only when the number of columns of $A$ equals the number of rows of $B$:

Each entry of $AB$ is a row of $A$ dotted with a column of $B$: multiply matching entries and add. Matrix multiplication is not commutative: usually $AB \ne BA$.

For a $2\times2$ matrix, the determinant is

For a $3\times3$ matrix, expand along the first row (note the alternating $+\,-\,+$ signs):

For $\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$ form the coefficient determinant

Replace the $x$-column with the constants to get $D_x$, and the $y$-column to get $D_y$:

Then, provided $D \ne 0$,   $x = \dfrac{D_x}{D}, \quad y = \dfrac{D_y}{D}.$ If $D=0$, Cramer's Rule does not apply (no unique solution).

To model a word problem:

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• Name each unknown with a variable (one per unknown quantity).
• Translate each sentence of information into one equation. You need as many equations as unknowns.
• Solve by any method, then interpret the answer in context and check it against the words.

A system is symmetric if swapping $x \leftrightarrow y$ leaves it unchanged, e.g. $\begin{cases} x+y = s \\ xy = p \end{cases}$. The two quantities

are the elementary symmetric functions. Any symmetric expression rewrites in terms of $s$ and $p$; the key identities are

Once $s$ and $p$ are known, $x$ and $y$ are the roots of the single quadratic

When at least one equation is a conic (circle, parabola, ellipse, hyperbola), count solutions geometrically:

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• Line meets a conic: substitute the line into the conic to get one quadratic. The discriminant tells you $2$ points (secant), $1$ point (tangent), or $0$ points (miss).
• Two conics: up to $4$ intersection points. Subtract the two equations to cancel the quadratic terms when possible --- this often leaves a line (the radical axis for two circles), reducing the problem to a line--conic case.

For a square matrix $A$, the inverse $A^{-1}$ satisfies $A A^{-1} = A^{-1} A = I$. It exists precisely when $\det A \ne 0$. The adjugate formula is

where $\operatorname{adj}(A)$ is the transpose of the cofactor matrix. Building it takes three steps:

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• Cofactors: $C_{ij} = (-1)^{i+j} M_{ij}$, where the minor $M_{ij}$ is the determinant of $A$ with row $i$ and column $j$ deleted.
• Adjugate: transpose the matrix of cofactors ($\operatorname{adj}(A)_{ij} = C_{ji}$).
• Scale: divide every entry by $\det A$.

Once $A^{-1}$ is known, the system $A\mathbf{x} = \mathbf{b}$ has the one-line solution $\mathbf{x} = A^{-1}\mathbf{b}$.

These rules let you evaluate determinants without brute force:

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• Row/column operations: adding a multiple of one row to another does not change $\det$; swapping two rows multiplies $\det$ by $-1$; scaling a row by $k$ scales $\det$ by $k$.
• Zero determinant: a repeated row, a zero row, or one row a multiple of another all force $\det = 0$.
• Products & transposes: $\det(AB) = \det A \,\det B$, $\det(A^{\mathsf T}) = \det A$, $\det(A^{-1}) = 1/\det A$, $\det(kA) = k^n \det A$ for an $n\times n$ matrix.
• Vandermonde: $\det\begin{bmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{bmatrix} = (b-a)(c-a)(c-b).$
• Rank-one update: $\det(I + aJ)$, where $J$ is the $n\times n$ all-ones matrix, equals $1 + na$ (since $J$ has eigenvalues $n$ and $0$).

A square matrix $A$ is singular (non-invertible) exactly when $\det A = 0$. For the system $A\mathbf{x}=\mathbf{b}$ this means no unique solution --- either no solution or infinitely many. Equivalent warning signs of singularity:

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• the rows (or columns) are linearly dependent --- one is a combination of the others;
• a row reduces to all zeros during elimination;
• Cramer's Rule breaks because $D = \det A = 0$.

To find the parameter that makes a matrix singular, set its determinant equal to zero and solve.

Repeated multiplication $A^n = A\cdot A \cdots A$ ($n$ factors) models step-by-step processes (Markov chains, Fibonacci-type recurrences). Two shortcuts:

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• Diagonal matrices power entrywise: $\operatorname{diag}(a,b)^n = \operatorname{diag}(a^n, b^n)$.
• Cayley--Hamilton: every matrix satisfies its own characteristic equation. For a $2\times2$ matrix, $A^2 = (\operatorname{tr}A)\,A - (\det A)\,I$, which lets you reduce any power $A^n$ to the form $\alpha A + \beta I$.

The Fibonacci matrix illustrates the recurrence link:

Sometimes a problem asks only for a combination of the unknowns --- not each one. Look for a shortcut before grinding out a full solution:

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• Add all equations. If the coefficients of $x,y,z$ each sum to the same value $c$, then adding gives $c(x+y+z) = (\text{sum of constants})$, so $x+y+z$ pops out directly.
• Take a linear combination of the equations designed to produce exactly the target combination.