Part I · Mathematical Language, Functions, and Algebraic Models

Chapter 4 — Linear Systems and Matrix Methods

Why This Matters

One equation describes one relationship. A circuit node, a network of constraints, or a polynomial fitted to data usually produces several relationships at once. A linear system lets us solve them together and, just as importantly, tells us when the data are inconsistent or leave freedom in the answer.

4.1 Linear systems as models

A linear equation in variables $x_1,\ldots,x_n$ has the form

a_1x_1+a_2x_2+\cdots+a_nx_n=b.

The variables occur only to the first power and are not multiplied by each other. A linear system consists of several such equations.

Engineering Connection — node voltages

At a circuit node, Kirchhoff’s current law says that the algebraic sum of currents is zero. If each branch current is written with Ohm’s law, unknown node voltages often appear linearly. The mathematics does not replace circuit reasoning: it organises the relationships produced by it.

Keep every constraint in view. A simultaneous solution is one ordered list of unknowns that makes every equation true at once. Each equation can be read as a line or plane of allowed points; solving a system finds their shared set rather than solving each relation independently.

x+y=3,\qquad2x-y=0

Write one equation for each independent constraint.
Use an operation that preserves the full common solution set.
Substitute the candidate back into every original equation, not only the last simplified row.

4.2 Elimination and back-substitution

Follow one row operation at a time. The augmented matrix below represents the same system throughout. Advance the stage only after reading the operation: the last column is transformed with the coefficient columns because it belongs to the equation, not because it is an answer column.

Start
[ 1  1 | 3 ]
[ 2 −1 | 0 ]

R₂ ← R₂ − 2R₁
[ 1  1 | 3 ]
[ 0 −3 | −6 ]

R₂ ← −⅓R₂
[ 1  1 | 3 ]
[ 0  1 | 2 ]

R₁ ← R₁ − R₂
[ 1  0 | 1 ]
[ 0  1 | 2 ]

A two-variable augmented matrix reduced by three reversible row operations.

Consider

\begin{aligned} 2x+y&=7,\\ x-y&=2. \end{aligned}

The $$y$$ -coefficients are already opposites, so add the two equations directly:

\[(2x+y)+(x-y)=7+2.\]

The $$y$$ terms cancel, leaving $$3x=9$$ , hence $$x=3$$ . Substituting into the second equation gives $$3-y=2$$ , so $$y=1$$ . The cancellation was deliberate: we chose an operation that removes one unknown while preserving the solution set.

The guiding idea is to replace equations by equivalent equations that remove one variable. Valid row operations are:

Exchange two equations.
Multiply one equation by a nonzero constant.
Add a multiple of one equation to another.

They preserve the solution set because each new equation is a logical consequence of the old system and the operation can be reversed.

From first principles. Replacing one equation by a non-zero multiple, or adding a multiple of one equation to another, does not change the pair of values that satisfies both equations. Elimination is therefore a solution-preserving rewrite, not a numerical trick.

Keep the first equation unchanged.
Replace the second equation by the second equation minus twice the first.
Solve the resulting one-variable equation, then substitute back.

Two linear equations as crossing lines and their shared solution.

4.3 Augmented matrices

The coefficients and right-hand sides can be recorded compactly as an augmented matrix:

\left[\begin{array}{cc|c} 2&1&7\\ 1&-1&2 \end{array}\right].

The vertical bar separates the coefficient matrix from the constants. Matrix notation reduces bookkeeping; it does not change the underlying equations.

The bar keeps constants attached to their equations. An augmented matrix is a compact transcription of a system, not a new problem. Every row operation changes all entries in its row, including the right-hand side, because the row continues to stand for one complete equation.

\left[\begin{array}{cc|c}1&1&3\\2&-1&0\end{array}\right]

Align coefficients in one fixed variable order.
Place the constants after the augmented bar.
Apply any row operation to the complete row, including its final entry.

4.4 Row-echelon form, pivots, and rank

After elimination, a system is easiest to read in row-echelon form: each successive nonzero row begins farther to the right. The first nonzero entry in a nonzero row is a pivot. The number of pivots in the coefficient matrix is its rank, denoted $$r$$ .

For a consistent system with $$n$$ unknowns:

$$r=n$$ means every variable is determined: there is one solution.
$$r<n$$ means $$n-r$$ variables are free: there are infinitely many solutions.

An augmented row of the form

[0\quad0\quad\cdots\quad0\mid b],\qquad b\ne0,

means $$0=b$$ , a contradiction. The system is inconsistent and has no solution.

Example — Free variable and vector form

Solve

\begin{aligned} x+2y-z&=4,\\ 2x+4y-2z&=8. \end{aligned}

The second equation is twice the first, so it adds no new restriction. Let $$y=s$$ and $$z=t$$ be free. Then

\[x=4-2s+t.\]

The complete solution is

\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}4\\0\\0\end{bmatrix} +s\begin{bmatrix}-2\\1\\0\end{bmatrix} +t\begin{bmatrix}1\\0\\1\end{bmatrix}.

The first vector is one particular solution. The other vectors describe allowed directions of change that keep the equations true.

What a pivot records. A pivot is not merely a leading number. It marks a variable direction that the equations constrain independently. A row of zeros contributes no new constraint; a row with only a non-zero right-hand side records an impossibility. Rank counts the independent constraint directions that remain.

Move a non-zero candidate into the next pivot position.
Use reversible row operations to clear entries below it.
Compare the pivots in the coefficient and augmented columns before declaring a solution type.

4.5 Matrix-vector form

For

\begin{aligned} a_{11}x_1+a_{12}x_2&=b_1,\\ a_{21}x_1+a_{22}x_2&=b_2, \end{aligned}

we write

\underbrace{\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}_{A} \underbrace{\begin{bmatrix}x_1\\x_2\end{bmatrix}}_{\mathbf x} =\underbrace{\begin{bmatrix}b_1\\b_2\end{bmatrix}}_{\mathbf b}.

Thus $A\mathbf x=\mathbf b$ . Matrix multiplication means that $A\mathbf x$ is a linear combination of the columns of $$A$$ . Solving the system asks whether $\mathbf b$ can be made from those columns, and if so, in how many ways.

Columns describe available directions. In A x=b, multiplying by the unknown vector forms a weighted combination of the columns of A. A solution exists precisely when the target vector can be assembled from those columns; multiple solutions occur when different weights assemble the same target.

A\mathbf{x}=\mathbf{b}

Two matrix columns combined to reach a target vector, showing the geometry behind A x equals b.

Common Mistakes

Swapping or scaling coefficients without doing the same operation to the right-hand side.
Dividing by a pivot before checking that it is nonzero; swap rows first when possible.
Saying “rank $$<n$$ means infinite solutions” without first checking consistency.
Calling a parameter an unknown to be solved. A free parameter labels a family of valid solutions.

4.6 Elimination is reversible bookkeeping

Replacing equation 2 by equation 2 minus three times equation 1 preserves every solution: any solution of the old equations satisfies the new equation, and adding three times equation 1 back reverses the operation. Scaling is reversible only by a nonzero number, which is why multiplying a row by zero is not allowed.

Example — complete elimination with a row swap

Solve

\begin{aligned} y+z&=3,\\ x-y+z&=1,\\ 2x+y-z&=4. \end{aligned}

Use the second equation as pivot row:

\left[\begin{array}{ccc|c} 1&-1&1&1\\ 0&1&1&3\\ 2&1&-1&4 \end{array}\right] \xrightarrow{R_3-2R_1} \left[\begin{array}{ccc|c} 1&-1&1&1\\ 0&1&1&3\\ 0&3&-3&2 \end{array}\right].

Then $R_3\leftarrow R_3-3R_2$ gives $$-6z=-7$$ , so $$z=7/6$$ , $$y=11/6$$ , and $$x=5/3$$ . Substitution into the original system is the final check, not optional decoration.

Example — rank distinguishes none from infinitely many

The forms

\left[\begin{array}{cc|c}1&0&2\\0&1&-1\end{array}\right],\quad \left[\begin{array}{cc|c}1&2&3\\0&0&0\end{array}\right],\quad \left[\begin{array}{cc|c}1&2&3\\0&0&1\end{array}\right]

mean, respectively, one solution, infinitely many solutions, and no solution. In the middle case rank is $$1<2$$ and the system is consistent, so one variable is free. In the last case the augmented column creates $$0=1$$ .

[Figure: Two lines crossing, coinciding, and parallel; a companion sketch of two planes intersecting in a line to motivate a free parameter.]

Each row operation has an undo. From x+y=3 and 2x-y=0, replace the second row by the second minus twice the first, scale its non-zero pivot, then clear the first row. Back-substitution is visible in the resulting identity matrix: the same solution satisfies the starting equations because every rewrite was reversible.

\left[\begin{array}{cc|c}1&1&3\\2&-1&0\end{array}\right]\to\left[\begin{array}{cc|c}1&1&3\\0&-3&-6\end{array}\right]\to\left[\begin{array}{cc|c}1&0&1\\0&1&2\end{array}\right]

R two becomes R two minus two R one, giving negative three y equals negative six.
Scale R two by negative one third, giving y equals two.
Replace R one by R one minus R two, giving x equals one, then verify both original equations.

Start
[ 1  1 | 3 ]
[ 2 −1 | 0 ]

R₂ ← R₂ − 2R₁
[ 1  1 | 3 ]
[ 0 −3 | −6 ]

R₂ ← −⅓R₂
[ 1  1 | 3 ]
[ 0  1 | 2 ]

R₁ ← R₁ − R₂
[ 1  0 | 1 ]
[ 0  1 | 2 ]

Textual augmented matrices show each reversible row operation from a two-equation system to x equals one and y equals two.

Chapter Summary

Linear systems model simultaneous constraints. Elimination produces a triangular or echelon form from which solutions can be read. Pivots reveal rank, consistency, and the number of free variables. Matrix-vector notation makes the geometry and bookkeeping of larger systems manageable.

Exercises

Solve by elimination: $$3x+2y=12$$ , $$x-y=1$$ .
Reduce the augmented matrix $\left[\begin{smallmatrix}1&2&-1&|&3\\2&4&1&|&8\end{smallmatrix}\right]$ and classify its solution set.
Write a two-node circuit model using node voltages $$v_1,v_2$$ if currents through resistors are specified by voltage differences divided by resistance.
A polynomial $$p(x)=a+bx+cx^2$$ passes through $$(0,1)$$ , $$(1,2)$$ , and $$(2,5)$$ . Form and solve the linear system for $$a,b,c$$ .
Solve the system in Section 4.6 by augmented-matrix operations.
Give an augmented matrix representing a $2\times2$ system with no solution.
A node at voltage $$v$$ connects through $1\ \mathrm{k}\Omega$ to $$5$$ V and through $2\ \mathrm{k}\Omega$ to ground. Form KCL and solve for $$v$$ .