Matrix and Determinants Formulas - Cheatsheet | Last Minute Notes

Basic Matrix Operations

Matrix Addition: $(A + B)_{ij} = A_{ij} + B_{ij}$

Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$, then $A + B = \begin{pmatrix} 1+5 & 2+6 \\ 3+7 & 4+8 \end{pmatrix} = \begin{pmatrix} 6 & 8 \\ 10 & 12 \end{pmatrix}$.

Matrix Subtraction: $(A - B)_{ij} = A_{ij} - B_{ij}$

Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$, then $A - B = \begin{pmatrix} 1-5 & 2-6 \\ 3-7 & 4-8 \end{pmatrix} = \begin{pmatrix} -4 & -4 \\ -4 & -4 \end{pmatrix}$.

Scalar Multiplication: $(cA)_{ij} = c \cdot A_{ij}$

Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $c = 3$, then $cA = 3 \cdot \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 3 \cdot 1 & 3 \cdot 2 \\ 3 \cdot 3 & 3 \cdot 4 \end{pmatrix} = \begin{pmatrix} 3 & 6 \\ 9 & 12 \end{pmatrix}$.

Matrix Multiplication: $(AB)_{ij} = \sum_{k} A_{ik} B_{kj}$

Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$, then $AB = \begin{pmatrix} 1 \cdot 5 + 2 \cdot 7 & 1 \cdot 6 + 2 \cdot 8 \\ 3 \cdot 5 + 4 \cdot 7 & 3 \cdot 6 + 4 \cdot 8 \end{pmatrix} = \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix}$.

Transpose of a Matrix: $(A^T)_{ij} = A_{ji}$

Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $A^T = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$.

Special Matrices

Identity Matrix: $I_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}$

Example: The 3x3 identity matrix is $I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.

Zero Matrix: All elements are zero, $0_{ij} = 0$

Example: A 2x2 zero matrix is $0 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.

Diagonal Matrix: $D_{ij} = \begin{cases} d_i & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}$

Example: A 3x3 diagonal matrix is $D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$.

Symmetric Matrix: $A = A^T$

Example: If $A = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$, then $A^T = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$, so $A$ is symmetric.

Skew-Symmetric Matrix: $A = -A^T$

Example: If $A = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}$, then $A^T = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}$ and $-A^T = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}$, so $A$ is skew-symmetric.

Determinants

Determinant of a 2x2 Matrix: $\det(A) = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$

Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $\det(A) = 1 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2$.

Determinant of a 3x3 Matrix: $\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$

Example: If $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$, then $\det(A) = 1(5 \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - 5 \cdot 7) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0$.

Properties of Determinants:
$\det(AB) = \det(A) \cdot \det(B)$
Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix}$, then $\det(A) = -2$, $\det(B) = 4$, and $\det(AB) = \det(\begin{pmatrix} 4 & 4 \\ 10 & 8 \end{pmatrix}) = 4 \cdot 8 - 4 \cdot 10 = 32 - 40 = -8$ which is equal to $-2 \cdot 4 = -8$.

$\det(A^T) = \det(A)$
Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $\det(A) = -2$ and $\det(A^T) = \det(\begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}) = 1 \cdot 4 - 3 \cdot 2 = 4 - 6 = -2$.

If $A$ is triangular, $\det(A)$ is the product of its diagonal elements.
Example: If $A = \begin{pmatrix} 1 & 0 & 0 \\ 4 & 5 & 0 \\ 7 & 8 & 9 \end{pmatrix}$, then $\det(A) = 1 \cdot 5 \cdot 9 = 45$.

Inverse of a Matrix

Inverse of a 2x2 Matrix: $A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $\det(A) = 1 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2$. Thus, $A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$.

General Formula: If $\det(A) \neq 0$, then $A^{-1}$ exists, and $AA^{-1} = A^{-1}A = I$

Example: Using the above matrix $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and its inverse $A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$:
$A \cdot A^{-1} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \cdot \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} = \begin{pmatrix} (1 \cdot -2 + 2 \cdot 1.5) \\ (1 \cdot 1 + 2 \cdot -0.5) \\ (3 \cdot -2 + 4 \cdot 1.5) \\ (3 \cdot 1 + 4 \cdot -0.5) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$.

Properties:
$(A^{-1})^{-1} = A$
Example: If $A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$, then $(A^{-1})^{-1} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = A$.

$(AB)^{-1} = B^{-1} A^{-1}$
Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix}$, then $A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$ and $B^{-1} = \begin{pmatrix} 0.5 & 0 \\ -0.25 & 0.5 \end{pmatrix}$. Thus, $(AB)^{-1} = B^{-1} A^{-1} = \begin{pmatrix} 0.5 & 0 \\ -0.25 & 0.5 \end{pmatrix} \cdot \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$.

$(A^T)^{-1} = (A^{-1})^T$
Example: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$ and $(A^{-1})^T = \begin{pmatrix} -2 & 1.5 \\ 1 & -0.5 \end{pmatrix}$ which is the inverse of $A^T = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$.

Trace of a Matrix

Definition: The trace of a square matrix is defined as the sum of its diagonal elements.
$\text{trace}(A) = \sum_{i} A_{ii}$

Example: Consider the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$. To find the trace of $A$, we sum its diagonal elements:
$\text{trace}(A) = 1 + 5 + 9 = 15$.

Eigenvalues and Eigenvectors

Eigenvalue Equation: $A \mathbf{v} = \lambda \mathbf{v}$

Example: If $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$, then $A \mathbf{v} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$. Thus, $\lambda = 3$.

Characteristic Polynomial: $\det(A - \lambda I) = 0$

Example: If $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, then $A - \lambda I = \begin{pmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{pmatrix}$. The characteristic polynomial is $\det(\begin{pmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{pmatrix}) = (2 - \lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0$. Solving for $\lambda$, we get $\lambda = 1, 3$.

Properties:
Sum of eigenvalues = Trace of $A$
Example: If $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, the eigenvalues are $1$ and $3$. The sum is $1 + 3 = 4$, which is equal to the trace of $A$ ($2 + 2 = 4$).

Product of eigenvalues = $\det(A)$
Example: If $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, the eigenvalues are $1$ and $3$. The product is $1 \cdot 3 = 3$, which is equal to the determinant of $A$ ($2 \cdot 2 - 1 \cdot 1 = 4 - 1 = 3$).

Rank of a Matrix

Definition: The rank of a matrix is the maximum number of linearly independent row vectors or column vectors in the matrix.

Example: Consider the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$. Its row vectors are $\begin{pmatrix} 1 & 2 & 3 \end{pmatrix}$, $\begin{pmatrix} 4 & 5 & 6 \end{pmatrix}$, and $\begin{pmatrix} 7 & 8 & 9 \end{pmatrix}$. Since the third row vector can be expressed as a linear combination of the first two, the rank of $A$ is $2$.

Details: The third row vector $\begin{pmatrix} 7 & 8 & 9 \end{pmatrix}$ can be written as $0 \cdot \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} + 1 \cdot \begin{pmatrix} 4 & 5 & 6 \end{pmatrix}$, demonstrating that it is indeed a linear combination of the first two row vectors.

Properties:

• $\text{rank}(A) \leq \min(m, n)$ for an $m \times n$ matrix $A$.
• Example: If $A$ is a $3 \times 4$ matrix, then $\text{rank}(A) \leq \min(3, 4) = 3$.


• $\text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B))$.
• Example: If $A$ is a $3 \times 2$ matrix with rank $2$, and $B$ is a $2 \times 4$ matrix with rank $2$, then $\text{rank}(AB) \leq \min(2, 2) = 2$.

Cofactor Matrix: The cofactor matrix of a matrix $A$ is defined as $C_{ij} = (-1)^{i+j} \det(M_{ij})$, where $M_{ij}$ is the minor of $A$ obtained by removing the $i$-th row and $j$-th column.

Example: Calculate the cofactor matrix of matrix $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.

Solution:$M_{11} = \begin{pmatrix} 4 \end{pmatrix}$,$M_{12} = \begin{pmatrix} 3 \end{pmatrix}$,$M_{21} = \begin{pmatrix} 2 \end{pmatrix}$,$M_{22} = \begin{pmatrix} 1 \end{pmatrix}$. Therefore,$C_{11} = (-1)^{1+1} \det(M_{11}) = 1 \cdot 4 = 4$,$C_{12} = (-1)^{1+2} \det(M_{12}) = -1 \cdot 3 = -3$,$C_{21} = (-1)^{2+1} \det(M_{21}) = -1 \cdot 2 = -2$,$C_{22} = (-1)^{2+2} \det(M_{22}) = 1 \cdot 1 = 1$. Thus, the cofactor matrix $C$ is $\begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix}$.

Adjugate Matrix: The adjugate matrix of $A$ is denoted as $\text{adj}(A) = C^T$, where $C$ is the cofactor matrix of $A$.

Example: Find the adjugate matrix of $B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.

Solution: The cofactor matrix $C$ calculated previously is $\begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix}$. Therefore, the adjugate matrix $\text{adj}(B)$ is $\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$.

Cramer's Rule: Cramer's Rule provides a method to solve systems of linear equations of the form $AX = B$. Specifically, it states that $x_i = \frac{\det(A_i)}{\det(A)}$, where $A_i$ is the matrix $A$ with its $i$-th column replaced by $B$.

Example: Solve the system of equations $2x + 3y = 8$ and $4x - y = 6$ using Cramer's Rule.

Solution:

• Calculate $\det(A)$ where $A = \begin{pmatrix} 2 & 3 \\ 4 & -1 \end{pmatrix}$: $\det(A) = 2 \cdot (-1) - 3 \cdot 4 = -2 - 12 = -14$.
• Calculate $\det(A_1)$ where $A_1 = \begin{pmatrix} 8 & 3 \\ 6 & -1 \end{pmatrix}$: $\det(A_1) = 8 \cdot (-1) - 3 \cdot 6 = -8 - 18 = -26$.
• Calculate $\det(A_2)$ where $A_2 = \begin{pmatrix} 2 & 8 \\ 4 & 6 \end{pmatrix}$: $\det(A_2) = 2 \cdot 6 - 8 \cdot 4 = 12 - 32 = -20$.
• Find $x$ and $y$:$x = \frac{\det(A_1)}{\det(A)} = \frac{-26}{-14} = \frac{13}{7}$,$y = \frac{\det(A_2)}{\det(A)} = \frac{-20}{-14} = \frac{10}{7}$.

Useful Properties

Distributive Property: $A(B + C) = AB + AC$

Example: Verify the distributive property for matrices with the following matrices:
$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$,$B = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$,$C = \begin{pmatrix} 0 & 1 \\ 4 & 2 \end{pmatrix}$.

Solution: Compute $A(B + C)$ and $AB + AC$ to show they are equal:
$A(B + C) = A\left(\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 4 & 2 \end{pmatrix}\right) = A\left(\begin{pmatrix} 2 & 2 \\ 5 & 5 \end{pmatrix}\right) = \begin{pmatrix} 12 & 12 \\ 26 & 26 \end{pmatrix}$,
$AB + AC = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} + \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 4 & 7 \\ 10 & 15 \end{pmatrix} + \begin{pmatrix} 8 & 5 \\ 20 & 11 \end{pmatrix} = \begin{pmatrix} 12 & 12 \\ 30 & 26 \end{pmatrix}$.
Therefore, $A(B + C) = AB + AC$.

Associative Property: $A(BC) = (AB)C$

Example: Demonstrate the associative property using the matrices:
$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$,$B = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$,$C = \begin{pmatrix} 0 & 1 \\ 4 & 2 \end{pmatrix}$.

Solution: Compute $A(BC)$ and $(AB)C$ to show they are equal:
$A(BC) = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \left(\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 4 & 2 \end{pmatrix}\right) = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 4 & 5 \\ 12 & 11 \end{pmatrix} = \begin{pmatrix} 28 & 27 \\ 60 & 57 \end{pmatrix}$,
$(AB)C = \left(\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}\right) \begin{pmatrix} 0 & 1 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 4 & 7 \\ 10 & 15 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 28 & 27 \\ 60 & 57 \end{pmatrix}$.
Therefore, $A(BC) = (AB)C$.

Commutative Property (not general, special cases only): $AB = BA$ for certain matrices.

Example: Find matrices $A$ and $B$ such that $AB = BA$.

Solution: Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$.
Compute $AB$ and $BA$:
$AB = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$,
$BA = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.
Therefore, $AB = BA$ in this case.