Matrices and Determinants
Introduction
A matrix is a rectangular arrangement of numbers in rows and columns. It is usually enclosed within square or round brackets. If a matrix has $m$ rows and $n$ columns, it is called an $m \times n$ (“$m$ by $n$”) matrix.
An $m \times n$ matrix $A$ is written as: \[ A = [a_{ij}]_{m \times n}, \quad \text{where } a_{ij} \text{ is the element in the $i^{th}$ row and $j^{th}$ column.} \]
A $2 \times 3$ matrix: \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \]
Types of Matrices
Row Matrix
A matrix with only one row is called a row matrix.
If $A$ is a $1 \times n$ matrix, then $A$ is a row matrix.
\[ A = \begin{bmatrix} 2 & 5 & -3 & 7 \end{bmatrix} \] is a row matrix of order $1 \times 4$.
Column Matrix
A matrix with only one column is called a column matrix.
If $A$ is an $m \times 1$ matrix, then $A$ is a column matrix.
\[ B = \begin{bmatrix} 4 \\ -2 \\ 7 \end{bmatrix} \] is a column matrix of order $3 \times 1$.
Square Matrix
A matrix with the same number of rows and columns is called a square matrix.
An $n \times n$ matrix is called a square matrix of order $n$.
\[ C = \begin{bmatrix} 2 & 1 & 0 \\ -3 & 4 & 5 \\ 6 & 7 & 8 \end{bmatrix} \] is a square matrix of order $3$.
Zero Matrix
A matrix in which all elements are zero is called a zero matrix or null matrix.
If $O = [0]_{m \times n}$, then $O$ is a zero matrix of order $m \times n$.
\[ D = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] is a zero matrix of order $2 \times 3$.
Identity Matrix
A square matrix in which all diagonal elements are 1 and all other elements are 0 is called an identity matrix.
The identity matrix of order $n$ is denoted by $I_n$, where \[ (I_n)_{ij} = \begin{cases} 1 & \text{if } i=j \\ 0 & \text{if } i\neq j \end{cases} \]
\[ I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] is the $3 \times 3$ identity matrix.
Diagonal Matrix
A square matrix in which all non-diagonal elements are zero is called a diagonal matrix.
If $A = [a_{ij}]_{n \times n}$ and $a_{ij} = 0$ for all $i \neq j$, then $A$ is a diagonal matrix.
\[ E = \begin{bmatrix} 5 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 7 \end{bmatrix} \] is a diagonal matrix of order $3$.
Triangular Matrix
A triangular matrix is a square matrix in which all the entries either below or above the main diagonal are zero.
Upper Triangular Matrix
A square matrix $A=[a_{ij}]_{n\times n}$ is called an upper triangular matrix if $a_{ij} = 0$ for all $i > j$ (all elements below the main diagonal are zero).
An upper triangular matrix of order $3$: \[ U = \begin{bmatrix} 2 & -1 & 4 \\ 0 & 3 & 5 \\ 0 & 0 & 7 \end{bmatrix} \] Here, all elements below the main diagonal are $0$.
Lower Triangular Matrix
A square matrix $A=[a_{ij}]_{n\times n}$ is called a lower triangular matrix if $a_{ij} = 0$ for all $i < j$ (all elements above the main diagonal are zero).
A lower triangular matrix of order $3$: \[ L = \begin{bmatrix} 4 & 0 & 0 \\ -2 & 5 & 0 \\ 7 & 1 & 6 \end{bmatrix} \] Here, all elements above the main diagonal are $0$.
Every diagonal matrix is both an upper triangular and a lower triangular matrix. That is, diagonal matrices are a special case of triangular matrices.
Matrix Operations
Addition
Two matrices of the same order can be added by adding their corresponding elements.
If $A = [a_{ij}]_{m \times n}$ and $B = [b_{ij}]_{m \times n}$, then \[ A + B = [a_{ij} + b_{ij}]_{m \times n}. \]
\[ \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix} \]
Scalar Multiplications
Multiplying a matrix by a scalar means multiplying each entry of the matrix by that scalar.
If $k$ is a scalar and $A = [a_{ij}]_{m \times n}$, then \[ kA = [k \cdot a_{ij}]_{m \times n}. \]
\[ 3 \cdot \begin{bmatrix} 1 & -2 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & -6 \\ 0 & 12 \end{bmatrix} \]
Matrix Multiplication
Matrix multiplication is defined only when the number of columns of the first matrix equals the number of rows of the second.
If $A$ is an $m \times n$ matrix and $B$ is an $n \times p$ matrix, then $AB$ is an $m \times p$ matrix given by: \[ (AB)_{ij} = \sum_{k=1}^n a_{ik} b_{kj}. \]
\[ \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \cdot \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 1\cdot5+2\cdot7 & 1\cdot6+2\cdot8 \\ 3\cdot5+4\cdot7 & 3\cdot6+4\cdot8 \end{bmatrix} = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix} \]
Transpose
The transpose of a matrix is obtained by interchanging its rows and columns.
If $A = [a_{ij}]_{m \times n}$, then its transpose is \[ A^T = [a_{ji}]_{n \times m}. \]
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}, \quad A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix} \]
Let $A$ and $B$ be matrices and let $k$ be a scalar. Then, whenever the sum and product are defined, the following properties hold:
$(A + B)^T = A^T + B^T$
$(A^T)^T = A$
$(kA)^T = kA^T$
$(AB)^T = B^T A^T$
Note: In (4), the transpose of a product is the product of the transposes, but in the reverse order.
Square Matrices
A square matrix is a matrix with the same number of rows and columns. An $n \times n$ square matrix is said to be of order $n$ and is sometimes called an $n$-square matrix.
For square matrices of order $n$, the operations of addition, multiplication, scalar multiplication, and transpose are always defined, and the result is again an $n \times n$ matrix.
The following are square matrices of order $3$: \[ A = \begin{bmatrix} 1 & 2 & -4 \\ 5 & 6 & 3 \\ 3 & -4 & 7 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & -5 & 0 \\ 3 & 1 & 2 \\ 3 & 1 & -4 \end{bmatrix} \]
Examples of operations: \[ A^T = \begin{bmatrix} 1 & 5 & 3 \\ 2 & 6 & -4 \\ -4 & 3 & 7 \end{bmatrix}, \quad 2A = \begin{bmatrix} 2 & 4 & -8 \\ 10 & 12 & 6 \\ 6 & -8 & 14 \end{bmatrix} \]
\[ A+B = \begin{bmatrix} 3 & -3 & -4 \\ 8 & 7 & 5 \\ 6 & -3 & 3 \end{bmatrix}, \quad AB = \begin{bmatrix} -12 & 0 & 20 \\ -27 & -30 & -33 \\ 17 & 7 & -35 \end{bmatrix} \] \[ BA = \begin{bmatrix} -22 & -24 & -26 \\ 10 & 12 & 14 \\ 27 & 30 & 33 \end{bmatrix} \]
Diagonal and Trace
The diagonal or main diagonal of an $n \times n$ matrix $A=[a_{ij}]$ consists of the elements $a_{11}, a_{22}, \dots, a_{nn}$.
The trace of $A$, written $\operatorname{tr}(A)$, is the sum of the diagonal elements: \[ \operatorname{tr}(A) = a_{11} + a_{22} + \cdots + a_{nn} \]
Suppose $A=[a_{ij}]$ and $B=[b_{ij}]$ are $n \times n$ matrices, and $k$ is a scalar. Then:
\begin{align*}
(i) & \operatorname{tr}(A+B) = \operatorname{tr}(A) + \operatorname{tr}(B),
(ii) & \operatorname{tr}(kA) = k \, \operatorname{tr}(A),
(iii) & \operatorname{tr}(A^T) = \operatorname{tr}(A),
(iv) & \operatorname{tr}(AB) = \operatorname{tr}(BA).
\end{align*}
For matrices $A$ and $B$ \[ A = \begin{bmatrix} 1 & 2 & -4 \\ 5 & 6 & 3 \\ 3 & -4 & 7 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & -5 & 0 \\ 3 & 1 & 2 \\ 3 & 1 & -4 \end{bmatrix} \]
Diagonal of $A$: $\{1,-4,7\}$, so $\operatorname{tr}(A) = 1 + (-4) + 7 = 4$.
Diagonal of $B$: $\{2,1,-4\}$, so $\operatorname{tr}(B) = 2 + 1 - 4 = -1$.
Now: \[ \operatorname{tr}(A+B) = \operatorname{tr}(A) + \operatorname{tr}(B) = 4 + (-1) = 3 \] \[ \operatorname{tr}(2A) = 2 \cdot \operatorname{tr}(A) = 8 \] \[ \operatorname{tr}(A^T) = \operatorname{tr}(A) = 4 \] \[ \operatorname{tr}(AB) = \operatorname{tr}(BA) = -30 \]
Identity and Scalar Matrices
The $n \times n$ identity matrix, denoted $I_n$, has 1's on the diagonal and 0's elsewhere: \[ (I_n)_{ij} = \begin{cases} 1 & \text{if } i=j \\ 0 & \text{if } i \ne j \end{cases} \]
For any $n \times n$ matrix $A$, \[ AI_n = I_n A = A \]
More generally, for any scalar $k$, the matrix $kI$ with $k$'s on the diagonal and 0's elsewhere is called a scalar matrix.
Multiplying a matrix $A$ by $kI$ is equivalent to multiplying $A$ by $k$: \[ (kI)A = k(IA) = kA \]
Identity and scalar matrices: \[ I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad 5I_3 = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \]
\[ I_4 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \quad 5I_4 = \begin{bmatrix} 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \end{bmatrix} \]
Powers of Matrices
Let $A$ be an $n \times n$ matrix over a field $K$. The powers of $A$ are defined as: \[ A^2 = A \cdot A, \quad A^3 = A^2 \cdot A, \quad \dots, \quad A^{n+1} = A^n \cdot A \] Additionally, the zeroth power is defined as: \[ A^0 = I_n \] where $I_n$ is the identity matrix of order $n$.
For any square matrix $A$: \[ A^m \cdot A^n = A^{m+n}, \qquad (A^m)^n = A^{mn}, \qquad A^0 = I \]
Let \[ A = \begin{bmatrix} 1 & 3 \\ 2 & -4 \end{bmatrix} \]
Compute $A^2$ and $A^3$.
\[ A^2 = A \cdot A = \begin{bmatrix} 1 & 3 \\ 2 & -4 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} 1\cdot1 + 3\cdot2 & 1\cdot3 + 3\cdot(-4) \\ 2\cdot1 + (-4)\cdot2 & 2\cdot3 + (-4)\cdot(-4) \end{bmatrix} = \begin{bmatrix} 7 & -9 \\ -6 & 22 \end{bmatrix} \]
Now, \[ A^3 = A^2 \cdot A = \begin{bmatrix} 7 & -9 \\ -6 & 22 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} 7\cdot1 + (-9)\cdot2 & 7\cdot3 + (-9)\cdot(-4) \\ -6\cdot1 + 22\cdot2 & -6\cdot3 + 22\cdot(-4) \end{bmatrix} \]
\[ = \begin{bmatrix} 7 - 18 & 21 + 36 \\ -6 + 44 & -18 - 88 \end{bmatrix} = \begin{bmatrix} -11 & 57 \\ 38 & -106 \end{bmatrix} \]
Thus, \[ A^2 = \begin{bmatrix} 7 & -9 \\ -6 & 22 \end{bmatrix}, \quad A^3 = \begin{bmatrix} -11 & 57 \\ 38 & -106 \end{bmatrix}. \]
Polynomials in Matrices
Let $f(x)$ be a polynomial in one variable: \[ f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_m x^m \] Then for a square matrix $A$, the polynomial in $A$ is defined as: \[ f(A) = a_0 I + a_1 A + a_2 A^2 + \cdots + a_m A^m \]
Matrix polynomials are obtained by replacing the variable $x$ with the matrix $A$, and replacing the scalar $1$ with the identity matrix $I$ of the same order.
Let \[ A = \begin{bmatrix} 1 & 3 \\ 2 & -4 \end{bmatrix}, \quad f(x) = 2 + x + x^2 \]
Then, \[ f(A) = 2I + A + A^2 \]
We already know: \[ A = \begin{bmatrix} 1 & 3 \\ 2 & -4 \end{bmatrix}, \quad A^2 = \begin{bmatrix} 7 & -9 \\ -6 & 22 \end{bmatrix}, \quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
So, \[ f(A) = 2\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} + \begin{bmatrix} 1 & 3 \\ 2 & -4 \end{bmatrix} + \begin{bmatrix} 7 & -9 \\ -6 & 22 \end{bmatrix} \]
\[ = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ 2 & -4 \end{bmatrix} + \begin{bmatrix} 7 & -9 \\ -6 & 22 \end{bmatrix} = \begin{bmatrix} 10 & -6 \\ -4 & 20 \end{bmatrix} \]
Thus, \[ f(A) = \begin{bmatrix} 10 & -6 \\ -4 & 20 \end{bmatrix}. \]
Definition of Determinant
The determinant of a square matrix $A = [a_{ij}] \in \mathbb{R}^{n \times n}$ is a scalar value, denoted by $\det(A)$ or $|A|$, which gives important information about the matrix, such as invertibility and volume scaling factor of the linear transformation defined by $A$.
Cofactor Expansion
The cofactor of an element $a_{ij}$ in matrix $A$ is defined as: \[ C_{ij} = (-1)^{i+j} M_{ij}, \] where $M_{ij}$ is the determinant of the $(n-1)\times(n-1)$ matrix obtained by deleting the $i$-th row and $j$-th column of $A$.
The determinant of $A$ can be computed by cofactor expansion along any row $i$ or column $j$: \[ \det(A) = \sum_{j=1}^{n} a_{ij} C_{ij} \quad \text{(expansion along row $i$)} \] \[ \det(A) = \sum_{i=1}^{n} a_{ij} C_{ij} \quad \text{(expansion along column $j$)} \]
For $n=1$, $A = [a_{11}]$, we define: \[ \det(A) = a_{11}. \]
For $n=2$, $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, we define: \[ \det(A) = ad - bc. \]
For $n=3$, $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$, we define: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg). \]
Compute the determinant of \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{bmatrix} \] using cofactor expansion along the first row. \[ \det(A) = 1 \cdot \det\begin{bmatrix} 4 & 5 \\ 0 & 6 \end{bmatrix} - 2 \cdot \det\begin{bmatrix} 0 & 5 \\ 1 & 6 \end{bmatrix} + 3 \cdot \det\begin{bmatrix} 0 & 4 \\ 1 & 0 \end{bmatrix}. \] \[ = 1(24 - 0) - 2(0-5) + 3(0-4) = 24 + 10 - 12 = 22. \]
Compute the determinant of \[ A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}. \] Solution: \[ \det(A) = 2 \cdot 4 - 3 \cdot 1 = 8 - 3 = 5. \]
Properties of Determinants
Some key properties of determinants are:
$\det(I_n) = 1$, where $I_n$ is the $n \times n$ identity matrix.
Swapping two rows (or columns) of a matrix multiplies its determinant by $-1$.
Multiplying a row (or column) by a scalar $k$ multiplies the determinant by $k$.
The determinant is linear in each row (or column) separately.
If a matrix has two identical rows (or columns), its determinant is zero.
$\det(A^\mathsf{T}) = \det(A)$.
$\det(AB) = \det(A)\det(B)$.
$A$ is invertible if and only if $\det(A) \neq 0$.
Property 1: $\det(I_n) = 1$.
For $I_3 = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$, \[ \det(I_3) = 1\cdot1\cdot1 = 1. \]
Property 2: Swapping rows changes the sign.
Let \[ A=\begin{bmatrix}1&2\\3&4\end{bmatrix},\quad \det(A) = 1\cdot4 - 2\cdot3 = -2. \]
Swap row 1 and row 2: \[ B=\begin{bmatrix}3&4\\1&2\end{bmatrix},\quad \det(B) = 3\cdot2 - 4\cdot1 = 2. \]
So $\det(B) = -\det(A)$.
Property 3: Multiplying a row scales the determinant.
Let \[ A=\begin{bmatrix}1&2\\3&4\end{bmatrix},\quad \det(A)=-2. \]
Multiply the first row by $5$: \[ B=\begin{bmatrix}5&10\\3&4\end{bmatrix},\quad \det(B)=5\cdot4 - 10\cdot3 = 20 - 30 = -10. \]
Thus $\det(B) = 5 \cdot \det(A)$.
Property 4: Determinant is linear in a row.
Let the first row be the sum of two row vectors: $[3,3]=[1,3]+[2,0]$. Consider \[ A=\begin{bmatrix}3 & 3\\[4pt] 4 & 5\end{bmatrix} =\begin{bmatrix}1+2 & 3+0\\[4pt] 4 & 5\end{bmatrix}. \] By linearity of the first row, \[ \det(A) = \det\begin{bmatrix}1 & 3\\[4pt] 4 & 5\end{bmatrix} + \det\begin{bmatrix}2 & 0\\[4pt] 4 & 5\end{bmatrix}. \] Compute each term: \[ \det\begin{bmatrix}1 & 3\\[4pt] 4 & 5\end{bmatrix} = 1\cdot5 - 3\cdot4 = -7, \qquad \det\begin{bmatrix}2 & 0\\[4pt] 4 & 5\end{bmatrix} = 2\cdot5 - 0\cdot4 = 10. \] Therefore \[ \det(A) = -7 + 10 = 3, \] which agrees with the direct computation \[ \det\begin{bmatrix}3 & 3\\[4pt] 4 & 5\end{bmatrix} = 3\cdot5 - 3\cdot4 = 15-12 = 3. \]
Property 5: Two identical rows $\implies$ determinant $=0$.
Let \[ A=\begin{bmatrix}1&2&3\\1&2&3\\4&5&6\end{bmatrix}. \]
Rows 1 and 2 are identical. Expanding: \[ \det(A)=1(2\cdot6-3\cdot5) -2(1\cdot6-3\cdot4) + 3(1\cdot5-2\cdot4). \]
\[ = 1(12-15) -2(6-12) + 3(5-8) = -3 -2(-6) + 3(-3). \]
\[ = -3 +12 -9 = 0. \]
Hence determinant is $0$.
Property 6: $\det(A^T) = \det(A)$.
Let \[ A=\begin{bmatrix}1&2&3\\0&4&5\\0&0&6\end{bmatrix},\quad \det(A) = 1\cdot4\cdot6 = 24. \]
Transpose: \[ A^T=\begin{bmatrix}1&0&0\\2&4&0\\3&5&6\end{bmatrix},\quad \det(A^T)=24. \]
Thus $\det(A^T)=\det(A)$.
Property 7: $\det(AB) = \det(A)\det(B)$.
Let \[ A=\begin{bmatrix}1&2\\3&4\end{bmatrix},\quad \det(A)=-2. \]
\[ B=\begin{bmatrix}0&1\\1&0\end{bmatrix},\quad \det(B)=-1. \]
Then \[ AB=\begin{bmatrix}2&1\\4&3\end{bmatrix},\quad \det(AB)=2\cdot3-1\cdot4=6-4=2. \]
Also $\det(A)\det(B)=(-2)(-1)=2$.
Property 8: $A$ invertible $\iff \det(A)\neq0$.
Let \[ A=\begin{bmatrix}2&1\\4&2\end{bmatrix}. \]
Compute determinant: \[ \det(A)=2\cdot2 - 1\cdot4 = 0. \]
Since $\det(A)=0$, $A$ is singular, hence not invertible.
Now take \[ B=\begin{bmatrix}1&2\\3&5\end{bmatrix},\quad \det(B)=1\cdot5 - 2\cdot3 = -1 \neq 0. \]
So $B$ is invertible.
Invertible (Nonsingular) Matrices
A square matrix $A$ is said to be invertible or nonsingular if there exists a matrix $B$ such that \[ AB = BA = I, \] where $I$ is the identity matrix of the same order. Such a matrix $B$ is called the inverse of $A$, and is denoted by $A^{-1}$.
Uniqueness of the Inverse: If $AB_1 = B_1A = I$ and $AB_2 = B_2A = I$, then \[ B_1 = B_1I = B_1(AB_2) = (B_1A)B_2 = IB_2 = B_2. \] Hence, the inverse of a matrix is unique.
Suppose \[ A = \begin{bmatrix} 6 & -5 \\ 3 & -3 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 \\ -5 & 6 \end{bmatrix}. \] Then \[ AB = \begin{bmatrix} 6 & -5 \\ 3 & -3 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -5 & 6 \end{bmatrix} = \begin{bmatrix} 31 & -30 \\ 18 & -18 \end{bmatrix}. \] After simplification (scaling), this reduces to the identity matrix $I_2$. Similarly, one can verify that $BA = I_2$. Thus, $A$ and $B$ are inverses of each other.
Property: For two invertible matrices $A$ and $B$, the product $AB$ is invertible and \[ (AB)^{-1} = B^{-1} A^{-1}. \] More generally, if $A_1, A_2, \dots, A_k$ are invertible, then \[ (A_1 A_2 \cdots A_k)^{-1} = A_k^{-1} \cdots A_2^{-1} A_1^{-1}. \]
Inverse of a $2 \times 2$ Matrix
Let \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}. \] The determinant of $A$ is \[ |A| = ad - bc. \]
If $|A| \neq 0$, then $A$ is invertible and \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}. \]
To find $A^{-1}$:
Interchange the diagonal elements $a$ and $d$.
Change the signs of the off-diagonal elements $b$ and $c$.
Divide each entry by the determinant $ad - bc$.
Find the inverses of the following matrices: \[ A = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 2 \\ 3 & 6 \end{bmatrix}. \]
Solution: First, compute $\det(A) = 2(5) - 3(4) = 10 - 12 = -2 \neq 0$. Thus $A$ is invertible and \[ A^{-1} = \frac{1}{-2} \begin{bmatrix} 5 & -4 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} -2.5 & 2 \\ 1.5 & -1 \end{bmatrix}. \]
Now, compute $\det(B) = 1(6) - 3(2) = 6 - 6 = 0$. Since $\det(B) = 0$, $B$ is not invertible.
Remark: A square matrix $A$ is invertible if and only if $\det(A) \neq 0$. This fact is valid for matrices of any order.
Inverse of an $n \times n$ Matrix
For an arbitrary square matrix $A$ of order $n$, the inverse $A^{-1}$ can be obtained by solving $n$ systems of $n$ linear equations or, equivalently, by using the adjoint formula: \[ A^{-1} = \frac{1}{\det(A)} \, \text{adj}(A), \quad \text{if } \det(A) \neq 0. \]
Find the inverse of \[ B = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}. \]
Solution: Step 1: Compute the determinant. \[ \det(B) = 1 \cdot \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2 \cdot \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix}. \]
\[ = 1(1\cdot 0 - 4\cdot 6) - 2(0\cdot 0 - 5\cdot 4) + 3(0\cdot 6 - 1\cdot 5). \]
\[ = 1(-24) - 2(-20) + 3(-5) = -24 + 40 - 15 = 1. \]
So $\det(B)=1$, hence $B$ is invertible.
Step 2: Compute cofactors. \[ C = \begin{bmatrix} \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} & -\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} & \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} \\[8pt] -\begin{vmatrix} 2 & 3 \\ 6 & 0 \end{vmatrix} & \begin{vmatrix} 1 & 3 \\ 5 & 0 \end{vmatrix} & -\begin{vmatrix} 1 & 2 \\ 5 & 6 \end{vmatrix} \\[8pt] \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} & -\begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} & \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} \end{bmatrix}. \]
\[ C = \begin{bmatrix} -24 & 20 & -5 \\ -18 & -15 & -7 \\ 5 & -4 & 1 \end{bmatrix}. \]
Step 3: Transpose to get the adjoint: \[ \text{adj}(B) = \begin{bmatrix} -24 & -18 & 5 \\ 20 & -15 & -4 \\ -5 & -7 & 1 \end{bmatrix}. \]
Step 4: Compute inverse: \[ B^{-1} = \frac{1}{\det(B)} \, \text{adj}(B) = \text{adj}(B). \]
\[ B^{-1} = \begin{bmatrix} -24 & -18 & 5 \\ 20 & -15 & -4 \\ -5 & -7 & 1 \end{bmatrix}. \]
Applications of Determinants
Determinants have several applications in linear algebra:
Solving linear systems: Using Cramer's rule.
Volume scaling: The absolute value of $\det(A)$ represents the scaling factor of volume under the transformation $x \mapsto Ax$.
Eigenvalues: Roots of $\det(A - \lambda I) = 0$ give eigenvalues.
Invertibility check: Quick test for non-singular matrices.
Area/Volume computation: Determinants of 2D/3D vectors give area/volume of parallelograms/parallelepipeds.
Find the area of the parallelogram formed by vectors $\mathbf{u} = (1,2)$ and $\mathbf{v} = (3,4)$. \[ \text{Area} = |\det \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}| = |1\cdot4 - 3\cdot2| = |-2| = 2. \]
Special Square Matrices
Symmetric and Skew-Symmetric
A square matrix $A$ is called symmetric if $A^T = A$. Equivalently, $A = [a_{ij}]$ is symmetric if $a_{ij} = a_{ji}$ for all $i,j$.
A square matrix $A$ is called skew-symmetric if $A^T = -A$. Equivalently, $a_{ij} = -a_{ji}$ for all $i,j$. Note that in a skew-symmetric matrix, all diagonal entries must be zero since $a_{ii} = -a_{ii}$.
\[ A = \begin{bmatrix} 4 & -3 & 5 \\ -3 & 6 & 7 \\ 5 & 7 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 3 & -4 \\ -3 & 0 & 5 \\ 4 & -5 & 0 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 5 & 5 \end{bmatrix} \]
- $A$ is symmetric because $A^T = A$. - $B$ is skew-symmetric because $B^T = -B$. - $C$ is neither symmetric nor skew-symmetric (not square).