Determinant of 4x4 Matrix | Examples and How to Find

Determinant of 4x4 Matrix: Determinant of a Matrix is a fundamental concept in linear algebra, essential for deriving a single scalar value from the matrix. 4x4 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 × 4 matrix and guide through the step-by-step process of calculating the determinant of 4x4 matrix. Additionally, it explores the practical applications of this mathematical operation.

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.

Determinant of 4x4 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:

\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}

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.

Read in Detail: Properties of Determinants

Determinant of 4 × 4 Matrix Formula

Determinant of any 4 × 4 Matrix i.e., \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} , 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 4x4 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 2x2 submatrix, for which you can directly compute the determinant. To understand how to find the determinant of a 4×4 matrix consider an example.

\begin{bmatrix} 2 & 1 & 3 & 4 \\ 0 & -1 & 2 & 1 \\ 3 & 2 & 0 & 5 \\ -1 & 3 & 2 & 1 \\ \end{bmatrix}

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 3x3 submatrix.

For A₁₁

A_{11} = \begin{bmatrix} -1 & 2 & 1 \\ 2 & 0 & 5 \\ 3 & 2 & 1 \\ \end{bmatrix}

\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)

⇒ |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

A_{12} = \begin{bmatrix} 0 & 2 & 1 \\ 3 & 0 & 5 \\ -1 & 2 & 1 \\ \end{bmatrix}

\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)

⇒ |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

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

\text{det}(A_{13}) = (0) \cdot \text{det}\left(\begin{bmatrix} 2 & 5 \\ 3 & 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)

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

⇒ |A₁₃| = (0)[(2)-(15)] - (-1)[(3)+(5)] + 2[(9)-(-2)]

⇒ |A₁₃| = 0 - (-1)(8) + 2(11)

⇒ |A₁₃| = 8 + 22= 30

For A₁₄

A_{14} = \begin{bmatrix} 0 & -1 & 2 \\ 3 & 2 & 0 \\ -1 & 3 & 2 \\ \end{bmatrix}

\text{det}(A_{14}) = (0) \cdot \text{det}\left(\begin{bmatrix} 2 & 0 \\ 3 & 2 \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)

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

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

⇒ |A₁₄| = 0 - (-1)(6) + 2(11)

⇒ |A₁₄| = 6 + 22 = 28

Step 3: Substitute the determinants of the 3x3 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 4x4 matrix is 48.

Also, Check

• Determinant of 2x2 Matrix
• Determinant of 3×3 Matrix

Determinant of 4×4 Matrix Examples

Example 1: A = \begin{bmatrix} 2 & 1 & 0 & 3 \\ 4 & -1 & 2 & 0 \\ -3 & 2 & 1 & 5 \\ 1 & 0 & -2 & 3 \end{bmatrix}

Solution:

First Expand along the first row:

\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})

Now, compute the determinant of each 3x3 submatrix.

For (A₁₁):

A_{11} = \begin{bmatrix} -1 & 2 & 0 \\ 2 & 1 & 5 \\ 0 & -2 & 3 \end{bmatrix}

\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)

= (-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):

A_{12} = \begin{bmatrix} 2 & 0 & 3 \\ -3 & 1 & 5 \\ 1 & 2 & 3 \end{bmatrix}

\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)

= (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):

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

\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)

= (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):

A_{14} = \begin{bmatrix} 2 & 1 & 0 \\ -3 & 2 & 1 \\ 1 & 0 & -2 \end{bmatrix}

\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)

= (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 3x3 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 A = \begin{bmatrix} 2 & 1 & -3 & 4 \\ -1 & 0 & 2 & 5 \\ 3 & 2 & 1 & 0 \\ 4 & -2 & 3 & 1 \end{bmatrix}

Solution:

To find the determinant of the matrix ( A ), we'll use the expansion by minors method along the first row:

\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}

Now, let's compute the determinants of the 3x3 submatrices:

\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)))

= 2 · (0 - 4 + 30) = 52

\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))

= -1 · (1 - 6 + 45) = 60

\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))

= -1 · (6 - 0 - 50) = 44

\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))

= -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 = \begin{bmatrix} -2 & 3 & 1 & 0 \\ 4 & 1 & -3 & 2 \\ 0 & -1 & 2 & 5 \\ 3 & 2 & 0 & -4 \end{bmatrix}

Solution:

To find the determinant of matrix ( B ), we'll use the expansion by minors method along the first row:

\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}

Now, let's compute the determinants of the 3x3 submatrices:

\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))

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

= -2 ⋅ (-8 + 18 - 8) = -2 ⋅ 2 = -4

\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))

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

= 3 ⋅ (-32 + 45 - 12) = 3 ⋅ 1 = 3

\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)

= -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

Determinant of 4×4 Matrix Practice Questions

Q1: Calculate the determinant of the following 4×4 matrix: A = \begin{bmatrix} 2 & 0 & 1 & 3 \\ -1 & 2 & 2 & 0 \\ 3 & -2 & 0 & 1 \\ 1 & 1 & 2 & -1 \\ \end{bmatrix}

Q2: Find the determinant of the matrix: B = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 2 & 3 & 4 & 5 \\ \end{bmatrix}

Q3: Calculate the determinant of the following 4×4 matrix: C = \begin{bmatrix} 2 & 1 & 0 & -1 \\ 3 & 2 & -1 & 0 \\ 0 & -3 & 2 & 1 \\ 1 & 0 & 3 & -2 \\ \end{bmatrix}

Q4: Determine the determinant of the matrix: D = \begin{bmatrix} 4 & 2 & 1 & 0 \\ -1 & 3 & 0 & 2 \\ 0 & 2 & 1 & -3 \\ 2 & 0 & -1 & 4 \\ \end{bmatrix}

Q5: Find the determinant of the matrix: E = \begin{bmatrix} 3 & 1 & -2 & 0 \\ 2 & 0 & 1 & 1 \\ -1 & 2 & 3 & -2 \\ 0 & 3 & -1 & 1 \\ \end{bmatrix}

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Article Tags :

• Mathematics
• School Learning
• Class 12
• Matrix and Determinant
• Algebra
• Maths-Class-12

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