Matrices | Mathematics - Wyatt's Notes

3.1 Matrix Operations

An $m \times n$ matrix $A$ over $F$ is a rectangular array of $mn$ elements from $F$ Arranged in $m$ rows and $n$ columns. The set of all such matrices is denoted $\mathcal{M}_{m \times n}(F)$ .

Addition. For $A, B \in \mathcal{M}_{m \times n}(F)$ , $(A + B)_{ij} = A_{ij} + B_{ij}$ .

Scalar multiplication. For $\alpha \in F$ and $A \in \mathcal{M}_{m \times n}(F)$ $(\alpha A)_{ij} = \alpha A_{ij}$ .

Matrix multiplication. For $A \in \mathcal{M}_{m \times n}(F)$ and $B \in \mathcal{M}_{n \times p}(F)$ The product $AB \in \mathcal{M}_{m \times p}(F)$ is defined by

$(AB)_{ij} = \sum_{k=1}^n A_{ik} B_{kj}$

Proposition 3.1. Matrix multiplication is associative but not commutative .

Proof. Associativity: $(AB)C$ has $(i,j)$ -entry $\sum_l (\sum_k A_{ik} B_{kl}) C_{lj} = \sum_k A_{ik} (\sum_l B_{kl} C_{lj}) = (A(BC))_{ij}$ By interchanging the order of summation (both sums are finite). For non-commutativity, $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ give $AB = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \neq \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = BA$ . $\blacksquare$

3.2 Transpose

The transpose of $A \in \mathcal{M}_{m \times n}(F)$ Denoted $A^T$ Is the $n \times m$ matrix With $(A^T)_{ij} = A_{ji}$ .

Properties of transpose:

$(A + B)^T = A^T + B^T$
$(\alpha A)^T = \alpha A^T$
$(AB)^T = B^T A^T$
$(A^T)^T = A$

3.3 Inverse Matrices

A square matrix $A \in \mathcal{M}_{n \times n}(F)$ is invertible if there exists a matrix $A^{-1} \in \mathcal{M}_{n \times n}(F)$ such that

$AA^{-1} = A^{-1}A = I_n$

Theorem 3.1. The following are equivalent for $A \in \mathcal{M}_{n \times n}(F)$ :

$A$ is invertible.
$\det(A) \neq 0$ .
The columns of $A$ are linearly independent.
The rows of $A$ are linearly independent.
$\mathrm{rank}(A) = n$ .
The equation $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$ .
The only solution to $A\mathbf{x} = \mathbf{0}$ is $\mathbf{x} = \mathbf{0}$ .

3.4 Determinants

The determinant is a function $\det : \mathcal{M}_{n \times n}(F) \to F$ defined recursively by Laplace expansion along the first row:

$\det(A) = \sum_{j=1}^n (-1)^{1+j} a_{1j} M_{1j}$

Where $M_{1j}$ is the $(1,j)$ -minor (the determinant of the $(n-1) \times (n-1)$ matrix obtained by Deleting row 1 and column $j$ ).

The $(i,j)$ -cofactor is $C_{ij} = (-1)^{i+j} M_{ij}$ So $\det(A) = \sum_{j=1}^n a_{ij} C_{ij}$ for any fixed row $i$ .

3.5 Properties of Determinants

Proposition 3.2 (Effect of Row Operations). Let $A \in \mathcal{M}_{n \times n}(F)$ .

Swapping two rows of $A$ multiplies the determinant by $-1$ .
Multiplying a row of $A$ by $\alpha \in F$ multiplies the determinant by $\alpha$ .
Adding a multiple of one row to another leaves the determinant unchanged.

Proof. (1) This follows from the antisymmetry of the Leibniz formula $\det(A) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}$ . Swapping two rows Changes the sign of every permutation, hence the sign of the sum.

(2) Multiplying row $i$ by $\alpha$ multiplies every term in the Leibniz expansion by $\alpha$ Hence $\det$ is multiplied by $\alpha$ .

(3) Adding $\alpha$ times row $j$ to row $i$ ( $i \neq j$ ): by multilinearity in row $i$

$\det(\mathrm{new}~A) = \det(A) + \alpha \cdot \det(\mathrm{matrix}~with~rows~i\mathrm{~and~j\mathrm}{~equal)}$

A matrix with two equal rows has determinant 0 (by antisymmetry: swapping them leaves the matrix Unchanged but multiplies $\det$ by $-1$ So $\det = -\det$ Hence $\det = 0$ ). Therefore $\det(\mathrm{new}~A) = \det(A)$ . $\blacksquare$

Theorem 3.3 (Multiplicativity). For $A, B \in \mathcal{M}_{n \times n}(F)$

$\det(AB) = \det(A)\det(B)$

Proof (via elementary matrices). Every matrix $B$ can be written as a product of elementary matrices Times an upper triangular matrix: $B = E_1 E_2 \cdots E_k U$ . For an elementary matrix $E$ :

If $E$ swaps rows, $\det(E) = -1$ and $\det(AE) = -\det(A) = \det(A)\det(E)$ .
If $E$ multiplies a row by $\alpha$ , $\det(E) = \alpha$ and $\det(AE) = \alpha\det(A) = \det(A)\det(E)$ .
If $E$ adds a multiple of one row to another, $\det(E) = 1$ and $\det(AE) = \det(A) = \det(A)\det(E)$ .

Thus $\det(AE) = \det(A)\det(E)$ for every elementary matrix. By induction,

$\det(AB) = \det(A \cdot E_1 \cdots E_k U) = \det(A) \cdot \det(E_1) \cdots \det(E_k) \cdot \det(U) = \det(A) \cdot \det(B)$

Since $\det(B) = \det(E_1)\cdots\det(E_k)\det(U)$ . $\blacksquare$

Corollary 3.4. $\det(A^T) = \det(A)$ And for invertible $A$ , $\det(A^{-1}) = 1/\det(A)$ .

Proof. $AA^{-1} = I$ So $\det(A)\det(A^{-1}) = \det(I) = 1$ Giving $\det(A^{-1}) = 1/\det(A)$ . For the transpose, use the Leibniz formula or observe that row Operations and column operations have the same effects on the determinant. $\blacksquare$

3.6 Adjugate and Inverse Formula

Definition. The adjugate (or adjoint) of $A \in \mathcal{M}_{n \times n}(F)$ is

$\mathrm{adj}(A) = (C_{ji})_{i,j=1}^n$

Where $C_{ij}$ is the $(i,j)$ -cofactor of $A$ . That is, $\mathrm{adj}(A)$ is the transpose of the Cofactor matrix.

Theorem 3.5. For any $A \in \mathcal{M}_{n \times n}(F)$

$A \cdot \mathrm{adj}(A) = \mathrm{adj}(A) \cdot A = \det(A) \cdot I_n$

In particular, if $\det(A) \neq 0$ Then $A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$ .

Proof. The $(i,j)$ -entry of $A \cdot \mathrm{adj}(A)$ is $\sum_{k=1}^n a_{ik} C_{jk}$ . When $i = j$ This is $\sum_{k=1}^n a_{ik} C_{ik} = \det(A)$ (cofactor expansion along row $i$ ). When $i \neq j$ This is the cofactor expansion of a matrix obtained from $A$ by replacing row $j$ With row $i$ Which has two equal rows and hence determinant 0. $\blacksquare$

3.7 Worked Examples

Problem. Compute $\det(A)$ where

$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$

Solution

Expanding along the first column:

$\det(A) = 1 \cdot \det\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} - 0 + 5 \cdot \det\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$

$= 1 \cdot (0 - 24) + 5 \cdot (8 - 3) = -24 + 25 = 1$

$\blacksquare$

Problem. Compute $\det(A)$ by row reduction where

$A = \begin{pmatrix} 2 & 1 & 3 & 1 \\ 4 & 2 & 5 & 3 \\ 6 & 3 & 8 & 5 \\ 1 & 1 & 1 & 1 \end{pmatrix}$

Solution

Apply row operations and track their effect on the determinant:

$\begin{pmatrix} 2 & 1 & 3 & 1 \\ 4 & 2 & 5 & 3 \\ 6 & 3 & 8 & 5 \\ 1 & 1 & 1 & 1 \end{pmatrix} \xrightarrow{R_2 - 2R_1, R_3 - 3R_1} \begin{pmatrix} 2 & 1 & 3 & 1 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & -1 & 2 \\ 0 & 1/2 & -1/2 & 1/2 \end{pmatrix}$

The determinant is unchanged (only type 3 operations). Now swap $R_2$ and $R_4$ (multiplies $\det$ by $-1$ ):

$\xrightarrow{R_2 \leftrightarrow R_4} \begin{pmatrix} 2 & 1 & 3 & 1 \\ 0 & 1/2 & -1/2 & 1/2 \\ 0 & 0 & -1 & 2 \\ 0 & 0 & -1 & 1 \end{pmatrix}$

Now $R_4 \to R_4 - R_3$ (determinant unchanged):

$\begin{pmatrix} 2 & 1 & 3 & 1 \\ 0 & 1/2 & -1/2 & 1/2 \\ 0 & 0 & -1 & 2 \\ 0 & 0 & 0 & -1 \end{pmatrix}$

The determinant is the product of diagonal entries, times $-1$ for the row swap:

$\det(A) = (-1) \cdot 2 \cdot \frac{1}{2} \cdot (-1) \cdot (-1) = -1$

$\blacksquare$

Problem. Find $A^{-1}$ using the adjugate formula, where

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$

Solution

$\det(A) = 1 \cdot 4 - 2 \cdot 3 = -2 \neq 0$ So $A$ is invertible.

Cofactors: $C_{11} = 4$ , $C_{12} = -3$ , $C_{21} = -2$ , $C_{22} = 1$ .

$\mathrm{adj}(A) = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$

$A^{-1} = \frac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}$

Verify: $AA^{-1} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix} = \begin{pmatrix} -2 + 3 & 1 - 1 \\ -6 + 6 & 3 - 2 \end{pmatrix} = I_2$ . $\blacksquare$

meaningful determinant for an $m \times n$ matrix with $m \neq n$ . Do not confuse $\det(AB) = \det(A)\det(B)$ with a Non-existent formula for non-square matrices.

3.8 Worked Example: Determinant via Row Reduction (Efficient Method)

Problem. Compute $\det(A)$ where

$A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & 20 \end{pmatrix}$

Solution

This matrix has Pascal-like entries. We use row operations:

$\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & 20 \end{pmatrix} \xrightarrow{R_i - R_{i-1}} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 3 & 6 \\ 0 & 1 & 4 & 10 \end{pmatrix} \xrightarrow{R_i - R_{i-1}} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 1 & 4 \end{pmatrix}$

$\xrightarrow{R_4 - R_3} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{pmatrix}$

All operations were type 3 (adding a multiple of one row to another), so the determinant is unchanged. The upper triangular matrix has diagonal entries $1, 1, 1, 1$ so $\det(A) = 1$ . $\blacksquare$

Proposition 3.6 (Determinant of a Triangular Matrix). If $A$ is upper or lower triangular, then $\det(A) = \prod_{i=1}^n a_{ii}$ .

Proof. By repeated cofactor expansion along the first column (for upper triangular), or induction. At each step, all terms involving off-diagonal entries vanish due to the zero structure, leaving only the product of diagonal entries. $\blacksquare$

3.9 Intuition: What Does the Determinant Measure Geometrically?

The determinant of an $n \times n$ matrix $A$ measures the factor by which $A$ scales $n$ -dimensional volume. Specifically, if $S$ is any measurable subset of $\mathbb{R}^n$ , then the image $A(S)$ has $n$ -dimensional volume equal to $|\det(A)|$ times the volume of $S$ .

For $2 \times 2$ matrices, $\det(A)$ is the signed area of the parallelogram formed by the column vectors of $A$ . For $3 \times 3$ matrices, $|\det(A)|$ is the volume of the parallelepiped formed by the three column vectors.

The sign of the determinant indicates orientation: $\det(A) > 0$ means $A$ preserves orientation (like a rotation), while $\det(A) < 0$ means $A$ reverses orientation (like a reflection). When $\det(A) = 0$ , the image of $A$ has zero $n$ -dimensional volume, meaning the columns of $A$ are linearly dependent and $A$ “squishes” $\mathbb{R}^n$ into a lower-dimensional subspace.

This geometric interpretation explains why $\det(AB) = \det(A)\det(B)$ : the volume scaling of the composition $AB$ is the product of the individual scalings. It also explains why $\det(A) = 0$ implies $A$ is not invertible: a map that squishes volume to zero cannot be reversed.

3.10 Worked Example: 3x3 Determinant by Cofactor Expansion

Problem. Compute $\det(A)$ where

$A = \begin{pmatrix} 2 & -1 & 0 \\ 3 & 4 & -2 \\ 1 & 0 & 5 \end{pmatrix}$

Solution

Expand along the first row:

$\det(A) = 2 \cdot \det\begin{pmatrix} 4 & -2 \\ 0 & 5 \end{pmatrix} - (-1) \cdot \det\begin{pmatrix} 3 & -2 \\ 1 & 5 \end{pmatrix} + 0 \cdot \det\begin{pmatrix} 3 & 4 \\ 1 & 0 \end{pmatrix}$

$= 2(20 - 0) + 1(15 + 2) + 0 = 40 + 17 = 57$

Alternatively, expand along the third row (which has a zero):

$\det(A) = 1 \cdot \det\begin{pmatrix} -1 & 0 \\ 4 & -2 \end{pmatrix} - 0 + 5 \cdot \det\begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$

$= 1(2 - 0) + 5(8 + 3) = 2 + 55 = 57 \quad \checkmark$

The second method is faster because the zero entry in the third row eliminates one $2 \times 2$ determinant computation. Always look for rows or columns with the most zeros before choosing which row/column to expand along.

Geometric check: Since $\det(A) = 57 \neq 0$ , the matrix $A$ is invertible, its columns form a basis for $\mathbb{R}^3$ , and the linear map $\mathbf{x} \mapsto A\mathbf{x}$ scales volume by a factor of 57. $\blacksquare$

3.11 Common Pitfalls

$\det(A + B) \neq \det(A) + \det(B)$ . For example, with $A = B = I_2$ , $\det(A + B) = \det(2I_2) = 4$ but $\det(A) + \det(B) = 2$ .
The adjugate formula is theoretically important but computationally inefficient. For large matrices, use Gaussian elimination or LU decomposition to compute inverses.
A matrix with $\det(A) = 0$ has no inverse. Do not attempt to divide by zero.
Cofactor expansion along different rows/columns gives the same answer, but some choices are faster. Always expand along the row or column with the most zeros.
$\det(A^T) = \det(A)$ , so row operations and column operations affect the determinant in the same way. You can compute the determinant by reducing along rows or columns.
The determinant of a product is the product of determinants, but $\det(A + B)$ has no simple formula. Do not distribute the determinant over addition.
For non-square matrices, the determinant is not defined. Do not attempt to compute $\det(A)$ for an $m \times n$ matrix with $m \neq n$ .

flowchart TD
    A[3_Matrices] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Cross-References

Eigenvalues and Eigenvectors: Eigenvalues are the roots of the characteristic polynomial defined via determinants.
Systems of Linear Equations: Gaussian elimination transforms matrices into row echelon form to solve linear systems.
Singular Value Decomposition: The SVD factorises any matrix using its singular values, generalising the eigendecomposition.

Mark this page as reviewed