Exam Practice

Comprehensive cheat sheet, quiz bank, and flashcards covering Lay's Linear Algebra (5th Ed) Chapters 1 & 2 (Sections 1.1-1.9 and 2.1-2.5).

Reference text: Linear Algebra and Its Applications (5th Edition), David C. Lay, Steven R. Lay, Judi J. McDonald

1. Systems & Row Echelon Forms (Lay §1.1 – §1.2)

Form	Defining Characteristics	Pivot Positions
Row Echelon Form (REF)	1. All non-zero rows are above any all-zero rows. 2. Each leading entry (pivot) of a row is strictly to the right of the leading entry of the row above it. 3. All entries in a column below a leading entry are zeros.	Leading non-zero entries in each non-zero row.
Reduced Row Echelon Form (RREF)	Satisfies all REF conditions PLUS: 1. The leading entry in each non-zero row is $1$ . 2. Each leading $1$ is the only non-zero entry in its entire column (zeros above and below).	Each pivot is $1$ , and pivot columns contain zeros in all other rows.

Linear System Solvability & Variables (Lay §1.2)

System State	Echelon Form Condition	Solution Set
Inconsistent (No Solution)	Rightmost column of augmented matrix contains a pivot position: $[0 \; 0 \; \cdots \; 0 \mid b]$ with $b \ne 0$ .	Empty set $\emptyset$
Consistent - Unique Solution	No row $[0 \; \cdots \; 0 \mid b]$ with $b \ne 0$ AND every column of coefficient matrix has a pivot ( $0$ free variables).	Exactly one unique solution vector $\mathbf{x}$
Consistent - Infinitely Many	Consistent system with at least one non-pivot column in coefficient matrix ( $\ge 1$ free variable).	Parametric solution set with free variables as parameters

Row Reduction Terminology (Lay §1.1 – §1.2)

Elementary Row Operations1. Replacement:

R_i \leftarrow R_i + c R_j

2. Interchange:

R_i \leftrightarrow R_j

3. Scaling:

R_i \leftarrow c R_i

(

c \ne 0

). All operations are reversible.

Row Equivalence (

A \sim B

)Two matrices are row equivalent if one can be transformed into the other by a sequence of elementary row operations. They share the identical solution set.

Basic vs Free VariablesBasic variables correspond to pivot columns in coefficient matrix

A

. Free variables correspond to non-pivot columns.

Uniqueness of RREF (Theorem 1)Each matrix is row equivalent to one and only one reduced echelon matrix (RREF is unique). REF is not unique.

2. Vector Equations & Span (Lay §1.3 – §1.4)

Linear CombinationGiven vectors

\mathbf{v}_1, \dots, \mathbf{v}_p

and scalars

c_1, \dots, c_p

, the vector

\mathbf{y} = c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p

is a linear combination.

\operatorname{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_p\}

The collection of all linear combinations

c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p

. In

\mathbb{R}^3

, span of 1 non-zero vector is a line through

\mathbf{0}

; span of 2 non-parallel vectors is a plane through

\mathbf{0}

Matrix Equation

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

A = [\mathbf{a}_1 \; \cdots \; \mathbf{a}_n]

, then

A\mathbf{x} = x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n

. Thus

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

has a solution iff

\mathbf{b} \in \operatorname{Span}\{\mathbf{a}_1, \dots, \mathbf{a}_n\}

Row-Vector Product RuleThe

i

-th entry of

A\mathbf{x}

is the dot product of row

i

A

and vector

\mathbf{x}

(A\mathbf{x})_i = a_{i1}x_1 + \cdots + a_{in}x_n

Theorem 4: Equivalence of Spanning $\mathbb{R}^m$ (Lay §1.4)

Statement (For any $m \times n$ matrix $A$ )	Meaning & Operational Test
Statement A	For each $\mathbf{b} \in \mathbb{R}^m$ , the equation $A\mathbf{x} = \mathbf{b}$ has a solution.
Statement B	Each $\mathbf{b} \in \mathbb{R}^m$ is a linear combination of the columns of $A$ .
Statement C	The columns of $A$ span $\mathbb{R}^m$ ( $\operatorname{Span}\{\mathbf{a}_1, \dots, \mathbf{a}_n\} = \mathbb{R}^m$ ).
Statement D (Operational Test)	$A$ has a pivot position in every row (i.e. every row has a leading entry in echelon form).

3. Solution Sets: Homogeneous vs Non-homogeneous (Lay §1.5 – §1.6)

Property	Homogeneous: $A\mathbf{x} = \mathbf{0}$	Non-homogeneous: $A\mathbf{x} = \mathbf{b}$ ( $\mathbf{b} \ne \mathbf{0}$ )
Trivial Solution	Always has $\mathbf{x} = \mathbf{0}$ (always consistent).	Never has $\mathbf{x} = \mathbf{0}$ as a solution.
Non-trivial Solutions	Exists if and only if the equation has at least one free variable.	May be inconsistent if augmented column $[A \mid \mathbf{b}]$ has a pivot.
Parametric Vector Form	$\mathbf{x} = s\mathbf{u} + t\mathbf{v}$ (passes through origin $\mathbf{0}$ ).	$\mathbf{x} = \mathbf{p} + s\mathbf{u} + t\mathbf{v}$ , where $\mathbf{p}$ is a particular solution ( $A\mathbf{p} = \mathbf{b}$ ) and $s\mathbf{u} + t\mathbf{v}$ solves $A\mathbf{x} = \mathbf{0}$ .
Geometric Meaning	Line or plane passing through the origin $\mathbf{0}$ .	Line or plane parallel to homogeneous solution set, shifted by vector $\mathbf{p}$ .

Applications of Linear Systems (Lay §1.6)

Network Flow (Junction Rule)At each intersection:

\sum \text{Flow In} = \sum \text{Flow Out}

. For the entire network:

\text{Total Flow In} = \text{Total Flow Out}

Chemical BalancingAtoms of each chemical element on reactant side = atoms on product side. Formulated as a homogeneous system

A\mathbf{x} = \mathbf{0}

for integer weights

x_i > 0

4. Linear Independence Quick Tests (Lay §1.7)

Condition / Scenario	Conclusion	Reason / Theorem
General set $\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$	Linearly independent iff $c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p = \mathbf{0}$ has ONLY the trivial solution $c_1 = \cdots = c_p = 0$ .	Definition: $A\mathbf{x} = \mathbf{0}$ has only the trivial solution $\iff A$ has a pivot in every column.
Two vectors $\{\mathbf{v}_1, \mathbf{v}_2\}$	Linearly dependent iff one is a scalar multiple of the other.	If neither vector is a scalar multiple of the other, they are linearly independent.
Set contains zero vector $\mathbf{0}$	Always linearly dependent.	Theorem 9: $1\mathbf{0} + 0\mathbf{v}_2 + \cdots + 0\mathbf{v}_p = \mathbf{0}$ provides a non-trivial combination.
More vectors than entries ( $p > n$ in $\mathbb{R}^n$ )	Always linearly dependent.	Theorem 8: In an $n \times p$ matrix with $p > n$ , at most $n$ pivots, leaving at least $p - n \ge 1$ free variables.

5. Linear Transformations (Lay §1.8 – §1.9)

Concept	Definition / Formula	Pivot Criterion on Standard Matrix $A$
Linearity Properties	1. $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ 2. $T(c\mathbf{u}) = c T(\mathbf{u})$ Consequently, $T(\mathbf{0}) = \mathbf{0}$ and $T(c\mathbf{u} + d\mathbf{v}) = c T(\mathbf{u}) + d T(\mathbf{v})$ .	Holds for any matrix transformation $T(\mathbf{x}) = A\mathbf{x}$ .
Standard Matrix $[T]$	$A = [T(\mathbf{e}_1) \; T(\mathbf{e}_2) \; \cdots \; T(\mathbf{e}_n)]$ , where $\mathbf{e}_j$ are columns of identity matrix $I_n$ .	Every linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is unique matrix multiplication $T(\mathbf{x}) = A\mathbf{x}$ .
One-to-One (Injective)	$T(\mathbf{u}) = T(\mathbf{v}) \implies \mathbf{u} = \mathbf{v}$ . Equivalently, $T(\mathbf{x}) = \mathbf{0}$ has only the trivial solution.	Columns of $A$ are linearly independent $\iff A$ has a pivot in EVERY COLUMN.
Onto $\mathbb{R}^m$ (Surjective)	For every $\mathbf{b} \in \mathbb{R}^m$ , there exists at least one $\mathbf{x}$ such that $T(\mathbf{x}) = \mathbf{b}$ .	Columns of $A$ span $\mathbb{R}^m \iff A$ has a pivot in EVERY ROW.

2D Geometric Linear Transformations (Lay §1.9)

Transformation	Standard Matrix $A$	Action on $(x, y)$
Reflection across $x$ -axis	$\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$	$(x, -y)$
Reflection across $y$ -axis	$\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$	$(-x, y)$
Reflection across line $y = x$	$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$	$(y, x)$
Counterclockwise Rotation by $\theta$	$\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$	Rotates vector counterclockwise by $\theta$
Horizontal Shear by factor $k$	$\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$	$(x + ky, y)$
Vertical Shear by factor $k$	$\begin{bmatrix} 1 & 0 \\ k & 1 \end{bmatrix}$	$(x, y + kx)$
Dilation ( $k > 1$ ) / Contraction ( $0 < k < 1$ )	$\begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$	$(kx, ky)$

6. Matrix Operations & Transpose (Lay §2.1)

Matrix Multiplication ConformityIf

A

m \times n

and

B

n \times p

, product

AB

m \times p

. Inner dimensions must match. In general,

AB \ne BA

(NOT commutative).

Row-Column Rule for

AB

The entry

(AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}

(dot product of row

i

A

and column

j

B

Transpose Properties1.

(A^T)^T = A

(A + B)^T = A^T + B^T

(rA)^T = r A^T

(AB)^T = B^T A^T

(REVERSE ORDER!)

Matrix PowersIf

A

is square

n \times n

A^k = A \cdot A \cdots A

(

k

times).

A^0 = I_n

7. Matrix Inverses & Inversion Algorithm (Lay §2.2)

Topic	Rule / Formula	Key Requirement
$2 \times 2$ Matrix Inverse	If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , then $A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$	Invertible if and only if $\det A = ad - bc \ne 0$ .
Inverse of a Product	$(AB)^{-1} = B^{-1} A^{-1}$	Reverse order! Both $A$ and $B$ must be invertible $n \times n$ square matrices.
Inverse of Transpose	$(A^T)^{-1} = (A^{-1})^T$	The inverse of the transpose equals the transpose of the inverse.
Algorithm for Finding $A^{-1}$	Row reduce augmented matrix $[A \mid I_n]$ . If $A$ is invertible, $[A \mid I_n] \sim [I_n \mid A^{-1}]$ .	If row reduction yields a row of zeros on the left side, $A$ is singular (not invertible).

8. The Invertible Matrix Theorem (IMT) Master Reference (Lay §2.3)

Statement (For any square $n \times n$ matrix $A$ , all 12 statements are equivalent)	Category
a. $A$ is an invertible matrix.	Invertibility
b. $A$ is row equivalent to the $n \times n$ identity matrix $I_n$ .	Row reduction
c. $A$ has $n$ pivot positions.	Row reduction
d. The equation $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.	Homogeneous System
e. The columns of $A$ form a linearly independent set.	Linear Independence
f. The linear transformation $\mathbf{x} \mapsto A\mathbf{x}$ is one-to-one.	Transformations
g. The equation $A\mathbf{x} = \mathbf{b}$ has at least one solution for each $\mathbf{b} \in \mathbb{R}^n$ .	Solvability
h. The columns of $A$ span $\mathbb{R}^n$ .	Vector Span
i. The linear transformation $\mathbf{x} \mapsto A\mathbf{x}$ maps $\mathbb{R}^n$ onto $\mathbb{R}^n$ .	Transformations
j. There is an $n \times n$ matrix $C$ such that $CA = I_n$ .	Left inverse
k. There is an $n \times n$ matrix $D$ such that $AD = I_n$ .	Right inverse
l. $A^T$ is an invertible matrix.	Transpose

9. Partitioned Matrices & LU Factorization (Lay §2.4 – §2.5)

Topic	Form / Algorithm	Procedure & Purpose
Block Matrix Multiplication	$\begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \begin{bmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{bmatrix} = \begin{bmatrix} A_{11}B_{11} + A_{12}B_{21} & A_{11}B_{12} + A_{12}B_{22} \\ A_{21}B_{11} + A_{22}B_{21} & A_{21}B_{12} + A_{22}B_{22} \end{bmatrix}$	Multiply blocks as if they were scalars, preserving block order.
Block Diagonal Inverse	$(\operatorname{diag}(A_1, \dots, A_k))^{-1} = \operatorname{diag}(A_1^{-1}, \dots, A_k^{-1})$	A block diagonal matrix is invertible iff each diagonal block is invertible.
LU Factorization: Structure	$A = LU$ , where $L$ is Unit Lower-Triangular ( $1$ s on main diagonal) and $U$ is Upper-Triangular (Row Echelon Form).	Applies when $A$ can be row-reduced to echelon form without row interchanges.
LU Factorization: Algorithm	1. Reduce $A$ to echelon form $U$ using only row replacements. 2. Divide each column of $L$ below diagonal by its leading entry, or place the multiplier $c$ in $L$ where $R_i \leftarrow R_i - c R_j$ was used.	$L$ records the multipliers needed to clear entries below pivots.
Solving $A\mathbf{x} = \mathbf{b}$ using $LU$	Step 1: Solve $L\mathbf{y} = \mathbf{b}$ for $\mathbf{y}$ via Forward Substitution. Step 2: Solve $U\mathbf{x} = \mathbf{y}$ for $\mathbf{x}$ via Back Substitution.	Significantly faster than full row reduction for multiple right-hand sides $\mathbf{b}$ .