Determinant of 4×4 Matrix

Determinant of a 4 × 4 Matrix is a fundamental concept in linear algebra, essential for deriving a single scalar value from the matrix. This article aims to elucidate the definition of a 4 × 4 matrix and guide through the step-by-step process of calculating its determinant. Additionally, it explores the practical applications of this mathematical operation. Whether you’re a student delving into linear algebra or an enthusiast aiming to grasp matrix operations thoroughly, mastering the determinant of a 4 × 4 matrix holds significant importance.

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 “det” symbol or vertical bars enclosing the matrix.

What is a 4×4 Matrix?

A 4×4 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 4×4 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 a_ij represents the element located in the i^th row and j^th column of the matrix.

4×4 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 4×4 Matrix

Here are some properties of a 4×4 matrix explained in simplified terms:

• Square Matrix: A 4×4 matrix has an equal number of rows and columns, making it a square matrix.
• Determinant: The determinant of a 4×4 matrix can be calculated using methods like cofactor expansion or row reduction. It provides information about the matrix’s invertibility and scaling factor for linear transformations.
• Inverse: A 4×4 matrix is invertible if its determinant is non-zero. The inverse of a 4×4 matrix allows solving systems of linear equations and undoing transformations represented by the matrix.
• Transpose: The transpose of a 4×4 matrix is obtained by interchanging its rows and columns. It can be useful in certain computations and transformations.
• Eigenvalues and Eigenvectors: 4×4 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 4×4 matrices following specific rules and properties.

Determinant of 4 × 4 Matrix Formula

Determinant of any 4 × 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 · det(A₁₁) – a₁₂ · det(A₁₂) + a₁₃ · det(A₁₃) – a₁₄ · det(A₁₄)

Where A_ij denotes the submatrix by deleting i^th row and j^th column.

How do you find the Determinant of a 4 × 4 Matrix?

To find the determinant of a 4×4 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 2×2 submatrix, for which you can directly compute the determinant. To understand how to find the determinant of a 4×4 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 · det(A₁₁) – 1 · det(A₁₂) + 3 · det(A₁₃) – 4 · det(A₁₄)

Where A_ij denotes the submatrix obtained by deleting the i-th row and j-th column.

Step 2: Compute the determinant of each 3×3 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) – 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]

⇒ |A₁₁| = (-1)[(0)(1)-(5)(2)] – 2[(2)(1)-(5)(3)] + 1[(2)(2)-(0)(3)]

⇒ |A₁₁| = (-1)[(-10)] – 2[(2)-(15)] + 1[(4)-(0)]

⇒ |A₁₁| = 10 – 2(-13) + 4

⇒ |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) – 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]

⇒ |A₁₂| = (0)[(0)(1)-(5)(2)] – 2[(3)(1)-(5)(-1)] + 1[(3)(2)-(0)(-1)]

⇒ |A₁₂| = (0)[(-10)] – 2[(3)+(5)] + 1[(6)-(0)]

⇒ |A₁₂| = 0 – 2(8) + 6

⇒ |A₁₂| = 0 – 16+ 6= 10

For A₁₃

[Tex]A_{13} = \begin{bmatrix} 0 & -1 & 2 \\ 3 & 2 & 5 \\ -1 & 3 & 1 \\ \end{bmatrix}[/Tex]

[Tex]\text{det}(A_{13}) = (0) \cdot \text{det}\left(\begin{bmatrix} -1 & 2 \\ 2 & 1 \end{bmatrix}\right) – (-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]

⇒ |A₁₃| = (0)[(2)(1)-(3)(5)] – (-1)[(3)(1)-(5)(-1)] + 2[(3)(3)-(2)(-1)]

⇒ |A₁₃| = (0)[(-1)-(4)] – (-1)[(3)+(5)] + 2[(9)-(-2)]

⇒ |A₁₃| = 0 – (-1)(8) + 2(11)

⇒ |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} -1 & 2 \\ 2 & 0 \end{bmatrix}\right) – (-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]

⇒ |A₁₄| = (0)[(-1)(0)-(2)(2)] – (-1)[(3)(2)-(0)(-1)] + 2[(3)(3)-(2)(-1)]

⇒ |A₁₄| = (0)[(0)-(4)] – (-1)[(6)-(0)] + 2[(9)-(-2)]

⇒ |A₁₄| = 0 – (-1)(6) + 2(11)

⇒ |A₁₄| = 6 + 22 = 28

Step 3: Substitute the determinants of the 3×3 submatrices into the expansion formula:

(A) = 2 · 40 – 1 · 10 + 3 · 30 – 4 · 28

Step 4: Compute the final determinant:

det(A) = 80 – 10 + 90 – 112

det(A) = 48

So, the determinant of the given 4×4 matrix is 48.

Determinant of 4×4 Matrix Solved 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}) – 1 \cdot \text{det}(A_{12}) + 0 \cdot \text{det}(A_{13}) – 3 \cdot \text{det}(A_{14})[/Tex]

Now, compute the determinant of each 3×3 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) – 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)) – 2((2)(3)-(5)(0)) + 0((2)(-2)-(1)(0))

= (-1)((3)+(10)) – 2((6)-(0)) + 0((-4)-(0))

= (-1)(13) – 2(6) + 0(-4)

= -13 – 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) – (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)) – (0)((-3)(3)-(5)(1)) + (3)((-3)(2)-(1)(1))

= (2)((3)-(10)) – (0)((-9)-(5)) + (3)((-6)-(1))

= (2)(-7) – (0)(-14) + (3)(-7)

= -14 – 0 – 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) – (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)) – (1)((-3)(3)-(5)(1)) + (3)((-3)(0)-(2)(1))

= (2)((6)-(0)) – (1)((-9)-(5)) + (3)((0)-(2))

= (2)(6) – (1)(-14) + (3)(-2)

= 12 + 14 – 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) – (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)) – (1)((-3)(-2)-(1)(1)) + (0)((-3)(0)-(2)(1))

= (2)((-4)-(0)) – (1)((6)-(1)) + (0)((0)-(2))

= (2)(-4) – (1)(5) + (0)(-2)

= -8 – 5 + 0

= -13

Now, substitute the determinants of the 3×3 submatrices into the expansion formula:

det(A) = 2 \cdot (-25) – 1 \cdot (-35) + 0 – 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 ), we’ll 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} – 1 \cdot \begin{vmatrix} -1 & 2 & 5 \\ 3 & 1 & 0 \\ 4 & 3 & 1 \end{vmatrix} – 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, let’s compute the determinants of the 3×3 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 – 0 \cdot 3) – 2 \cdot (2 \cdot 1 – 0 \cdot (-2)) + 5 \cdot (2 \cdot 3 – 2 \cdot (-2)))[/Tex]

= 2 · (0 – 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 – 0 \cdot 3) – 2 \cdot (3 \cdot 1 – 0 \cdot 4) + 5 \cdot (3 \cdot 3 – 1 \cdot 4))[/Tex]

= -1 · (1 – 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 – 0 \cdot (-2)) – 0 \cdot (3 \cdot 5 – 0 \cdot 4) + 5 \cdot (3 \cdot (-2) – 2 \cdot 4))[/Tex]

= -1 · (6 – 0 – 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 – 1 \cdot (-2)) – 0 \cdot (2 \cdot 3 – 1 \cdot 4) + 2 \cdot (3 \cdot (-2) – 2 \cdot 4))[/Tex]

= -1 · (8 – 0 + 0) = -8

Now, substitute these determinants back into the expansion formula:

det(A) = 2 · 52 – 1 · 60 – 3 · 44 + 4 · (-8) = 104 – 60 – 132 – 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 ), we’ll 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} – 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, let’s compute the determinants of the 3×3 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) – 5 \cdot 0) – (-3) \cdot (-1 \cdot (-4) – 5 \cdot 2) + 2 \cdot (-1 \cdot 0 – 2 \cdot 2))[/Tex]

= -2 ⋅ (1 ⋅ (-8) – (-3) ⋅ (4 – 10) + 2 ⋅ (-4))

= -2 ⋅ (-8 + 18 – 8) = -2 ⋅ 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) – 5 \cdot 0) – (-3) \cdot (0 \cdot (-4) – 5 \cdot 3) + 2 \cdot (0 \cdot 0 – 2 \cdot 3))[/Tex]

= 3 ⋅ (4 ⋅ (-8) – (-3) ⋅ (0 – 15) + 2 ⋅ (0 – 6))

= 3 ⋅ (-32 + 45 – 12) = 3 ⋅ 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) – 2 \cdot 4) – 1 \cdot (0 \cdot (-4) – 2 \cdot 3) + 2 \cdot (0 \cdot 4 – (-1) \cdot 3)[/Tex]

= -1 ⋅ (-16 – 8) – 1 ⋅ (0 – 6) + 2 ⋅ (0 + 3)

= -1 ⋅ (-24) – 1 ⋅ (-6) + 2 ⋅ 3

= 24 + 6 + 6

= 36

Now, substitute these determinants back into the expansion formula:

det(B) = -2 ⋅ (-4) + 3 ⋅ 3 – 1 ⋅ 36 + 0 ⋅ anything

= 8 + 9 – 36 + 0

= -19

So, the determinant of matrix ( B ) is det(B) = -19

Practice Questions on Determinant of 4×4 Matrix

Q1: Calculate the determinant of the following 4×4 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 4×4 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]

Determinant of 4×4 Matrix: FAQs

How do you Find the Determinant of a 4×4 Matrix?

To find the determinant of a 4×4 matrix, you can use various methods like cofactor expansion or row reduction techniques.

What is the Determinant of a 4×4 Identity Matrix?

The determinant of a 4×4 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 4×4 Matrix Using Cofactor Expansion?

Determining the determinant of a 4×4 matrix using cofactor expansion involves breaking it down into smaller 3×3 matrices, applying the cofactor formula, and summing the products.

What is the Formula of Determinants?

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 4×4 Matrix Have an Inverse?

A 4×4 matrix can have an inverse if its determinant is nonzero; otherwise, it is singular and lacks an inverse.

How do You Show a 4×4 Matrix is Invertible?

To show a 4×4 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.