Determinant of 4x4 Matrix (original) (raw)

Last Updated : 17 Nov, 2025

The determinant of a matrix is a single numerical value that can be calculated from a square matrix (a matrix with the same number of rows and columns).

It tells us important information about the matrix, such as whether it is invertible (has an inverse) and how it scales space in geometry.
A 4×4 matrix is an array of numbers with four rows and four columns (a square matrix).

Determinant-of-4-into-4-Matrix_

Properties of 4×4 Matrix

Here are some properties of a 4×4 matrix:

**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., \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) = a 11 · det(A 11 ) - a 12 · det(A 12 ) + a 13 · det(A 13 ) - a 14 · det(A 14 )

Where Aij is the 3×3 submatrix formed by deleting row 1 and column j. (The signs follow (−1)1+j.)

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(A 11 ) - 1 · det(A 12 ) + 3 · det(A 13 ) - 4 · det(A 14 )

Where Aij 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) + 1 \cdot \text{det}\left(\begin{bmatrix} 3 & 2 \\ -1 & 3 \end{bmatrix}\right)

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

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

⇒ |A13| = 0 - (-1)(8) + 1(11)

⇒ |A13| = 8 + 11 = 19

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 · 19 - 4 · 28

**Step 4: Compute the final determinant:

det(A) = 80 + 10 + 57 - 112

det(A) = 35

So, the determinant of the given 4x4 matrix is 35.

**Also Check:

Determinant of 2x2 Matrix

Determinant of 3×3 Matrix

Special Cases for Zero Determinant

Here are the special cases (or situations) where the determinant of a matrix becomes zero (0):

1. **One Row or Column Is All Zeros

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

Here, the second row consist entirely of zeros.

When the row( or column ) in determinant is made of all zeros,the value of the determinant is always zero.

|A| = 0

2. **Two Rows or Columns Are Identical

**Example : \begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 3 \\ 5 & 6 & 5 \end{bmatrix}

Notice that Column 1 and Column 3 are exactly the same.

If any two columns ( or rows) of determinants are identical , its determinants is zero.

|B| = 0

3. Two Rows or Columns Are Proportional

**Example: \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 7 & 8 & 9 \end{bmatrix} \Longrightarrow Row \ 2 = 2 \times Row \ 1 \Longrightarrow det = 0

Here, the second row is excatly twice the first row.

That means the two rows are proportional, and whenever any two rows( or column) are proportional, the determinants is zero.

Determinant of 4×4 Matrix Solved Examples

**Example 1: Calculate the determinant of the matrix 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 (A 11 ):

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 (A 12 ):

A_{12} = \begin{bmatrix} 4 & 2 & 0 \\ -3 & 1 & 5 \\ 1 & 2 & 3 \end{bmatrix}

\text{det}(A_{12}) = (4) \cdot \text{det}\left(\begin{bmatrix} 1 & 5 \\ 2 & 3 \end{bmatrix}\right) - (2) \cdot \text{det}\left(\begin{bmatrix} -3 & 5 \\ 1 & 3 \end{bmatrix}\right) + (0) \cdot \text{det}\left(\begin{bmatrix} -3 & 1 \\ 1 & 2 \end{bmatrix}\right)

= (4)((1)(3)-(5)(2)) - (2)((-3)(3)-(5)(1)) + (0)((-3)(2)-(1)(1))

= (4)((3)-(10)) - (2)((-9)-(5)) + 0

= 4 (-7) - 2(- 14 )

= =28 + 28 = 0

**For (A 13 ):

Since a13 = 0, the contribution of C13 will be zero. (Computation unnecessary.)

**For (A 14 ):

A_{14} = \begin{bmatrix} 4 & -1 & 2 \\ -3 & 2 & 1 \\ 1 & 0 & -2 \end{bmatrix}

\text{det}(A_{14}) = (4) \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) + (2) \cdot \text{det}\left(\begin{bmatrix} -3 & 2 \\ 1 & 0 \end{bmatrix}\right)

=4[(2)(−2)−(1)(0)]−(−1)[(−3)(−2)−(1)(1)]+2[(−3)(0)−(2)(1)]

=4 [− 4 − 0] + 1[6 − 1] + 2[0 − 2]

= 4(-4) + 5 + (- 4)

=− 16 + 5 − 4

= - 15

Now, substitute the determinants of the 3x3 submatrices into the expansion formula:

det(A) = 2 ( - 25 ) - 1( 80 ) - 3 ( - 15)

= -50 - 80 + 45 + 0

= - 130 + 45

= - 85

So, the determinant of matrix (A) is - 85 .

**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 & 1\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) = (0 \cdot (1 \cdot 1 - 0 \cdot 3) - 2 \cdot (2 \cdot 1 - 0 \cdot (-2)) + 5 \cdot (2 \cdot 3 - 2 \cdot (-2)))

= 0 - 2 ( 2 - 0) + 5 ( 6 + 2)

= -4 + 5(8) = - 4 + 40 = 36

\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 - 0) - 2 ( 3- 0) + 5(9 - 4)

= - 1 - 6 + 5(5)

= - 7 + 25 = 18

\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( 2- 0) - 0 + 5 (-6 - 8 )

= - 72

\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 · (6 + 2) - 0 + 2( - 6 - 8)

= (- 1)(8) + 2 (-14) = -8 - 28 = - 36

Now, substitute these determinants back into the expansion formula:

det(A) = 72 - 18 + 216 + 144 = 54 + 360 = 414

So, the determinant of matrix ( A ) is det(A) = 414.

**Example 3: Let C = \begin{bmatrix}3 & 0 & -2 & 1 \\0 & 4 & 1 & -3 \\2 & -1 & 3 & 0 \\1 & 2 & 0 & 5\end{bmatrix} find its determinants using the cofactor expansion method.

**Solution:

C_{11} =\begin{vmatrix}4 & 1 & -3\\-1 & 3 & 0\\2 & 0 & 5\end{vmatrix}

Expand along the first row [3, 0, 2, 1]. The expansion is

det⁡(C) = 3 C11 + 0 C12 + 2 C13 + 1C14

where C1j = (−1)1+j det⁡(M1j) (cofactor).

=4(15) - 1(-5) - 3(-6)
= 60 + 5 + 18
= 83.

C_{13} =\begin{vmatrix}0 & 4 & -3\\2 & -1 & 0\\1 & 2 & 5\end{vmatrix}
= 0 - 4(10) - 3(5)
= -40 - 15
= -55

C_{14} =\begin{vmatrix}0 & 4 & 1\\2 & -1 & 3\\1 & 2 & 0\end{vmatrix}

=0 - 4(-3) + 1(5)
=12 + 5
= 17.

Now substitute:

|C| = 3(83) - 2(-55) + (-17)

|C| = 249 + 110 - 17

|C| = 342

Determinant of 4×4 Matrix Practice Questions

**Question 1: 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}

**Question 2: 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}

**Question 3: 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}

**Question 4: 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}

**Question 5: 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}

**Question 6: Find the determinant of the following lower triangular matrix: B = \begin{bmatrix}5 & 0 & 0 & 0 \\-2 & 3 & 0 & 0 \\4 & 1 & 7 & 0 \\6 & -3 & 2 & 9\end{bmatrix}