# Nakafa Framework: LLM

URL: /en/subject/high-school/11/mathematics/matrix/matrix-determinant
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---

export const metadata = {
  title: "Matrix Determinant",
  description: "Calculate 2x2 matrix determinants and solve linear equations using Cramer's Rule. Master determinant formulas with step-by-step examples and practice.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "06/05/2025",
  subject: "Matrix",
};

## 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:

<BlockMath math="A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}" />

The determinant of matrix A, usually written as <InlineMath math="\det(A)" /> or <InlineMath math="|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:

<BlockMath math="\det(A) = |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc" />

Note that the notation <InlineMath math="\begin{vmatrix} a & b \\ c & d \end{vmatrix}" /> uses straight lines, which denote the determinant, as opposed to square brackets <InlineMath math="\begin{bmatrix} a & b \\ c & d \end{bmatrix}" /> which denote the matrix itself.

### Calculation of a 2x2 Matrix Determinant

Suppose we have matrix B:

<BlockMath math="B = \begin{bmatrix} -1 & 3 \\ -7 & -5 \end{bmatrix}" />

To calculate its determinant, we identify <InlineMath math="a = -1" />, <InlineMath math="b = 3" />, <InlineMath math="c = -7" />, and <InlineMath math="d = -5" />.

Then, the determinant of matrix B is:

<div className="flex flex-col gap-4">
  <BlockMath math="\det(B) = (-1 \times -5) - (3 \times -7)" />
  <BlockMath math="\det(B) = 5 - (-21)" />
  <BlockMath math="\det(B) = 5 + 21" />
  <BlockMath math="\det(B) = 26" />
</div>

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):

<div className="flex flex-col gap-4">
  <BlockMath math="a_{11}x + a_{12}y = b_1" />
  <BlockMath math="a_{21}x + a_{22}y = b_2" />
</div>

In this system, <InlineMath math="x" /> and <InlineMath math="y" /> are the variables whose values we want to find. The coefficients <InlineMath math="a_{11}, a_{12}, a_{21}, a_{22}" /> and constants <InlineMath math="b_1, b_2" /> are known numbers.

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

<BlockMath math="\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}" />

The first step is to calculate the determinant of the coefficient matrix, which we call <InlineMath math="D" />:

<BlockMath math="D = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}" />

A system of linear equations will have a unique solution if and only if <InlineMath math="D \neq 0" />.

Next, we calculate two other determinants:

1.  <InlineMath math="D_x" />, which is the determinant of the coefficient
    matrix where the first column (coefficients of <InlineMath math="x" />) is replaced
    by the constant column (<InlineMath math="b_1, b_2" />
    ):

    <BlockMath math="D_x = \begin{vmatrix} b_1 & a_{12} \\ b_2 & a_{22} \end{vmatrix} = b_1a_{22} - a_{12}b_2" />

2.  <InlineMath math="D_y" />, which is the determinant of the coefficient
    matrix where the second column (coefficients of <InlineMath math="y" />) is replaced
    by the constant column (<InlineMath math="b_1, b_2" />
    ):

    <BlockMath math="D_y = \begin{vmatrix} a_{11} & b_1 \\ a_{21} & b_2 \end{vmatrix} = a_{11}b_2 - b_1a_{21}" />

After obtaining the values of <InlineMath math="D" />, <InlineMath math="D_x" />, and <InlineMath math="D_y" />, we can find the values of <InlineMath math="x" /> and <InlineMath math="y" /> using the formulas:

<div className="flex flex-col gap-4">
  <BlockMath math="x = \frac{D_x}{D}" />
  <BlockMath math="y = \frac{D_y}{D}" />
</div>

These formulas are only valid if <InlineMath math="D \neq 0" />.

### Solving SPLDV with Determinants

Determine the solution of the following system of linear equations:

<div className="flex flex-col gap-4">
  <BlockMath math="2x - y = 7" />
  <BlockMath math="x - 4y = 14" />
</div>

From the system above, we get:

<InlineMath math="a_{11} = 2" />, <InlineMath math="a_{12} = -1" />, <InlineMath math="b_1 = 7" />

<InlineMath math="a_{21} = 1" />, <InlineMath math="a_{22} = -4" />, <InlineMath math="b_2 = 14" />

**Step 1**: Calculate the determinant <InlineMath math="D" />.

<BlockMath math="D = \begin{vmatrix} 2 & -1 \\ 1 & -4 \end{vmatrix} = (2 \times -4) - (-1 \times 1) = -8 - (-1) = -8 + 1 = -7" />
Since <InlineMath math="D = -7 \neq 0" />, this system has a unique solution.

**Step 2**: Calculate the determinant <InlineMath math="D_x" />.

<BlockMath math="D_x = \begin{vmatrix} 7 & -1 \\ 14 & -4 \end{vmatrix} = (7 \times -4) - (-1 \times 14) = -28 - (-14) = -28 + 14 = -14" />

**Step 3**: Calculate the determinant <InlineMath math="D_y" />.

<BlockMath math="D_y = \begin{vmatrix} 2 & 7 \\ 1 & 14 \end{vmatrix} = (2 \times 14) - (7 \times 1) = 28 - 7 = 21" />

**Step 4**: Calculate the values of <InlineMath math="x" /> and <InlineMath math="y" />.

<div className="flex flex-col gap-4">
  <BlockMath math="x = \frac{D_x}{D} = \frac{-14}{-7} = 2" />
  <BlockMath math="y = \frac{D_y}{D} = \frac{21}{-7} = -3" />
</div>

So, the solution set of the system of linear equations is <InlineMath math="x=2" /> and <InlineMath math="y=-3" />, or can be written as the ordered pair <InlineMath math="(2, -3)" />.

## Exercises

1.  Given matrix <InlineMath math="M = \begin{bmatrix} 9 & x \\ 8 & -7 \end{bmatrix}" /> and <InlineMath math="\det(M) = 9" />. Determine the value of <InlineMath math="x" />.
2.  Determine the solution of the following system of linear equations:

    <BlockMath math="\begin{cases} 2x - y = 8 \\ x + 3y = -10 \end{cases}" />

### Answer Key

1.  For matrix <InlineMath math="M = \begin{bmatrix} 9 & x \\ 8 & -7 \end{bmatrix}" />, its determinant is:

    <BlockMath math="\det(M) = (9 \times -7) - (x \times 8) = -63 - 8x" />

    Given <InlineMath math="\det(M) = 9" />, then:

    <div className="flex flex-col gap-4">
      <BlockMath math="-63 - 8x = 9" />
      <BlockMath math="-8x = 9 + 63" />
      <BlockMath math="-8x = 72" />
      <BlockMath math="x = \frac{72}{-8}" />
      <BlockMath math="x = -9" />
    </div>

    So, the value of <InlineMath math="x" /> is -9.

2.  System of linear equations:

    <div className="flex flex-col gap-4">
      <BlockMath math="2x - y = 8" />
      <BlockMath math="x + 3y = -10" />
    </div>

    We determine <InlineMath math="D, D_x," /> and <InlineMath math="D_y" />.

    <div className="flex flex-col gap-4">
      <BlockMath math="D = \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} = (2 \times 3) - (-1 \times 1) = 6 - (-1) = 6 + 1 = 7" />
      <BlockMath math="D_x = \begin{vmatrix} 8 & -1 \\ -10 & 3 \end{vmatrix} = (8 \times 3) - (-1 \times -10) = 24 - 10 = 14" />
      <BlockMath math="D_y = \begin{vmatrix} 2 & 8 \\ 1 & -10 \end{vmatrix} = (2 \times -10) - (8 \times 1) = -20 - 8 = -28" />
    </div>

    Then, the values of <InlineMath math="x" /> and <InlineMath math="y" /> are:

    <div className="flex flex-col gap-4">
      <BlockMath math="x = \frac{D_x}{D} = \frac{14}{7} = 2" />
      <BlockMath math="y = \frac{D_y}{D} = \frac{-28}{7} = -4" />
    </div>

    So, the solution of the system of linear equations is <InlineMath math="x=2" /> and <InlineMath math="y=-4" />, or the ordered pair <InlineMath math="(2, -4)" />.