Determinant of 4Γ4 Matrix | Examples and How to Find
Determinant of 4Γ4 Matrix: Determinant of a Matrix is a fundamental concept in linear algebra, essential for deriving a single scalar value from the matrix. 4Γ4 is a square matrix with 4 rows and 4 columns whose determinant can be found by a formula which we will discuss.
This article will explore the definition of a 4 Γ 4 matrix and guide through the step-by-step process of calculating the determinant of 4Γ4 matrix. Additionally, it explores the practical applications of this mathematical operation.
Table of Content
- What is the Determinant of a Matrix?
- Determinant of 4Γ4 Matrix
- Properties of 4Γ4 Matrix
- Determinant of 4 Γ 4 Matrix Formula
- How do you find the Determinant of a 4 Γ 4 Matrix?
- Determinant of 4Γ4 Matrix Examples
- Determinant of 4Γ4 Matrix Practice Questions
What is the Determinant of a Matrix?
The determinant of a matrix is a scalar value that can be calculated from the elements of a square matrix. It provides important information about the matrix, such as whether it is invertible and the scaling factor of linear transformations represented by the matrix.
Various methods, such as cofactor expansion or row reduction, can be employed to find the determinant of a matrix, depending on the size and structure of the matrix. Once calculated, the determinant is denoted by the βdetβ symbol or vertical bars enclosing the matrix.
Determinant of 4Γ4 Matrix
A 4Γ4 matrix is a rectangular array of numbers arranged in four rows and four columns. Each element in the matrix is identified by its row and column position. The general form of a 4Γ4 matrix looks like this:
[Tex]\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} [/Tex]
Where aij represents the element located in the ith row and jth column of the matrix.
4Γ4 matrices are commonly encountered in various fields such as computer graphics, physics, engineering, and mathematics. They are used to represent transformations, solve systems of linear equations, and perform operations in linear algebra.
Properties of 4Γ4 Matrix
Here are some properties of a 4Γ4 matrix explained in simplified terms:
- Square Matrix: A 4Γ4 matrix has an equal number of rows and columns, making it a square matrix.
- Determinant: The determinant of a 4Γ4 matrix can be calculated using methods like cofactor expansion or row reduction. It provides information about the matrixβs invertibility and scaling factor for linear transformations.
- Inverse: A 4Γ4 matrix is invertible if its determinant is non-zero. The inverse of a 4Γ4 matrix allows solving systems of linear equations and undoing transformations represented by the matrix.
- Transpose: The transpose of a 4Γ4 matrix is obtained by interchanging its rows and columns. It can be useful in certain computations and transformations.
- Eigenvalues and Eigenvectors: 4Γ4 matrices can be analysed to find their eigenvalues and eigenvectors, which represent properties of the matrix under linear transformations.
- Symmetry: Depending on the specific matrix, it may exhibit properties of symmetry such as being symmetric, skew-symmetric, or neither.
- Matrix Operations: Various operations such as addition, subtraction, multiplication, and scalar multiplication can be performed on 4Γ4 matrices following specific rules and properties.
Read in Detail: Properties of Determinants
Determinant of 4 Γ 4 Matrix Formula
Determinant of any 4 Γ 4 Matrix i.e., [Tex]\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} [/Tex], can be calculated using the following formula:
det(A) = a11 Β· det(A11) β a12 Β· det(A12) + a13 Β· det(A13) β a14 Β· det(A14)
Where Aij denotes the submatrix by deleting ith row and jth column.
How do you find the Determinant of a 4 Γ 4 Matrix?
To find the determinant of a 4Γ4 matrix, you can use various methods such as expansion by minors, row reduction, or applying specific properties.
One common method is to use expansion by minors, where you expand along a row or column by multiplying each element by its cofactor and summing the results. This process continues recursively until you reach a 2Γ2 submatrix, for which you can directly compute the determinant. To understand how to find the determinant of a 4Γ4 matrix consider an example.
[Tex]\begin{bmatrix} 2 & 1 & 3 & 4 \\ 0 & -1 & 2 & 1 \\ 3 & 2 & 0 & 5 \\ -1 & 3 & 2 & 1 \\ \end{bmatrix}[/Tex]
Step 1: Expand along the first row:
det(A) = 2 Β· det(A11) β 1 Β· det(A12) + 3 Β· det(A13) β 4 Β· det(A14)
Where Aij denotes the submatrix obtained by deleting the i-th row and j-th column.
Step 2: Compute the determinant of each 3Γ3 submatrix.
For A11
[Tex]A_{11} = \begin{bmatrix} -1 & 2 & 1 \\ 2 & 0 & 5 \\ 3 & 2 & 1 \\ \end{bmatrix}[/Tex]
[Tex]\text{det}(A_{11}) = (-1) \cdot \text{det}\left(\begin{bmatrix} 0 & 5 \\ 2 & 1 \end{bmatrix}\right) β 2 \cdot \text{det}\left(\begin{bmatrix} 2 & 5 \\ 3 & 1 \end{bmatrix}\right) + 1 \cdot \text{det}\left(\begin{bmatrix} 2 & 0 \\ 3 & 2 \end{bmatrix}\right)[/Tex]
β |A11| = (-1)[(0)(1)-(5)(2)] β 2[(2)(1)-(5)(3)] + 1[(2)(2)-(0)(3)]
β |A11| = (-1)[(-10)] β 2[(2)-(15)] + 1[(4)-(0)]
β |A11| = 10 β 2(-13) + 4
β |A11| = 10 + 26 + 4= 40
For A12
[Tex]A_{12} = \begin{bmatrix} 0 & 2 & 1 \\ 3 & 0 & 5 \\ -1 & 2 & 1 \\ \end{bmatrix}[/Tex]
[Tex]\text{det}(A_{12}) = (0) \cdot \text{det}\left(\begin{bmatrix} 0 & 5 \\ 2 & 1 \end{bmatrix}\right) β 2 \cdot \text{det}\left(\begin{bmatrix} 3 & 5 \\ -1 & 1 \end{bmatrix}\right) + 1 \cdot \text{det}\left(\begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix}\right)[/Tex]
β |A12| = (0)[(0)(1)-(5)(2)] β 2[(3)(1)-(5)(-1)] + 1[(3)(2)-(0)(-1)]
β |A12| = (0)[(-10)] β 2[(3)+(5)] + 1[(6)-(0)]
β |A12| = 0 β 2(8) + 6
β |A12| = 0 β 16+ 6= 10
For A13
[Tex]A_{13} = \begin{bmatrix} 0 & -1 & 1 \\ 3 & 2 & 5 \\ -1 & 3 & 1 \\ \end{bmatrix}[/Tex]
[Tex]\text{det}(A_{13}) = (0) \cdot \text{det}\left(\begin{bmatrix} 2 & 5 \\ 3 & 1 \end{bmatrix}\right) β (-1) \cdot \text{det}\left(\begin{bmatrix} 3 & 5 \\ -1 & 1 \end{bmatrix}\right) + 2 \cdot \text{det}\left(\begin{bmatrix} 3 & 2 \\ -1 & 3 \end{bmatrix}\right)[/Tex]
β |A13| = (0)[(2)(1)-(3)(5)] β (-1)[(3)(1)-(5)(-1)] + 2[(3)(3)-(2)(-1)]
β |A13| = (0)[(2)-(15)] β (-1)[(3)+(5)] + 2[(9)-(-2)]
β |A13| = 0 β (-1)(8) + 2(11)
β |A13| = 8 + 22= 30
For A14
[Tex]A_{14} = \begin{bmatrix} 0 & -1 & 2 \\ 3 & 2 & 0 \\ -1 & 3 & 2 \\ \end{bmatrix}[/Tex]
[Tex]\text{det}(A_{14}) = (0) \cdot \text{det}\left(\begin{bmatrix} 2 & 0 \\ 3 & 2 \end{bmatrix}\right) β (-1) \cdot \text{det}\left(\begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix}\right) + 2 \cdot \text{det}\left(\begin{bmatrix} 3 & 2 \\ -1 & 3 \end{bmatrix}\right)[/Tex]
β |A14| = (0)[(2)(2)-(3)(0)] β (-1)[(3)(2)-(0)(-1)] + 2[(3)(3)-(2)(-1)]
β |A14| = (0)[(4)-(0)] β (-1)[(6)-(0)] + 2[(9)-(-2)]
β |A14| = 0 β (-1)(6) + 2(11)
β |A14| = 6 + 22 = 28
Step 3: Substitute the determinants of the 3Γ3 submatrices into the expansion formula:
(A) = 2 Β· 40 β 1 Β· 10 + 3 Β· 30 β 4 Β· 28
Step 4: Compute the final determinant:
det(A) = 80 β 10 + 90 β 112
det(A) = 48
So, the determinant of the given 4Γ4 matrix is 48.
Also, Check
Determinant of 4Γ4 Matrix Examples
Example 1: A = [Tex]\begin{bmatrix} 2 & 1 & 0 & 3 \\ 4 & -1 & 2 & 0 \\ -3 & 2 & 1 & 5 \\ 1 & 0 & -2 & 3 \end{bmatrix}[/Tex]
Solution:
First Expand along the first row:
[Tex]\text{det}(A) = 2 \cdot \text{det}(A_{11}) β 1 \cdot \text{det}(A_{12}) + 0 \cdot \text{det}(A_{13}) β 3 \cdot \text{det}(A_{14})[/Tex]
Now, compute the determinant of each 3Γ3 submatrix.
For (A11):
[Tex]A_{11} = \begin{bmatrix} -1 & 2 & 0 \\ 2 & 1 & 5 \\ 0 & -2 & 3 \end{bmatrix}[/Tex]
[Tex]\text{det}(A_{11}) = (-1) \cdot \text{det}\left(\begin{bmatrix} 1 & 5 \\ -2 & 3 \end{bmatrix}\right) β 2 \cdot \text{det}\left(\begin{bmatrix} 2 & 5 \\ 0 & 3 \end{bmatrix}\right) + 0 \cdot \text{det}\left(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\right)[/Tex]
= (-1)((1)(3)-(5)(-2)) β 2((2)(3)-(5)(0)) + 0((2)(-2)-(1)(0))
= (-1)((3)+(10)) β 2((6)-(0)) + 0((-4)-(0))
= (-1)(13) β 2(6) + 0(-4)
= -13 β 12
= -25
For (A12):
[Tex]A_{12} = \begin{bmatrix} 2 & 0 & 3 \\ -3 & 1 & 5 \\ 1 & 2 & 3 \end{bmatrix}[/Tex]
[Tex]\text{det}(A_{12}) = (2) \cdot \text{det}\left(\begin{bmatrix} 1 & 5 \\ 2 & 3 \end{bmatrix}\right) β (0) \cdot \text{det}\left(\begin{bmatrix} -3 & 5 \\ 1 & 3 \end{bmatrix}\right) + (3) \cdot \text{det}\left(\begin{bmatrix} -3 & 1 \\ 1 & 2 \end{bmatrix}\right)[/Tex]
= (2)((1)(3)-(5)(2)) β (0)((-3)(3)-(5)(1)) + (3)((-3)(2)-(1)(1))
= (2)((3)-(10)) β (0)((-9)-(5)) + (3)((-6)-(1))
= (2)(-7) β (0)(-14) + (3)(-7)
= -14 β 0 β 21
= -35
For (A13):
[Tex]A_{13} = \begin{bmatrix} 2 & 1 & 3 \\ -3 & 2 & 5 \\ 1 & 0 & 3 \end{bmatrix}[/Tex]
[Tex]\text{det}(A_{13}) = (2) \cdot \text{det}\left(\begin{bmatrix} 2 & 5 \\ 0 & 3 \end{bmatrix}\right) β (1) \cdot \text{det}\left(\begin{bmatrix} -3 & 5 \\ 1 & 3 \end{bmatrix}\right) + (3) \cdot \text{det}\left(\begin{bmatrix} -3 & 2 \\ 1 & 0 \end{bmatrix}\right)[/Tex]
= (2)((2)(3)-(5)(0)) β (1)((-3)(3)-(5)(1)) + (3)((-3)(0)-(2)(1))
= (2)((6)-(0)) β (1)((-9)-(5)) + (3)((0)-(2))
= (2)(6) β (1)(-14) + (3)(-2)
= 12 + 14 β 6
= 20
For (A14):
[Tex]A_{14} = \begin{bmatrix} 2 & 1 & 0 \\ -3 & 2 & 1 \\ 1 & 0 & -2 \end{bmatrix}[/Tex]
[Tex]\text{det}(A_{14}) = (2) \cdot \text{det}\left(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\right) β (1) \cdot \text{det}\left(\begin{bmatrix} -3 & 1 \\ 1 & -2 \end{bmatrix}\right) + (0) \cdot \text{det}\left(\begin{bmatrix} -3 & 2 \\ 1 & 0 \end{bmatrix}\right)[/Tex]
= (2)((2)(-2)-(1)(0)) β (1)((-3)(-2)-(1)(1)) + (0)((-3)(0)-(2)(1))
= (2)((-4)-(0)) β (1)((6)-(1)) + (0)((0)-(2))
= (2)(-4) β (1)(5) + (0)(-2)
= -8 β 5 + 0
= -13
Now, substitute the determinants of the 3Γ3 submatrices into the expansion formula:
det(A) = 2 \cdot (-25) β 1 \cdot (-35) + 0 β 3 \cdot (-13)
= -50 + 35 + 0 + 39
= -50 + 35 + 39
= 24
So, the determinant of matrix (A) is 24.
Example 2: Calculate the determinant of the matrix [Tex]A = \begin{bmatrix} 2 & 1 & -3 & 4 \\ -1 & 0 & 2 & 5 \\ 3 & 2 & 1 & 0 \\ 4 & -2 & 3 & 1 \end{bmatrix}[/Tex]
Solution:
To find the determinant of the matrix ( A ), weβll use the expansion by minors method along the first row:
[Tex]\text{det}(A) = 2 \cdot \begin{vmatrix} 0 & 2 & 5 \\ 2 & 1 & 0 \\ -2 & 3 & 1 \end{vmatrix} β 1 \cdot \begin{vmatrix} -1 & 2 & 5 \\ 3 & 1 & 0 \\ 4 & 3 & 1 \end{vmatrix} β 3 \cdot \begin{vmatrix} -1 & 0 & 5 \\ 3 & 2 & 0 \\ 4 & -2 & 3 \end{vmatrix} + 4 \cdot \begin{vmatrix} -1 & 0 & 2 \\ 3 & 2 & 1 \\ 4 & -2 & 3 \end{vmatrix} [/Tex]
Now, letβs compute the determinants of the 3Γ3 submatrices:
[Tex]\text{det}\left( \begin{vmatrix} 0 & 2 & 5 \\ 2 & 1 & 0 \\ -2 & 3 & 1 \end{vmatrix} \right) = 2 \cdot (0 \cdot (1 \cdot 1 β 0 \cdot 3) β 2 \cdot (2 \cdot 1 β 0 \cdot (-2)) + 5 \cdot (2 \cdot 3 β 2 \cdot (-2)))[/Tex]
= 2 Β· (0 β 4 + 30) = 52
[Tex]\text{det}\left( \begin{vmatrix} -1 & 2 & 5 \\ 3 & 1 & 0 \\ 4 & 3 & 1 \end{vmatrix} \right) = -1 \cdot ((1 \cdot 1 β 0 \cdot 3) β 2 \cdot (3 \cdot 1 β 0 \cdot 4) + 5 \cdot (3 \cdot 3 β 1 \cdot 4))[/Tex]
= -1 Β· (1 β 6 + 45) = 60
[Tex]\text{det}\left( \begin{vmatrix} -1 & 0 & 5 \\ 3 & 2 & 0 \\ 4 & -2 & 3 \end{vmatrix} \right) = -1 \cdot ((2 \cdot 3 β 0 \cdot (-2)) β 0 \cdot (3 \cdot 5 β 0 \cdot 4) + 5 \cdot (3 \cdot (-2) β 2 \cdot 4))[/Tex]
= -1 Β· (6 β 0 β 50) = 44
[Tex]\text{det}\left( \begin{vmatrix} -1 & 0 & 2 \\ 3 & 2 & 1 \\ 4 & -2 & 3 \end{vmatrix} \right) = -1 \cdot ((2 \cdot 3 β 1 \cdot (-2)) β 0 \cdot (2 \cdot 3 β 1 \cdot 4) + 2 \cdot (3 \cdot (-2) β 2 \cdot 4))[/Tex]
= -1 Β· (8 β 0 + 0) = -8
Now, substitute these determinants back into the expansion formula:
det(A) = 2 Β· 52 β 1 Β· 60 β 3 Β· 44 + 4 Β· (-8) = 104 β 60 β 132 β 32 = -120
So, the determinant of matrix ( A ) is det(A) = -120.
Example 3: Find the determinant of the matrix B = [Tex]\begin{bmatrix} -2 & 3 & 1 & 0 \\ 4 & 1 & -3 & 2 \\ 0 & -1 & 2 & 5 \\ 3 & 2 & 0 & -4 \end{bmatrix}[/Tex]
Solution:
To find the determinant of matrix ( B ), weβll use the expansion by minors method along the first row:
[Tex]\text{det}(B) = -2 \cdot \begin{vmatrix} 1 & -3 & 2 \\ -1 & 2 & 5 \\ 2 & 0 & -4 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & -3 & 2 \\ 0 & 2 & 5 \\ 3 & 0 & -4 \end{vmatrix} β 1 \cdot \begin{vmatrix} 4 & 1 & 0 \\ 0 & -1 & 2 \\ 3 & 4 & -4 \end{vmatrix} + 0 \cdot \begin{vmatrix} 4 & 1 & 0 \\ 0 & -1 & 2 \\ 3 & 4 & 1 \end{vmatrix}[/Tex]
Now, letβs compute the determinants of the 3Γ3 submatrices:
[Tex]\text{det}\left( \begin{vmatrix} 1 & -3 & 2 \\ -1 & 2 & 5 \\ 2 & 0 & -4 \end{vmatrix} \right) = -2 \cdot (1 \cdot (2 \cdot (-4) β 5 \cdot 0) β (-3) \cdot (-1 \cdot (-4) β 5 \cdot 2) + 2 \cdot (-1 \cdot 0 β 2 \cdot 2))[/Tex]
= -2 β (1 β (-8) β (-3) β (4 β 10) + 2 β (-4))
= -2 β (-8 + 18 β 8) = -2 β 2 = -4
[Tex]\text{det}\left( \begin{vmatrix} 4 & -3 & 2 \\ 0 & 2 & 5 \\ 3 & 0 & -4 \end{vmatrix} \right) = 3 \cdot (4 \cdot (2 \cdot (-4) β 5 \cdot 0) β (-3) \cdot (0 \cdot (-4) β 5 \cdot 3) + 2 \cdot (0 \cdot 0 β 2 \cdot 3))[/Tex]
= 3 β (4 β (-8) β (-3) β (0 β 15) + 2 β (0 β 6))
= 3 β (-32 + 45 β 12) = 3 β 1 = 3
[Tex]\text{det}\left( \begin{vmatrix} 4 & 1 & 0 \\ 0 & -1 & 2 \\ 3 & 4 & -4 \end{vmatrix} \right) = -1 \cdot (4 \cdot (-4) β 2 \cdot 4) β 1 \cdot (0 \cdot (-4) β 2 \cdot 3) + 2 \cdot (0 \cdot 4 β (-1) \cdot 3)[/Tex]
= -1 β (-16 β 8) β 1 β (0 β 6) + 2 β (0 + 3)
= -1 β (-24) β 1 β (-6) + 2 β 3
= 24 + 6 + 6
= 36
Now, substitute these determinants back into the expansion formula:
det(B) = -2 β (-4) + 3 β 3 β 1 β 36 + 0 β anything
= 8 + 9 β 36 + 0
= -19
So, the determinant of matrix ( B ) is det(B) = -19
Determinant of 4Γ4 Matrix Practice Questions
Q1: Calculate the determinant of the following 4Γ4 matrix: [Tex]A = \begin{bmatrix} 2 & 0 & 1 & 3 \\ -1 & 2 & 2 & 0 \\ 3 & -2 & 0 & 1 \\ 1 & 1 & 2 & -1 \\ \end{bmatrix}[/Tex]
Q2: Find the determinant of the matrix: [Tex]B = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 2 & 3 & 4 & 5 \\ \end{bmatrix}[/Tex]
Q3: Calculate the determinant of the following 4Γ4 matrix: [Tex]C = \begin{bmatrix} 2 & 1 & 0 & -1 \\ 3 & 2 & -1 & 0 \\ 0 & -3 & 2 & 1 \\ 1 & 0 & 3 & -2 \\ \end{bmatrix}[/Tex]
Q4: Determine the determinant of the matrix: [Tex]D = \begin{bmatrix} 4 & 2 & 1 & 0 \\ -1 & 3 & 0 & 2 \\ 0 & 2 & 1 & -3 \\ 2 & 0 & -1 & 4 \\ \end{bmatrix} [/Tex]
Q5: Find the determinant of the matrix: [Tex] E = \begin{bmatrix} 3 & 1 & -2 & 0 \\ 2 & 0 & 1 & 1 \\ -1 & 2 & 3 & -2 \\ 0 & 3 & -1 & 1 \\ \end{bmatrix} [/Tex]
FAQs on Determinant of 4Γ4 Matrix
How do you find the determinant of a 4Γ4 matrix?
To find the determinant of a 4Γ4 matrix, you can use various methods like cofactor expansion or row reduction techniques.
What is the determinant of a 4Γ4 identity matrix?
The determinant of a 4Γ4 identity matrix is 1, as it is a special case where all diagonal elements are 1, and the rest are 0.
How to find the determinant of a 4Γ4 matrix using cofactor expansion?
Determining the determinant of a 4Γ4 matrix using cofactor expansion involves breaking it down into smaller 3Γ3 matrices, applying the cofactor formula, and summing the products.
What is the formula of determinant?
The formula for the determinant involves summing the products of elements and their cofactors in each row or column, considering their signs.
Can a determinant be negative?
Yes, determinants can be negative, positive, or zero, depending on the specific matrix and its properties.
Can a 4Γ4 matrix have an inverse?
A 4Γ4 matrix can have an inverse if its determinant is nonzero; otherwise, it is singular and lacks an inverse.
How do you show a 4Γ4 matrix is invertible?
To show a 4Γ4 matrix is invertible, confirm that its determinant is nonzero, indicating the existence of an inverse, and use additional criteria like row reduction to verify invertibility.