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Matrix

Matrix Determinant

Understanding Matrix Determinants

The determinant of a matrix is a scalar value (a single number) that can be calculated from the elements of a square matrix. The concept of a determinant is very important in linear algebra, one of its uses being to help solve systems of linear equations. Every square matrix has a unique determinant value.

Calculating the Determinant of a 2x2 Order Matrix

A 2x2 order matrix is a matrix that has two rows and two columns. Suppose we have matrix A as follows:

A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}A=[ac​bd​]

The determinant of matrix A, usually written as det⁡(A)\det(A)det(A) or ∣A∣|A|∣A∣, is calculated by subtracting the product of the elements of the main diagonal from the product of the elements of the second diagonal.

The formula is:

det⁡(A)=∣A∣=∣abcd∣=ad−bc\det(A) = |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bcdet(A)=∣A∣=​ac​bd​​=ad−bc

Note that the notation ∣abcd∣\begin{vmatrix} a & b \\ c & d \end{vmatrix}​ac​bd​​ uses straight lines, which denote the determinant, as opposed to square brackets [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}[ac​bd​] which denote the matrix itself.

Calculation of a 2x2 Matrix Determinant

Suppose we have matrix B:

B=[−13−7−5]B = \begin{bmatrix} -1 & 3 \\ -7 & -5 \end{bmatrix}B=[−1−7​3−5​]

To calculate its determinant, we identify a=−1a = -1a=−1, b=3b = 3b=3, c=−7c = -7c=−7, and d=−5d = -5d=−5.

Then, the determinant of matrix B is:

det⁡(B)=(−1×−5)−(3×−7)\det(B) = (-1 \times -5) - (3 \times -7)det(B)=(−1×−5)−(3×−7)
det⁡(B)=5−(−21)\det(B) = 5 - (-21)det(B)=5−(−21)
det⁡(B)=5+21\det(B) = 5 + 21det(B)=5+21
det⁡(B)=26\det(B) = 26det(B)=26

So, the determinant value of matrix B is 26.

Solving Systems of Linear Equations with Two Variables (SPLDV) using Determinants

One important application of determinants is to solve systems of linear equations. This method is often called Cramer's Rule.

Consider the following system of linear equations with two variables (SPLDV):

a11x+a12y=b1a_{11}x + a_{12}y = b_1a11​x+a12​y=b1​
a21x+a22y=b2a_{21}x + a_{22}y = b_2a21​x+a22​y=b2​

In this system, xxx and yyy are the variables whose values we want to find. The coefficients a11,a12,a21,a22a_{11}, a_{12}, a_{21}, a_{22}a11​,a12​,a21​,a22​ and constants b1,b2b_1, b_2b1​,b2​ are known numbers.

This system of equations can be converted into matrix multiplication form:

[a11a12a21a22][xy]=[b1b2]\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}[a11​a21​​a12​a22​​][xy​]=[b1​b2​​]

The first step is to calculate the determinant of the coefficient matrix, which we call DDD:

D=∣a11a12a21a22∣=a11a22−a12a21D = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}D=​a11​a21​​a12​a22​​​=a11​a22​−a12​a21​

A system of linear equations will have a unique solution if and only if D≠0D \neq 0D=0.

Next, we calculate two other determinants:

  1. DxD_xDx​, which is the determinant of the coefficient matrix where the first column (coefficients of xxx) is replaced by the constant column (b1,b2b_1, b_2b1​,b2​ ):

    Dx=∣b1a12b2a22∣=b1a22−a12b2D_x = \begin{vmatrix} b_1 & a_{12} \\ b_2 & a_{22} \end{vmatrix} = b_1a_{22} - a_{12}b_2Dx​=​b1​b2​​a12​a22​​​=b1​a22​−a12​b2​

  2. DyD_yDy​, which is the determinant of the coefficient matrix where the second column (coefficients of yyy) is replaced by the constant column (b1,b2b_1, b_2b1​,b2​ ):

    Dy=∣a11b1a21b2∣=a11b2−b1a21D_y = \begin{vmatrix} a_{11} & b_1 \\ a_{21} & b_2 \end{vmatrix} = a_{11}b_2 - b_1a_{21}Dy​=​a11​a21​​b1​b2​​​=a11​b2​−b1​a21​

After obtaining the values of DDD, DxD_xDx​, and DyD_yDy​, we can find the values of xxx and yyy using the formulas:

x=DxDx = \frac{D_x}{D}x=DDx​​
y=DyDy = \frac{D_y}{D}y=DDy​​

These formulas are only valid if D≠0D \neq 0D=0.

Solving SPLDV with Determinants

Determine the solution of the following system of linear equations:

2x−y=72x - y = 72x−y=7
x−4y=14x - 4y = 14x−4y=14

From the system above, we get:

a11=2a_{11} = 2a11​=2, a12=−1a_{12} = -1a12​=−1, b1=7b_1 = 7b1​=7

a21=1a_{21} = 1a21​=1, a22=−4a_{22} = -4a22​=−4, b2=14b_2 = 14b2​=14

Step 1: Calculate the determinant DDD.

D=∣2−11−4∣=(2×−4)−(−1×1)=−8−(−1)=−8+1=−7D = \begin{vmatrix} 2 & -1 \\ 1 & -4 \end{vmatrix} = (2 \times -4) - (-1 \times 1) = -8 - (-1) = -8 + 1 = -7D=​21​−1−4​​=(2×−4)−(−1×1)=−8−(−1)=−8+1=−7

Since D=−7≠0D = -7 \neq 0D=−7=0, this system has a unique solution.

Step 2: Calculate the determinant DxD_xDx​.

Dx=∣7−114−4∣=(7×−4)−(−1×14)=−28−(−14)=−28+14=−14D_x = \begin{vmatrix} 7 & -1 \\ 14 & -4 \end{vmatrix} = (7 \times -4) - (-1 \times 14) = -28 - (-14) = -28 + 14 = -14Dx​=​714​−1−4​​=(7×−4)−(−1×14)=−28−(−14)=−28+14=−14

Step 3: Calculate the determinant DyD_yDy​.

Dy=∣27114∣=(2×14)−(7×1)=28−7=21D_y = \begin{vmatrix} 2 & 7 \\ 1 & 14 \end{vmatrix} = (2 \times 14) - (7 \times 1) = 28 - 7 = 21Dy​=​21​714​​=(2×14)−(7×1)=28−7=21

Step 4: Calculate the values of xxx and yyy.

x=DxD=−14−7=2x = \frac{D_x}{D} = \frac{-14}{-7} = 2x=DDx​​=−7−14​=2
y=DyD=21−7=−3y = \frac{D_y}{D} = \frac{21}{-7} = -3y=DDy​​=−721​=−3

So, the solution set of the system of linear equations is x=2x=2x=2 and y=−3y=-3y=−3, or can be written as the ordered pair (2,−3)(2, -3)(2,−3).

Exercises

  1. Given matrix M=[9x8−7]M = \begin{bmatrix} 9 & x \\ 8 & -7 \end{bmatrix}M=[98​x−7​] and det⁡(M)=9\det(M) = 9det(M)=9. Determine the value of xxx.

  2. Determine the solution of the following system of linear equations:

    {2x−y=8x+3y=−10\begin{cases} 2x - y = 8 \\ x + 3y = -10 \end{cases}{2x−y=8x+3y=−10​

Answer Key

  1. For matrix M=[9x8−7]M = \begin{bmatrix} 9 & x \\ 8 & -7 \end{bmatrix}M=[98​x−7​], its determinant is:

    det⁡(M)=(9×−7)−(x×8)=−63−8x\det(M) = (9 \times -7) - (x \times 8) = -63 - 8xdet(M)=(9×−7)−(x×8)=−63−8x

    Given det⁡(M)=9\det(M) = 9det(M)=9, then:

    −63−8x=9-63 - 8x = 9−63−8x=9
    −8x=9+63-8x = 9 + 63−8x=9+63
    −8x=72-8x = 72−8x=72
    x=72−8x = \frac{72}{-8}x=−872​
    x=−9x = -9x=−9

    So, the value of xxx is -9.

  2. System of linear equations:

    2x−y=82x - y = 82x−y=8
    x+3y=−10x + 3y = -10x+3y=−10

    We determine D,Dx,D, D_x,D,Dx​, and DyD_yDy​.

    D=∣2−113∣=(2×3)−(−1×1)=6−(−1)=6+1=7D = \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} = (2 \times 3) - (-1 \times 1) = 6 - (-1) = 6 + 1 = 7D=​21​−13​​=(2×3)−(−1×1)=6−(−1)=6+1=7
    Dx=∣8−1−103∣=(8×3)−(−1×−10)=24−10=14D_x = \begin{vmatrix} 8 & -1 \\ -10 & 3 \end{vmatrix} = (8 \times 3) - (-1 \times -10) = 24 - 10 = 14Dx​=​8−10​−13​​=(8×3)−(−1×−10)=24−10=14
    Dy=∣281−10∣=(2×−10)−(8×1)=−20−8=−28D_y = \begin{vmatrix} 2 & 8 \\ 1 & -10 \end{vmatrix} = (2 \times -10) - (8 \times 1) = -20 - 8 = -28Dy​=​21​8−10​​=(2×−10)−(8×1)=−20−8=−28

    Then, the values of xxx and yyy are:

    x=DxD=147=2x = \frac{D_x}{D} = \frac{14}{7} = 2x=DDx​​=714​=2
    y=DyD=−287=−4y = \frac{D_y}{D} = \frac{-28}{7} = -4y=DDy​​=7−28​=−4

    So, the solution of the system of linear equations is x=2x=2x=2 and y=−4y=-4y=−4, or the ordered pair (2,−4)(2, -4)(2,−4).

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  • Matrix DeterminantCalculate 2x2 matrix determinants and solve linear equations using Cramer's Rule. Master determinant formulas with step-by-step examples and practice.
On this page
  • Understanding Matrix Determinants
  • Calculating the Determinant of a 2x2 Order Matrix
    • Calculation of a 2x2 Matrix Determinant
  • Solving Systems of Linear Equations with Two Variables (SPLDV) using Determinants
    • Solving SPLDV with Determinants
  • Exercises
    • Answer Key
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