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 $\det(A) = 1 \cdot 5 \cdot 9 = 45$.

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 $\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$.

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, $A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$.

Inverse of a 3x3 Matrix: To find the inverse of a 3x3 matrix, we follow these steps:

1. Calculate the determinant of the matrix.
2. Find the matrix of minors.
3. Convert the matrix of minors to the matrix of cofactors.
4. Transpose the matrix of cofactors to get the adjugate matrix.
5. Divide each element of the adjugate matrix by the determinant.

If $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, then:

$A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} \det(A_{11}) & -\det(A_{12}) & \det(A_{13}) \\ -\det(A_{21}) & \det(A_{22}) & -\det(A_{23}) \\ \det(A_{31}) & -\det(A_{32}) & \det(A_{33}) \end{pmatrix}^T$

where $A_{ij}$ is the minor matrix obtained by deleting the i-th row and j-th column from A.

Example: If $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$, then:
$\det(A) = 1\cdot(1\cdot0 - 4\cdot6) - 2\cdot(0\cdot0 - 4\cdot5) + 3\cdot(0\cdot6 - 1\cdot5) = -24 + 40 - 15 = 1$.
The minors of A are:
$\begin{pmatrix} -24 & 20 & -5 \\ 12 & -15 & 5 \\ -2 & 3 & -1 \end{pmatrix}$.
The cofactors of A are:
$\begin{pmatrix} -24 & -20 & -5 \\ -12 & -15 & -5 \\ -2 & -3 & -1 \end{pmatrix}$.
The adjugate (transpose of cofactors) is:
$\begin{pmatrix} -24 & -12 & -2 \\ -20 & -15 & -3 \\ -5 & -5 & -1 \end{pmatrix}$.
Thus, the inverse of A is:
$A^{-1} = \frac{1}{1} \begin{pmatrix} -24 & -12 & -2 \\ -20 & -15 & -3 \\ -5 & -5 & -1 \end{pmatrix} = \begin{pmatrix} -24 & -12 & -2 \\ -20 & -15 & -3 \\ -5 & -5 & -1 \end{pmatrix}$.

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 $\text{rank}(AB) \leq \min(2, 2) = 2$.

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.