Determinant of 4 x 4 Matrix

The determinant of a 4 × 4 matrix is a scalar value that provides important information about the matrix, such as whether it's invertible. It can be computed using cofactor expansion, where the determinant is broken down into smaller 3 × 3 matrices. The determinant is useful in various applications, including solving systems of linear equations, determining matrix invertibility, and calculating eigenvalues. For a 4 × 4 matrix, cofactor expansion involves calculating the minors and applying alternating signs to sum the results.

1.0What is Determinant?

A determinant is a unique value that can be calculated from a square matrix (a matrix with the same number of rows and columns). It gives important information about the matrix, such as whether the matrix has an inverse. If the determinant is zero, the matrix is singular (non-invertible). If the determinant is non-zero, the matrix is invertible.

For example, for a 2 × 2 matrix:

$det\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=ad-bc$

The determinant helps in solving systems of linear equations, finding eigenvalues, and other applications in mathematics and engineering.

2.0Determinant of a 4 × 4 Matrix 

The determinant of a 4 × 4 matrix is a scalar value that can be calculated from the elements of a 4 × 4 square matrix. It provides important information about the matrix, such as whether it is invertible. If the determinant is zero, the matrix does not have an inverse; if it's non-zero, the matrix has an inverse. 

To calculate the determinant of a 4 × 4 matrix, you can use the cofactor expansion method, which breaks the matrix down into smaller matrices. This process involves expanding the determinant along a row or column and calculating the determinants of 3 × 3 matrices that are formed by removing one row and one column from the 4 × 4 matrix. The same process continues recursively until you reach 2 × 2 matrices, which can be directly calculated.

3.0Formula for Determinant of a 4 × 4 Matrix

Let’s consider a 4 × 4 matrix A:

$A=\left[\begin{array}{cccc}a& b& c& d\\ e& f& g& h\\ i& j& k& l\\ m& n& o& p\end{array}\right]$

The determinant of matrix A, denoted , can be calculated by expanding along the first row (or any row/column):

$det(A)=a.det(M_{11})-b.det(M_{12})+c.det(M_{13})-d.det(M_{14})$

Where:

• $M_{11},M_{12},M_{13},M_{14}$ ​are the 3 × 3 matrices obtained by removing the corresponding row and column from the original matrix.

4.0Step-by-Step Method to Find the Determinant of a 4 × 4 Matrix

The determinant of a 4 × 4 matrix is a bit tricky compared to smaller matrices (like 2 × 2 or 3 × 3), but you can solve it by using cofactor expansion. Here’s how it works:

Let’s say we have the following 4 × 4 matrix A: 

$A=\left[\begin{array}{cccc}a& b& c& d\\ e& f& g& h\\ i& j& k& l\\ m& n& o& p\end{array}\right]$

Step 1: Choose a Row or Column

To make calculations easier, we typically choose the first row (but you can choose any row or column). The determinant of the matrix is calculated by expanding along the first row like this:

$det(A)=a.det(M_{11})-b.det(M_{12})+c.det(M_{13})-d.det(M_{14})$

Where:

• $M_{11},M_{12},M_{13},M_{14}$ are 3 × 3 matrices formed by removing the row and column of each element in the first row.
• The signs alternate: +, −, +, −.

Step 2: Find the Determinants of the Smaller Matrices

Now, you need to find the determinants of the 3 × 3 matrices formed in the previous step. To do that, you can use the same method of cofactor expansion, but this time, you’ll be working with 3 × 3 matrices instead of 4 × 4 matrices.

Step 3: Continue Until You Reach 2 × 2 Matrices

When you get to 2 × 2 matrices, you can directly calculate their determinants using the simple formula:

$det\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=ad-bc$

5.0Solved Example For Finding The Determinant of a 4 × 4 Matrix

Let’s look at a simple example:

$A=\left[\begin{array}{cccc}1& 0& 4& -6\\ 2& 5& 0& 3\\ -1& 2& 3& 5\\ 2& 1& -2& 3\end{array}\right]$   

Solution

To find the determinant of A, we expand along the first row:

$det(A)=1.det(M_{11})-0.det(M_{12})+4.det(M_{13})-(-6).det(M_{14})$

Now, we calculate the 3 × 3 matrices:

• $M_{11}=\left[\begin{array}{ccc}5& 0& 3\\ 2& 3& 5\\ 1& -2& 3\end{array}\right]$

$=5\left[\begin{array}{cc}3& 5\\ -2& 3\end{array}\right]-0\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]+3\left[\begin{array}{cc}2& 3\\ 1& -2\end{array}\right]$

$=5(9+10)–0+3(–4–3)$

$=5(19)+3(–7)$

$=95–21$

$=74$

• $M_{12}=\left[\begin{array}{ccc}2& 0& 3\\ -1& 3& 5\\ 2& -2& 3\end{array}\right]$

$=2\left[\begin{array}{cc}3& 5\\ -2& 3\end{array}\right]-0\left[\begin{array}{cc}-1& 5\\ 2& 3\end{array}\right]+3\left[\begin{array}{cc}-1& 3\\ 2& -2\end{array}\right]$

$=2(9+10)–0+3(2–6)$

$=2(19)–0+3(–4)$

$=38–12$

$=26$

• $M_{13}=\left[\begin{array}{ccc}2& 5& 3\\ -1& 2& 5\\ 2& 1& 3\end{array}\right]$

$=2\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]-5\left[\begin{array}{cc}-1& 5\\ 2& 3\end{array}\right]+3\left[\begin{array}{cc}-1& 2\\ 2& -1\end{array}\right]$

$=2(6–5)–5(–3–10)+3(–1–4)$

$=2(1)-5(–13)+3(–5)$

$=2+65–15$

$=52$

• $M_{14}=\left[\begin{array}{ccc}2& 5& 0\\ -1& 2& 3\\ 2& 1& -2\end{array}\right]$

$=2\left[\begin{array}{cc}2& 3\\ 1& -2\end{array}\right]-5\left[\begin{array}{cc}-1& 3\\ 2& -2\end{array}\right]+0\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right]$

$=2(–4–3)–5(2–6)+0$

$=2(–7)–5(–4)$

$=–14+20$

$=6$

$det(A)=1.det(M_{11})-0.det(M_{12})+4.det(M_{13})-(-6).det(M_{14})$

$det(A)=(1\times 74)–(0\times 26)+(4\times 52)+(6\times 6)$

$det(A)=74–0+208+36$

$det(A)=318$

Example 2: Find the determinant of the 4 × 4 matrix

$A=\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 5& 6& 7\\ 2& 4& 6& 7\\ 3& 6& 7& 8\end{array}\right]$

To find the determination of A, we expand of along the first row

Det (A) = 1·det (M₁₁) – 2. det (M₁₂) + 3·det(M₁₃) –4·det(M₁₄)

$M_{11}=\left[\begin{array}{ccc}5& 6& 7\\ 4& 6& 7\\ 6& 7& 8\end{array}\right]$

$=5\left[\begin{array}{cc}6& 7\\ 7& 8\end{array}\right]-6\left[\begin{array}{cc}4& 7\\ 6& 8\end{array}\right]+7\left[\begin{array}{cc}4& 6\\ 6& 7\end{array}\right]$

= 5 [48 – 49] – 6 [32 – 42] + 7 [28 – 36] 

= 5 (–1) –6 (–10) + 7 (–8)

= – 5 + 60 – 56 = – 1.

$M_{12}=\left[\begin{array}{ccc}0& 6& 7\\ 2& 6& 7\\ 3& 7& 8\end{array}\right]$

$=0\left[\begin{array}{cc}6& 7\\ 7& 8\end{array}\right]-6\left[\begin{array}{cc}2& 7\\ 3& 8\end{array}\right]+7\left[\begin{array}{cc}2& 6\\ 3& 7\end{array}\right]$

= 0 – 6 [16 – 21] + 7 [14 – 18]

= –6 [– 5] + 7 [– 4]

= + 20 – 28 = – 8.

$M_{13}=\left[\begin{array}{ccc}0& 5& 7\\ 2& 4& 7\\ 3& 6& 8\end{array}\right]$

$=0\left[\begin{array}{cc}4& 7\\ 6& 8\end{array}\right]-5\left[\begin{array}{cc}2& 7\\ 3& 8\end{array}\right]+7\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]$

= 0 – 5(16 – 21) + 7 (12 – 120)

= – 5 (–5) + 0 = +25

$M_{14}=\left[\begin{array}{ccc}0& 5& 6\\ 2& 4& 6\\ 3& 6& 7\end{array}\right]$

$=0\left[\begin{array}{cc}4& 6\\ 6& 7\end{array}\right]-5\left[\begin{array}{cc}2& 6\\ 3& 7\end{array}\right]+6\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]$

= 0 – 5 [14 – 18] + 6 [12 – 12] 

= – 5 [–4] = 20.

$|A|=1\cdot \det (M_{11})-2\cdot \det (M_{12})+3\cdot \det (M_{13})-4\cdot \det (M_{14})$

$|A|=1\cdot (-1)-2\cdot (-8)+3\cdot (25)-4\cdot (20)$

= -1 + 16 + 75 - 80
= 10.

Example 3: Consider the block matrix $C=\left[\begin{array}{cc}A& B\\ 0& D\end{array}\right]$

Where A is a 2 × 2 matrix, B is a 2 × 2 matrix, D is a 2 × 2 matrix, and O is a 2 × 2 zero matrix, if:

$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right],D=\left[\begin{array}{cc}9& 10\\ 11& 12\end{array}\right]$

Find the determinant of the block matrix C.

Solution: 

For a block matrix.$\left[\begin{array}{cc}A& B\\ 0& D\end{array}\right]$

The determinant can be calculated using the formula

Det (C) = det (A). det (D)

$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$

$|A|=1\cdot 4-3\cdot 2=4-6=-2$

$D=\left[\begin{array}{cc}9& 10\\ 11& 12\end{array}\right]$

$|D|=9\cdot 12-10\cdot 11=108-110=-2$

$\det (C)=\det (A)\cdot \det (D)$

$=(-2)\cdot (-2)=4$

Example 4: Calculate the determinant of 4 × 4 matrix

$A=\left[\begin{array}{cccc}2& 1& 0& -1\\ 3& 2& -1& 0\\ 0& -3& 2& 1\\ 1& 0& 3& -2\end{array}\right]$

Solution:

To find the determinant of A, we expand along the first row.

Det (A) = 2. Det (M₁₁) –1 det (M₁₂) + 0·det (M₁₃) – (–1) det (M₁₄)

$M_{11}=\left[\begin{array}{ccc}2& 1& 0\\ -3& 2& 1\\ 0& 3& -2\end{array}\right]$

$M_{11}=2\left[\begin{array}{cc}2& 1\\ 3& -2\end{array}\right]-(-1)\left[\begin{array}{cc}-3& 1\\ 0& -2\end{array}\right]+0\left[\begin{array}{cc}-3& 2\\ 0& 3\end{array}\right]$

= 2 (–4 –3) + 1 (+6 – 0) + 0

= 2 (–7) + 1 (6) = – 14 + 6 = – 8.

$M_{12}=\left[\begin{array}{ccc}3& -1& 0\\ 0& 2& 1\\ 1& 3& -2\end{array}\right]$

$M_{12}=3\left[\begin{array}{cc}2& 1\\ 3& -2\end{array}\right]-(-1)\left[\begin{array}{cc}0& 1\\ 1& -2\end{array}\right]+0\left[\begin{array}{cc}0& 2\\ 1& 3\end{array}\right]$

= 3 (4 – 3) + 1 (–1) + 0

= 3 (1) – 1 (–1) = 3 + 1 = 4.

$M_{13}=\left[\begin{array}{ccc}3& 2& 0\\ 0& -3& 1\\ 1& 0& -2\end{array}\right]$

$M_{13}=3\left[\begin{array}{cc}-3& 1\\ 0& -2\end{array}\right]-2\left[\begin{array}{cc}0& 1\\ 1& -2\end{array}\right]+0\left[\begin{array}{cc}0& -3\\ 1& 0\end{array}\right]$

= 3 (+ 6 – 0) – 2 (0 – 1) + 0

= 3 (6) –2 (–1) = 18 + 2 = 20

$M_{14}=\left[\begin{array}{ccc}3& 2& -1\\ 0& -3& 2\\ 1& 0& 3\end{array}\right]$

$M_{14}=3\left[\begin{array}{cc}-3& 2\\ 0& 3\end{array}\right]-2\left[\begin{array}{cc}0& 2\\ 1& 3\end{array}\right]+(-1)\left[\begin{array}{cc}0& -3\\ 1& 0\end{array}\right]$

= 3 (– 9 – 0) –2 (0 – 2) – 1 (0 +3)  

= – 27 + 4 – 3

= – 26.

Det (A) = 2 (–8) – 1 (4) + 0 (20) – (–1) (–26)

= – 16 – 4 + 0 – 26 

= – 46.