Systems of Linear Equations and Matrices
Anton, Rorres, & Kaul §§1.1–1.3, 1.10
Assignment 1 due 9/7
October 9, 2026
\[\begin{multline*} \left\{ \begin{aligned} x+2y-z&=2\\ 2x-y+z&=-3\\ 3x+y+2z&=1 \end{aligned} \right. \;\longrightarrow\; [\mat{A}\mid\vct{b}]= \left[\begin{array}{rrr|r} 1&2&-1&2\\ 2&-1&1&-3\\ 3&1&2&1 \end{array}\right] \\[-0.2em] \longrightarrow\; [\mat{U}\mid\vct{c}]= \left[\begin{array}{rrr|r} 1&2&-1&2\\ 0&-5&3&-7\\ 0&0&2&2 \end{array}\right] \;\longrightarrow\; \begin{aligned} x&=-1\\ y&=2\\ z&=1 \end{aligned} \end{multline*}\]
\[ 2x=-1 \]
One solution
\[x=-\tfrac12\]
\[ 0x=-1 \]
No solutions
\[0=-1\]
\[ 0x=0 \]
\(\infty\) many solutions
\[x\text{ arbitrary}\]
We transform equations in ways that are reversible and so preserve the solution set
For the second and third examples, no reversible transformation can isolate \(x\): a zero coefficient remains zero
A system of \(m\) linear equations in \(n\) unknowns has exactly one of three outcomes:
Solve a system by replacing it with an easier system having exactly the same solutions
\[\begin{align*} x+y&=3\\ 3x-y&=5 \end{align*}\]
Exactly one
\[\{(2,1)\}\]
\[\begin{align*} x+y&=3\\ x+y&=4 \end{align*}\]
None
\[\varnothing\]
\[\begin{align*} x+y&=3\\ 2x+2y&=6 \end{align*}\]
Infinitely many
\[\{(3-t,t):t\in\reals\}\]
The relationship between the two equations explains each outcome
\(\begin{aligned}x+y&=3\\3x-y&=5\end{aligned}\)
\(\begin{aligned}x+y&=3\\x+y&=4\end{aligned}\)
\(\begin{aligned}x+y&=3\\2x+2y&=6\end{aligned}\)
Suppose \((x,y)\) satisfies
\[\begin{align*} x+y&=3 &(E_1)\\ 3x-y&=5 & (E_2) \end{align*}\]
Then it also satisfies
\[\begin{align*} x+y&=3 &(E_1)\\ -4y&=-4 & (E_2)-3 (E_1) \end{align*}\]
And the original system of equations can be recovered because
\[E_2^{\mathrm{old}}=E_2^{\mathrm{new}}+3E_1\]
The new system contains the same information because the step can be undone, i.e., it is reversible
| Transformation of the equations | Inverse transformation |
|---|---|
| Interchange \(E_i\) and \(E_j\) | interchange them again |
| Replace \(E_i\) by \(E_i+cE_j\), \(i\ne j\) | replace it by \(E_i-cE_j\) |
| Replace \(E_i\) by \(cE_i\), \(c\ne0\) | multiply by \(1/c\) |
These elementary row operations preserve the entire solution set because they are reversible
The restriction \(c\ne0\) for the last move matters: multiplying an equation by \(0\) discards information
The last move is not really needed for solving equations
\(\exstar\) Decide whether each transformation preserves the solution set for every system
\(E_3\leftarrow 0E_3\)
\(E_1\leftrightarrow E_3\)
\(E_2\leftarrow E_2-2E_1\)
\(E_2\leftarrow E_1+E_2\), then discard \(E_1\)
\(E_1\leftarrow -3E_1\)
\(E_1 \leftarrow E_1^2\)
For each reversible step, name its inverse
We will return to one system throughout the lecture:
\[\begin{align*} x+2y-z&=2 \qquad &(E_1)\\ 2x-y+z&=-3 &(E_2)\\ 3x+y+2z&=1 &(E_3) \end{align*}\]
Goal: create equations with fewer unknowns while preserving every solution
Elimination produces the equivalent triangular system, which is easy to solve
\[\begin{align*} x+2y-z&=2\\ -5y+3z&=-7\\ 2z&=2 \end{align*}\]
Keep \(E_1\) and use it to remove \(x\) from the equations below:
\[ \begin{aligned} x+2y-z&=2\\ 2x-y+z&=-3\\ 3x+y+2z&=1 \end{aligned} \quad \xrightarrow[\ E_3\leftarrow E_3-3E_1\ ]{\ E_2\leftarrow E_2-2E_1\ } \quad \begin{aligned} x+2y-z&=2\\ -5y+3z&=-7\\ -5y+5z&=-5 \end{aligned} \]
Each arrow abbreviates a reversible transformation
\[ E_2^{\mathrm{old}}=E_2^{\mathrm{new}}+2E_1, \qquad E_3^{\mathrm{old}}=E_3^{\mathrm{new}}+3E_1 \]
\[ \begin{aligned} x+2y-z&=2\\ -5y+3z&=-7\\ -5y+5z&=-5 \end{aligned} \quad \xrightarrow{\ E_3\leftarrow E_3-E_2\ } \quad \begin{aligned} x+2y-z&=2\\ -5y+3z&=-7\\ 2z&=2 \end{aligned} \]
\[\begin{align*} 2z&=2 & \implies z& = \frac{2}{2} = 1 \\ -5y+3z& =-7 &\implies y &= \frac{-7 - 3\times 1}{-5} = 2 \\ x+2y-z & =2 & \implies x & = 2 - (2\times 2 - 1)=-1 \end{align*}\]
The original system has the unique solution \((x,y,z)=(-1,2,1)\)
Eliminate \(x\) below the first equation
\(\implies\) an equivalent system
Eliminate \(y\) below the second equation
\(\implies\) an equivalent triangular system
Back substitution reveals the solution
Elimination does not
Elimination does expose information already present
Each system at each step has exactly the same solution set because every elimination step is reversible
The dot product connects algebraic computation with geometric length, angle, and orthogonality
For \(\vct{a},\vct{x}\in\reals^n\), the dot product
\[ \vct{a}^{\top}\vct{x} =a_1x_1+\cdots+a_nx_n \]
Example:
\[ \begin{bmatrix}2 & -1 & 3\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} =2x-y+3z \]
A dot product turns two vectors into one scalar by pairing corresponding entries
The standard dot product on \(\reals^n\) is our first inner product—the model for measuring length, angle, and orthogonality
For nonzero \(\vct{a},\vct{x}\in\reals^n\), the angle between them is \(\theta\in[0,\pi]\)
The dot product defines the Euclidean norm and measures alignment:
\[\begin{align*} \lVert\vct{a}\rVert &= \sqrt{\vct{a}^{\top}\vct{a}} = \sqrt{a_1^2+\cdots+a_n^2},\\ \vct{a}^{\top}\vct{x} &= \lVert\vct{a}\rVert\,\lVert\vct{x}\rVert\cos(\theta) \end{align*}\]
Two vectors are orthogonal when \(\vct{a}^{\top}\vct{x}=0\)
Nonzero orthogonal vectors are perpendicular
The three points \(\vct{0},\vct{a},\vct{x}\) lie in a plane or along a line
The planar Law of Cosines gives
\[ \lVert\vct{a}-\vct{x}\rVert^2 =\lVert\vct{a}\rVert^2+\lVert\vct{x}\rVert^2 -2\lVert\vct{a}\rVert\lVert\vct{x}\rVert\cos(\theta) \]
Coordinate algebra gives
\[\begin{align*} \lVert\vct{a}-\vct{x}\rVert^2 &=(\vct{a}-\vct{x})^{\top}(\vct{a}-\vct{x})\\ &=\lVert\vct{a}\rVert^2+\lVert\vct{x}\rVert^2 -2\vct{a}^{\top}\vct{x} \end{align*}\]
Comparing the two expressions proves
\[ \vct{a}^{\top}\vct{x} =\lVert\vct{a}\rVert\lVert\vct{x}\rVert\cos(\theta) \]
In \(\reals^n\), geometry explains the formula
In a general real inner-product space, the formula defines the angle
Nonzero vectors \(\vct{a}\) and \(\vct{x}\) are parallel when there is a scalar \(\lambda\ne0\) such that
\[ \vct{x}=\lambda\vct{a} \]
Parallel vectors are nonzero scalar multiples of one another
For nonzero \(\vct{a}\in\reals^n\), let \(\vct{x}\) range over \(\reals^n\). The solution set
\[ H=\{\vct{x}\in\reals^n:\vct{a}^{\top}\vct{x}=b\} \]
is an \((n-1)\)-dimensional hyperplane with normal vector \(\vct{a}\)
Choose two particular points \(\vct{u},\vct{t}\in H\). Then
\[ \vct{a}^{\top}(\vct{u}-\vct{t}) =\vct{a}^{\top}\vct{u}-\vct{a}^{\top}\vct{t} =b-b=0 \]
In \(\reals^2\), a hyperplane is a line
The normal vector \(\vct{a}\) is orthogonal to every direction \(\vct{u}-\vct{t}\) within the hyperplane
\[ \va_1^\top \vx = b_1 \qquad \va_2^\top \vx = b_2 \]
Suppose \(\vct{a}_2=\lambda\vct{a}_1\) with \(\lambda\ne0\)
Same plane is common to both equations
Distinct parallel planes: no point satisfies both equations
Write each matrix row as the transpose of a coefficient vector:
\[\begin{align*} \begin{matrix} x + 2y -z = 2 \\ 2x - y + z = -3 \\ 3x + y + 2z = 1 \end{matrix} \iff \begin{matrix} \vct{a}_1^{\mathsf T}\vct{x}=2 &\vct{a}_1^{\mathsf T}=\begin{bmatrix}1&2&-1\end{bmatrix}\\ \vct{a}_2^{\mathsf T}\vct{x}=-3 &\vct{a}_2^{\mathsf T}=\begin{bmatrix}2&-1&1\end{bmatrix}\\ \vct{a}_3^{\mathsf T}\vct{x}=1 &\vct{a}_3^{\mathsf T}=\begin{bmatrix}3&1&2\end{bmatrix} \end{matrix} \quad \vct{x}=\begin{bmatrix}x\\y\\z\end{bmatrix} \end{align*}\]
Stack the coefficient rows to get
\[ \mA \vx = \vb \]
where
\[ \mat{A}= \begin{bmatrix} 1&2&-1\\ 2&-1&1\\ 3&1&2 \end{bmatrix}, \qquad \vct{b}=\begin{bmatrix}2\\-3\\1\end{bmatrix} \]
\[ \mat{A}\vct{x} = \begin{bmatrix} \vct{a}_1^{\mathsf T}\vct{x}\\ \vct{a}_2^{\mathsf T}\vct{x}\\ \vdots\\ \vct{a}_m^{\mathsf T}\vct{x} \end{bmatrix} \qquad \vct{a}_i^{\mathsf T}=\text{row }i\text{ of }\mat{A} \]
Thus
\[ \mat{A}\vct{x}=\vct{b} \qquad\Longleftrightarrow\qquad \vct{a}_i^{\mathsf T}\vct{x}=b_i \quad(i=1,\ldots,m) \]
Matrix–vector multiplication: every matrix row takes a dot product with the vector
\(\mat{A}=[\vct{A}_1,\ldots,\vct{A}_n]\)
Entries of \(\vct{x}\) — weights on the columns
\[ \mat{A}\vct{x} =x_1\vct{A}_1+\cdots+x_n\vct{A}_n \]
Running solution \((x,y,z)=(-1,2,1)\)
\[ -\begin{bmatrix}1\\2\\3\end{bmatrix} +2\begin{bmatrix}2\\-1\\1\end{bmatrix} +\begin{bmatrix}-1\\1\\2\end{bmatrix} =\begin{bmatrix}2\\-3\\1\end{bmatrix} =\vct{b} \]
Given \(\vct{b}\), solve \(\mat{A}\vct{x}=\vct{b}\) for column weights that add up to \(\vct{b}\)
For \(\mat{A}\in\reals^{m\times n}\), use uppercase vector symbols for columns and lowercase vector symbols for rows
\[\begin{gather*} \mat{A}=[\vct{A}_1,\ldots,\vct{A}_n] \qquad \mat{A}=\begin{bmatrix}\vct{a}_1^{\mathsf T}\\ \vdots\\ \vct{a}_m^{\mathsf T}\end{bmatrix} \end{gather*}\]
and likewise for \(\mat{B}\in\reals^{n\times p}\) and \(\mat{C}=\mat{A}\mat{B}\in\reals^{m\times p}\)
\[\begin{align*} [\vct{C}_1,\ldots,\vct{C}_p] = \mat{C} = \mat{A}\mat{B} &= \mat{A}[\vct{B}_1,\ldots,\vct{B}_p] = [\mat{A}\vct{B}_1,\ldots,\mat{A}\vct{B}_p] \end{align*}\]
\[\begin{align*} \begin{bmatrix}\vct{c}_1^{\mathsf T}\\ \vdots\\ \vct{c}_m^{\mathsf T}\end{bmatrix} = \mat{C} = \mat{A}\mat{B} &= \begin{bmatrix}\vct{a}_1^{\mathsf T}\\ \vdots\\ \vct{a}_m^{\mathsf T}\end{bmatrix}\mat{B} = \begin{bmatrix}\vct{a}_1^{\mathsf T}\mat{B}\\ \vdots\\ \vct{a}_m^{\mathsf T}\mat{B}\end{bmatrix} \end{align*}\]
\[\begin{align*} \begin{bmatrix}c_{11}&\cdots&c_{1p}\\ \vdots&&\vdots\\ c_{m1}&\cdots&c_{mp}\end{bmatrix} = \mat{C} = \mat{A}\mat{B} &= \begin{bmatrix} \vct{a}_1^{\mathsf T}\vct{B}_1 & \cdots & \vct{a}_1^{\mathsf T}\vct{B}_p\\ \vdots & & \vdots\\ \vct{a}_m^{\mathsf T}\vct{B}_1 & \cdots & \vct{a}_m^{\mathsf T}\vct{B}_p \end{bmatrix} \end{align*}\]
Matrix–matrix multiplication collects all dot products of rows of the first matrix with columns of the second
It is associative but generally not commutative: compatible factors may be regrouped, not reordered
\[\begin{align*} a_{11}x_1+\cdots+a_{1n}x_n&=b_1\\ &\ \vdots\\ a_{m1}x_1+\cdots+a_{mn}x_n&=b_m \end{align*}\]
becomes
\[ \mat{A}\vct{x}=\vct{b}, \qquad \mat{A}\in\reals^{m\times n}, \quad \vct{x}\in\reals^n, \quad \vct{b}\in\reals^m \]
The compact notation records all \(m\) equations at once
\[ \left. \begin{aligned} x+2y-z&=2\\ 2x-y+z&=-3\\ 3x+y+2z&=1 \end{aligned} \quad\right\} \qquad\longleftrightarrow\qquad \left[\mat{A}\mid\vct{b}\right] = \left[ \begin{array}{rrr|r} 1&2&-1&2\\ 2&-1&1&-3\\ 3&1&2&1 \end{array} \right] \]
Row operations on \([\mat{A}\mid\vct{b}]\) are compact versions of reversible operations on the equations
Compute with rows first; reinterpret the same steps as reversible matrix actions later
\[ [\,\mat{A}\mid\vct{b}\,] = \left[\begin{array}{rrr|r} 1&2&-1&2\\2&-1&1&-3\\3&1&2&1 \end{array}\right] \;\longrightarrow\; \left[\begin{array}{rrr|r} 1&2&-1&2\\0&-5&3&-7\\0&0&2&2 \end{array}\right] =[\,\mat{U}\mid\vct{c}\,] \]
\([\,\mat{U}\mid\vct{c}\,]\) represents the triangular system \(\mat{U}\vct{x}=\vct{c}\)
Back substitution then solves for \(\vct{x}\)
Start with
\[\begin{align*} x+2y-z&=2 &&(E_1)\\ 2x-y+z&=-3 &&(E_2)\\ 3x+y+2z&=1 &&(E_3) \end{align*}\]
Apply \(E_2\leftarrow E_2-2E_1\):
\[ -5y+3z=-7 \]
The same coefficients and right-hand sides give
\[\begin{gather*} \left[\begin{array}{rrr|r} 1&2&-1&2\\2&-1&1&-3\\3&1&2&1 \end{array}\right] \\[-0.2em] \overset{R_2\leftarrow R_2-2R_1}{\Downarrow} \\[-0.2em] \left[\begin{array}{rrr|r} 1&2&-1&2\\0&-5&3&-7\\3&1&2&1 \end{array}\right] \end{gather*}\]
\(E_i\) names an equation; \(R_i\) names the corresponding matrix row
Many sequences of reversible operations produce a correct echelon form
For each candidate column, moving left to right:
\[ E_i\leftarrow E_i-mE_j \qquad\text{or}\qquad R_i\leftarrow R_i-mR_j \]
where \(m\) is the entry to clear divided by the pivot
You may choose any correct sequence of reversible operations
Valid choices include unnecessary row exchanges or a different combination of scaling and replacement steps
Our examples follow the same recipe every time so the steps are predictable rather than matters of preference
For a solve, operate directly on \([\,\mat{A}\mid\vct{b}\,]\)
This is closest to the computation carried out by software
Continue the same elimination with compact matrix notation:
\[\begin{align*} &\left[ \begin{array}{rrr|r} 1&2&-1&2\\2&-1&1&-3\\3&1&2&1 \end{array} \right] \xrightarrow[\ R_3\leftarrow R_3-3R_1\ ]{\ R_2\leftarrow R_2-2R_1\ } \left[ \begin{array}{rrr|r} 1&2&-1&2\\0&-5&3&-7\\0&-5&5&-5 \end{array} \right]\\[0.8em] &\hspace{7em}\xrightarrow{\ R_3\leftarrow R_3-R_2\ } \left[ \begin{array}{rrr|r} 1&2&-1&2\\0&-5&3&-7\\0&0&2&2 \end{array} \right] \end{align*}\]
For any \(m\times n\) matrix, including an augmented matrix:
\[ \left[ \begin{array}{rrr|r} \alert{1}&2&-1&2\\ 0&\alert{-5}&3&-7\\ 0&0&\alert{2}&2 \end{array} \right] \]
This is row echelon form
For a consistent system, pivots in the variable columns count independent constraints; each pivot column identifies one pivot variable
Pivot entries need not equal \(1\) or lie on the main diagonal
An \(m\times n\) matrix is in reduced row-echelon form (RREF) when
\[ \begin{bmatrix} 1&2&1&3\\ 0&0&1&1\\ 0&0&0&0 \end{bmatrix} \xrightarrow{R_1\leftarrow R_1-R_2} \left[ \begin{array}{cccc} \bbox[3px,border:2px solid teal]{\begin{matrix}1\\0\\0\end{matrix}} &\begin{matrix}2\\0\\0\end{matrix} &\bbox[3px,border:2px solid teal]{\begin{matrix}0\\1\\0\end{matrix}} &\begin{matrix}2\\1\\0\end{matrix} \end{array} \right] \]
Here the boxed pivot columns are \(1,3\); nonpivot columns \(2,4\) remain nonzero
Gauss–Jordan is the reduction process; RREF is its resulting form
Scale pivots to \(1\); clear above and below each pivot in its column
The first entry cannot serve as a pivot
\[\begin{align*} \left[ \begin{array}{rrr|r} 0&1&1&3\\ 1&1&-1&0\\ 2&-1&1&3 \end{array} \right] &\xrightarrow{\ R_1\leftrightarrow R_2\ } \left[ \begin{array}{rrr|r} 1&1&-1&0\\ 0&1&1&3\\ 2&-1&1&3 \end{array} \right] \end{align*}\]
Now the entry in position \((1,1)\) can organize elimination below it
When a pivot candidate is zero, exchange rows if a nonzero entry lies below it
Otherwise skip that column
\(\exstar\) After the exchange, which row operation eliminates the \(2\) below the new first pivot?
\[\begin{align*} \left[ \begin{array}{rrr|r} 1&2&-1&2\\0&-5&3&-7\\0&0&2&2 \end{array} \right] &\qquad\longleftrightarrow\qquad \begin{aligned} x+2y-z&=2\\ -5y+3z&=-7\\ 2z&=2 \end{aligned}\\ &\qquad\implies\qquad z=1,\quad y=2,\quad x=-1 \end{align*}\]
Every variable column contains a pivot
No variable can be chosen freely, so the consistent system has exactly one solution
Suppose elimination instead produced one of these echelon forms
\[ \left[ \begin{array}{rrr|r} 1&2&-1&2\\ 0&-5&3&-7\\ 0&0&0&2 \end{array} \right] \]
A bottom row represents the equation \(0=2\)
Inconsistent — no solutions
\[ \left[ \begin{array}{rrr|r} 1&2&-1&2\\ 0&-5&3&-7\\ 0&0&0&0 \end{array} \right] \]
A bottom row represents the equation \(0=0\)
Consistent, but one variable has no pivot
Undoing the same elimination steps gives two possible original systems
\[\begin{align*} x+2y-z&=2\\ 2x-y+z&=-3\\ 3x+y&=1 \end{align*}\]
The first two equations imply \(3x+y=-1\), which conflicts with the third
\[\begin{align*} x+2y-z&=2\\ 2x-y+z&=-3\\ 3x+y&=-1 \end{align*}\]
The third equation is the sum of the first two
Reversible row operations let us move between the original planes and the bottom-row evidence
\(P_1\) and \(P_2\) meet along \(L\), but \(P_3\cap L=\varnothing\)
All three planes contain the same solution line \(L\)
For the consistent echelon system
\[\begin{align*} x+2y-z&=2\\ -5y+3z&=-7\\ 0&=0, \end{align*}\]
The \(z\) column has no pivot, so \(z\) is the free variable:
\[ y=\frac{7+3z}{5}, \qquad x=\frac{-4-z}{5} \]
\[ \left\{ \left(\frac{-4-z}{5},\frac{7+3z}{5},z\right):z\in\reals \right\} \]
Different values of \(z\) give infinitely many solutions
After elimination:
| Row structure | What it means | Number of solutions |
|---|---|---|
| \([\,0\ \cdots\ 0\mid c\,]\), \(c\ne0\) | contradiction | none |
| consistent; pivot in every variable column | every variable determined | exactly one |
| consistent; fewer pivots than variables | one or more parameters | infinitely many |
A linear system has zero, one, or infinitely many solutions—never exactly two
\(\exstar\) Why would two distinct solutions force infinitely many?
Let \(r\) be the number of pivots in the \(n\) variable columns. For a consistent system, there are
\[ \underbrace{r}_{\text{pivot variables}} \qquad\text{and}\qquad \underbrace{n-r}_{\text{free variables}} \]
The pivot variables are solved in terms of the free variables, so every variable is uniquely determined exactly when \(r=n\)
With \(m\) equations, at most \(m\) variable columns can contain pivots. Hence
\[ m<n \quad\implies\quad r\le m<n \quad\implies\quad n-r>0 \]
If \(m<n\), the system has no solutions or infinitely many solutions—never exactly one
A \(9\,\mathrm{V}\) source drives three coupled current loops
Current is conserved at junctions
Add the voltage drops around each clockwise loop, with current in \(\mathrm{mA}\)
For the first loop: \(1I_1+3(I_1-I_2)=9\)
\[ \begin{matrix} 4I_1 & -3I_2 & & =9\\ -3I_1 & +7I_2 & -2I_3 & =0\\ & -2I_2 & +4I_3 & =0 \end{matrix} \]
\(\exstar\) Verify by Gaussian elimination that
\[ (I_1,I_2,I_3)=(3.6,1.8,0.9)\,\mathrm{mA} \]
Three loop equations determine three mesh currents
Set the first top resistance and the left vertical resistance to zero
The first loop now contains a \(9\,\mathrm{V}\) source but no resistance
\(0I_1+0(I_1-I_2)=9\)
\[ \begin{matrix} 0I_1 & & & =9\\ & 4I_2 & -2I_3 & =0\\ & -2I_2 & +4I_3 & =0 \end{matrix} \]
The first row is already the contradiction
\[ [\,0\quad0\quad0\mid9\,] \]
The idealized short-circuit model is inconsistent
Matrix multiplication gives an algebraic representation of the same row operations in the language of vectors and matrices
For our example, the two elimination stages are
\[\begin{align*} \mat{M}_2\mat{M}_1[\,\mat{A}\mid\vct{b}\,] &= \begin{bmatrix} 1&0&0\\0&1&0\\0&-1&1 \end{bmatrix} \begin{bmatrix} 1&0&0\\-2&1&0\\-3&0&1 \end{bmatrix} \left[\begin{array}{rrr|r} 1&2&-1&2\\2&-1&1&-3\\3&1&2&1 \end{array}\right]\\[0.6em] &= \left[\begin{array}{rrr|r} 1&2&-1&2\\0&-5&3&-7\\0&0&2&2 \end{array}\right] =[\,\mat{U}\mid\vct{c}\,] \end{align*}\]
Let an arbitrary \(3\times3\) matrix be written in terms of its rows and its nine entries:
\[ \mat{A}= \begin{bmatrix} \vct{a}_1^{\mathsf T}\\ \vct{a}_2^{\mathsf T}\\ \vct{a}_3^{\mathsf T} \end{bmatrix} = \begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{bmatrix} \]
To perform \(R_2\leftarrow R_2-mR_1\), multiply on the left:
\[ \begin{bmatrix} 1&0&0\\ -m&1&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} \vct{a}_1^{\mathsf T}\\ \vct{a}_2^{\mathsf T}\\ \vct{a}_3^{\mathsf T} \end{bmatrix} = \begin{bmatrix} \vct{a}_1^{\mathsf T}\\ \vct{a}_2^{\mathsf T}-m\vct{a}_1^{\mathsf T}\\ \vct{a}_3^{\mathsf T} \end{bmatrix} \]
An elementary matrix performs one elementary row operation by left multiplication
Here it changes row \(2\) and leaves rows \(1\) and \(3\) unchanged
The same left multiplication changes every entry in row \(2\) and no other row:
\[ \begin{bmatrix} 1&0&0\\ -m&1&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{bmatrix} = \begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}-ma_{11}&a_{22}-ma_{12}&a_{23}-ma_{13}\\ a_{31}&a_{32}&a_{33} \end{bmatrix} \]
The row-vector and entrywise views describe the same row operation
For our augmented matrix, the first elimination stage combines two row-operation matrices:
\[ \overset{R_3\leftarrow R_3-3R_1}{ \begin{bmatrix} 1&0&0\\0&1&0\\-3&0&1 \end{bmatrix} } \overset{R_2\leftarrow R_2-2R_1}{ \begin{bmatrix} 1&0&0\\-2&1&0\\0&0&1 \end{bmatrix} } = \mat{M}_1 = \begin{bmatrix} 1&0&0\\-2&1&0\\-3&0&1 \end{bmatrix} \]
Left multiplication now performs both row operations:
\[ \mat{M}_1[\,\mat{A}\mid\vct{b}\,] = \left[\begin{array}{rrr|r} 1&2&-1&2\\0&-5&3&-7\\0&-5&5&-5 \end{array}\right] \]
\(\mat{M}_1\) is an elimination matrix: it groups the row operations that zero the first column below the pivot
The second elimination stage uses \(R_3\leftarrow R_3-R_2\):
\[ \mat{M}_2= \begin{bmatrix} 1&0&0\\0&1&0\\0&-1&1 \end{bmatrix} \]
Therefore
\[ \mat{M}_2\mat{M}_1[\,\mat{A}\mid\vct{b}\,] = \left[\begin{array}{rrr|r} 1&2&-1&2\\0&-5&3&-7\\0&0&2&2 \end{array}\right] =[\,\mat{U}\mid\vct{c}\,] \]
\(\mat{M}_j\) groups all row additions that zero column \(j\) below its pivot
\([\,\mat{U}\mid\vct{c}\,]\) represents \(\mat{U}\vct{x}=\vct{c}\); back substitution now solves for \(\vct{x}\)
Undo the first elimination stage with \(R_2\leftarrow R_2+2R_1\) and \(R_3\leftarrow R_3+3R_1\):
\[ \mat{N}_1= \begin{bmatrix} 1&0&0\\2&1&0\\3&0&1 \end{bmatrix} \]
Indeed, \(\mat{I}\) is the identity matrix, and
\[ \mat{N}_1\mat{M}_1=\mat{I} = \begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix} \]
Associativity shows that the whole system is restored:
\[ \mat{N}_1\bigl(\mat{M}_1[\,\mat{A}\mid\vct{b}\,]\bigr) =(\mat{N}_1\mat{M}_1)[\,\mat{A}\mid\vct{b}\,] =[\,\mat{A}\mid\vct{b}\,] \]
Inverses will call \(\mat{N}_1\) the inverse of \(\mat{M}_1\)
First suppose every successive diagonal pivot candidate is nonzero, so no row exchanges or skipped pivot columns are needed. For \(\mat{A}\in\reals^{m\times n}\), the last possible elimination stage is
\[ q=\min(m-1,n) \]
Because \(\mat{M}_j\) zeros column \(j\) below row \(j\),
\[ \mat{M}_q\cdots\mat{M}_1 [\,\mat{A}\mid\vct{b}\,] = [\,\mat{U}\mid\vct{c}\,] \]
This product records the elimination algebraically
It need not be formed to solve the system
The rightmost matrix \(\mat{M}_1\) acts first
For a square or wide matrix (\(m\le n\)), \(q=m-1\)
A tall matrix (\(m > n\)) may also need \(\mat{M}_n\) because column \(n\) can have entries below row \(n\)
Matrix multiplication is associative: factors may be regrouped
It is generally not commutative: factors may not be reordered
Equations show the details
Augmented matrices compress the same information for computation
Elimination matrices group row operations algebraically
Reversible row operations preserve the solution set
Dot products express individual equations
Matrix multiplication collects many dot products and composes actions
Pivots organize elimination and reveal whether a system has zero, one, or infinitely many solutions
Compute only what the problem requires
Use computation to carry out row operations—not to replace interpretation
Use the Lecture 01 companion notebook for executable examples
Companion notebook workflow
Standard kernel: qmcpy
Elimination answers a concrete question:
\[ \text{Which }\vct{x}\text{ satisfy }\mat{A}\vct{x}=\vct{b}\text{?} \]
It also opens structural questions:
© 2026 Fred J. Hickernell · Illinois Tech · assisted by ChatGPT and Codex · Systems and Matrices · MATH 332 — Fall 2026 Website · \(\exstar\) = exercise