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Mathematics LibreTexts

4.7: Solve Systems of Equations Using Determinants

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    5143
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    Learning Objectives

    By the end of this section, you will be able to:

    • Evaluate the determinant of a 2×2 matrix
    • Evaluate the determinant of a 3×3 matrix
    • Use Cramer’s Rule to solve systems of equations
    • Solve applications using determinants

    Before you get started, take this readiness quiz.

    1. Simplify: \(5(−2)−(−4)(1)\).
      If you missed this problem, review [link].
    2. Simplify: \(−3(8−10)+(−2)(6−3)−4(−3−(−4))\).
      If you missed this problem, review [link].
    3. Simplify: \(\frac{−12}{−8}\).
      If you missed this problem, review [link].

    In this section we will learn of another method to solve systems of linear equations called Cramer’s rule. Before we can begin to use the rule, we need to learn some new definitions and notation.

    Evaluate the Determinant of a \(2×2\) Matrix

    If a matrix has the same number of rows and columns, we call it a square matrix. Each square matrix has a real number associated with it called its determinant. To find the determinant of the square matrix \(\left[ \begin{matrix} a &b \\ c&d \end{matrix} \right] \), we first write it as \(\left| \begin{matrix} a &b \\ c&d \end{matrix} \right| \). To get the real number value of the determinate we subtract the products of the diagonals, as shown.

    A 2 by 2 determinant is show, with its first row being a, b and second one being c, d. These values are written between two vertical lines instead of brackets as in the case of matrices. Two arrows are shown, one from a to d, the other from c to b. This determinant is equal to ad minus bc.

    DETERMINANT

    The determinant of any square matrix \(\left[ \begin{matrix} a &b \\ c&d \end{matrix} \right] \), where a, b, c, and d are real numbers, is

    \[\left| \begin{matrix} a &b \\ c&d \end{matrix} \right| =ad−bc \nonumber \]

    Example \(\PageIndex{1}\)

    Evaluate the determinate of ⓐ \(\left[ \begin{matrix} 4 &-2 \\ 3&-1 \end{matrix} \right] \) ⓑ \(\left[ \begin{matrix} -3 &-4 \\ -2&0 \end{matrix} \right] \).

    Answer

      .
    Write the determinant. .
    Subtract the products of the diagonals. .
    Simplify. .
    Simplify. .

      .
    Write the determinant. .
    Subtract the products of the diagonals. .
    Simplify. .
    Simplify. .
    Example \(\PageIndex{2}\)

    Evaluate the determinate of ⓐ \(\left[ \begin{matrix} 5&−3\\2&−4 \end{matrix} \right] \) ⓑ \(\left[ \begin{matrix} −4&−6\\0&7 \end{matrix} \right] \).

    Answer

    ⓐ \(−14\); ⓑ \(−28\)

    Example \(\PageIndex{3}\)

    Evaluate the determinate of ⓐ \(\left[ \begin{matrix} −1&3\\−2&4 \end{matrix} \right] \) ⓑ \(\left[ \begin{matrix} −7&−3\\−5&0 \end{matrix} \right] \).

    Answer

    ⓐ 2 ⓑ \(−15\)

    Evaluate the Determinant of a \(3×3\) Matrix

    To evaluate the determinant of a \(3×3\) matrix, we have to be able to evaluate the minor of an entry in the determinant. The minor of an entry is the \(2×2\) determinant found by eliminating the row and column in the \(3×3\) determinant that contains the entry.

    MINOR OF AN ENTRY IN \(3×3\) A DETERMINANT

    The minor of an entry in a \(3×3\) determinant is the \(2×2\) determinant found by eliminating the row and column in the \(3×3\) determinant that contains the entry.

    To find the minor of entry \(a_1\), we eliminate the row and column which contain it. So we eliminate the first row and first column. Then we write the \(2×2\) determinant that remains.

    The first row of the 3 by 3 determinant is a1, b1, c1. Row 2 is a2, b2, c2. Row 3 is a3, b3, c3. a1 is highlighted. Lines strike out the first row and the first column. What remains is called minor of a1. It is shown as a separate determinant whose first row is b2, c2 and second row is b3, c3.

    To find the minor of entry \(b_2\), we eliminate the row and column that contain it. So we eliminate the \(2^{nd}\) row and \(2^{nd}\) column. Then we write the \(2×2\) determinant that remains.

    The first row of the 3 by 3 determinant is a1, b1, c1. Row 2 is a2, b2, c2. Row 3 is a3, b3, c3. b2 is highlighted. Lines strike out the second row and second column. What remains is minor of b2. It is written as a separate determinant whose first row is a1, c1 and second row is a3, c3.

    Example \(\PageIndex{4}\)

    For the determinant \(\left| \begin{matrix} 4&−2&3\\1&0&−3\\−2&−4&2 \end{matrix} \right|\), find and then evaluate the minor of ⓐ \(a_1\) ⓑ \(b_3\) ⓒ \(c_2\).

    Answer

      .
    Eliminate the row and column that contains \(a_1\). .
    Write the \(2×2\) determinant that remains. .
    Evaluate. .
    Simplify. .

    Eliminate the row and column that contains \(b_3\). .
    Write the \(2×2\) determinant that remains. .
    Evaluate. .
    Simplify. .

      .
    Eliminate the row and column that contains \(c_2\). .
    Write the \(2×2\) determinant that remains. .
    Evaluate. .
    Simplify. .
    Example \(\PageIndex{5}\)

    For the determinant \(\left| \begin{matrix} 1&−1&4\\0&2&−1\\−2&−3&3 \end{matrix} \right|\), find and then evaluate the minor of ⓐ \(a_1\) ⓑ \(b_2\) ⓒ \(c_3\).

    Answer

    ⓐ 3 ⓑ 11 ⓒ 2

    Example \(\PageIndex{6}\)

    For the determinant \(\left| \begin{matrix} −2&−1&0\\3&0&−1\\−1&−2&3 \end{matrix} \right|\), find and then evaluate the minor of ⓐ \(a_2\) ⓑ \(b_3\) ⓒ \(c_2\).

    Answer

    ⓐ \(−3\) ⓑ 2 ⓒ 3

    We are now ready to evaluate a \(3×3\) determinant. To do this we expand by minors, which allows us to evaluate the \(3×3\)determinant using \(2×2\) determinants—which we already know how to evaluate!

    To evaluate a \(3×3\) determinant by expanding by minors along the first row, we use the following pattern:

    A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.

    Remember, to find the minor of an entry we eliminate the row and column that contains the entry.

    EXPANDING BY MINORS ALONG THE FIRST ROW TO EVALUATE A \(3×3\) DETERMINANT

    To evaluate a \(3×3\) determinant by expanding by minors along the first row, the following pattern:

    A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.

    Example \(\PageIndex{7}\)

    Evaluate the determinant \(\left| \begin{matrix} 2&−3&−1\\3&2&0\\−1&−1&−2 \end{matrix} \right|\) by expanding by minors along the first row.

    Answer
      .
    Expand by minors along the first row .
    Evaluate each determinant. .
    Simplify. .
    Simplify. .
    Simplify. .
    Example \(\PageIndex{8}\)

    Evaluate the determinant \(\left| \begin{matrix} 3&−2&4\\0&−1&−2\\2&3&−1 \end{matrix} \right|\), by expanding by minors along the first row.

    Answer

    37

    Example \(\PageIndex{9}\)

    Evaluate the determinant \(\left| \begin{matrix} 3&−2&−2\\2&−1&4\\−1&0&−3 \end{matrix} \right|\), by expanding by minors along the first row.

    Answer

    7

    To evaluate a \(3×3\) determinant we can expand by minors using any row or column. Choosing a row or column other than the first row sometimes makes the work easier.

    When we expand by any row or column, we must be careful about the sign of the terms in the expansion. To determine the sign of the terms, we use the following sign pattern chart.

    \[\left| \begin{matrix} +&−&+\\−&+&−\\+&−&+ \end{matrix} \right|\nonumber\]

    SIGN PATTERN

    When expanding by minors using a row or column, the sign of the terms in the expansion follow the following pattern.\[\left| \begin{matrix} +&−&+\\−&+&−\\+&−&+ \end{matrix} \right|\nonumber\]

    Notice that the sign pattern in the first row matches the signs between the terms in the expansion by the first row.

    A 3 by 3 determinant has row 1: plus, minus, plus, row 2: minus, plus, minus and row 3: plus, minus, plus. The three signs in the first row each point to a minor determinant in the expansion of a 3 by 3 determinant. Plus points to minor of a1, minus to the minor of b1 and plus to the minor of c1.

    Since we can expand by any row or column, how do we decide which row or column to use? Usually we try to pick a row or column that will make our calculation easier. If the determinant contains a 0, using the row or column that contains the 0 will make the calculations easier.

    Example \(\PageIndex{10}\)

    Evaluate the determinant \(\left| \begin{matrix} 4&−1&−3\\3&0&2\\5&−4&−3 \end{matrix} \right|\) by expanding by minors.

    Answer

    To expand by minors, we look for a row or column that will make our calculations easier. Since 0 is in the second row and second column, expanding by either of those is a good choice. Since the second row has fewer negatives than the second column, we will expand by the second row.

      .
    Expand using the second row.  
    Be careful of the signs. .
    Evaluate each determinant. .
    Simplify. .
    Simplify. .
    Add. .
    Example \(\PageIndex{11}\)

    Evaluate the determinant \(\left| \begin{matrix} 2&−1&−3\\0&3&−4\\3&−4&−3 \end{matrix} \right|\) by expanding by minors.

    Answer

    \(−11\)

    Example \(\PageIndex{12}\)

    Evaluate the determinant \(\left| \begin{matrix} −2&−1&−3\\−1&2&2\\4&−4&0 \end{matrix} \right|\) by expanding by minors.

    Answer

    8

    Use Cramer’s Rule to Solve Systems of Equations

    Cramer’s Rule is a method of solving systems of equations using determinants. It can be derived by solving the general form of the systems of equations by elimination. Here we will demonstrate the rule for both systems of two equations with two variables and for systems of three equations with three variables.

    Let’s start with the systems of two equations with two variables.

    CRAMER’S RULE FOR SOLVING A SYSTEM OF TWO EQUATIONS

    For the system of equations \(\left\{\begin{array} {l} a_1x+b_1y=k_1 \\ a_2x+b_2y=k_2\end{array}\right.\), the solution \((x,y)\) can be determined by

    x is Dx upon D and y is Dy upon D where D is determinant with row 1: a1, b1 and row 2 a2, b2, use coefficients of the variables; Dx is determinant with row 1: k1, b1 and row 2: k2, b2, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1 and row 2: a2, k2, replace the y coefficients with constants

    Notice that to form the determinant D, we use take the coefficients of the variables.

    The equations are a1x plus b1y equals k1 and a2x plus b2y equals k2. Here, a1, a2, b1, b2 are coefficients. The determinant is D with row 1: a1, b1 and row 2: a2, b2. Column 1 has coefficients of x and column 2 has coefficients of

    Notice that to form the determinant \(D_x\) and \(D_y\), we substitute the constants for the coefficients of the variable we are finding.

    The equations are a1x plus b1y equals k1 and a2x plus b2y equals k2. Here, a1, a2, b1, b2 are coefficients. The determinant is Dx has row 1: k1, b1 and row 2: k2, b2. Here columns 1 and 2 re constants and coefficients of y respectively. Determinant Dy has row 1: a1, k1 and row 2: a2, k2. Here, columns 1 and 2 are coefficients of x and constants respectively.

    Example \(\PageIndex{13}\): How to Solve a System of Equations Using Cramer’s Rule

    Solve using Cramer’s Rule: \(\left\{ \begin{array} {l} 2x+y=−4\\3x−2y=−6\end{array}\right.\)

    Answer

    The equations are 2x plus y equals minus 4 and 3x minus 2y equals minus 6. Step 1. Evaluate the determinant D, using the coefficients of the variables. Determinant D has row 1: 2, 1 and row 2: 3, minus 2. So, D is minus 7.Step 2. Evaluate the determinant Dx. Use the constants in place of the x coefficients. We replace the coefficients of x, 2 and 3, with the constants, negative 4 and negative 6. We get Dx equal to 14.Step 3. Evaluate the determinant Dy. Use the constants in place of the y coefficients. We replace the coefficients of y, 1 and 2, with the constants, negative 4 and negative 6. We get Dy equal to 0.Step 4. Find x and y. Substituting values of D, Dx and Dy in the equations x equal to Dx upon D and y equal to Dy upon D, we get x equal to minus 2 and y equal to 0.Step 5. Write the solution as an ordered pair minus 2, 0.Step 6. Check that the ordered pair is a solution to both original equations.

    Example \(\PageIndex{14}\)

    Solve using Cramer’s rule: \(\left\{\begin{array} {l} 3x+y=−3 \\ 2x+3y=6 \end{array} \right.\)

    Answer

    \((−\frac{15}{7},\frac{24}{7})\)

    Example \(\PageIndex{15}\)

    Solve using Cramer’s rule: \(\left\{\begin{array} {l} −x+y=2\\2x+y=−4 \end{array} \right.\)

    Answer

    \((−2,0)\)

    SOLVE A SYSTEM OF TWO EQUATIONS USING CRAMER’S RULE.
    1. Evaluate the determinant D, using the coefficients of the variables.
    2. Evaluate the determinant \(D_x\). Use the constants in place of the x coefficients.
    3. Evaluate the determinant \(D_y\). Use the constants in place of the y coefficients.
    4. Find x and y. \(x=\frac{D_x}{D}\), \(y=\frac{D_y}{D}\)
    5. Write the solution as an ordered pair.
    6. Check that the ordered pair is a solution to both original equations.

    To solve a system of three equations with three variables with Cramer’s Rule, we basically do what we did for a system of two equations. However, we now have to solve for three variables to get the solution. The determinants are also going to be \(3×3\) which will make our work more interesting!

    CRAMER’S RULE FOR SOLVING A SYSTEM OF THREE EQUATIONS

    For the system of equations \(\left\{\begin{array} {l} a_1x+b_1y+c_1z=k_1\\a_2x+b_2y+c_2z=k_2\\a_3x+b_3y+c_3z=k_3\end{array}\right.\), the solution \((x,y,z)\) can be determined by

    x is Dx upon D, y is Dy upon D and z is Dz upon D, where D is determinant with row 1: a1, b1, c1, row 2: a2, b2, c2, row 3: a3, b3, c3, use coefficients of the variables; Dx is determinant with row 1: k1, b1, c1, row 2: k2, b2, c2 and rwo 3: k3, b3, c3, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1, c1, row 2: a2, k2, c2 and row 3: a3, k3, c3, replace the y coefficients with constants; Dz is determinant with row 1: a1, b1, k1; row 2: a2, b2, k2, row 3: a3, b3, k3; replace the z coefficients with constants.

    Example \(\PageIndex{16}\)

    Solve the system of equations using Cramer’s Rule: \(\left\{\begin{array} {l} 3x−5y+4z=5\\5x+2y+z=0\\2x+3y−2z=3 \end{array} \right.\)

    Answer
    Evaluate the determinant D. .
    Expand by minors using column 1.  
    . .
    Evaluate the determinants. .
    Simplify. .
    Simplify. .
    Simplify. .
    Evaluate the determinant \(D_x\). Use the
    constants to replace the coefficients of x.
    .
    Expand by minors using column 1. .
    Evaluate the determinants. .
    Simplify. .
    Simplify. .
    Evaluate the determinant Dy.Dy. Use the
    constants to replace the coefficients of y.
    .
    . .
    Evaluate the determinants. .
    Simplify. .
    Simplify. .
    Simplify. .
    Evaluate the determinant Dz.Dz. Use the
    constants to replace the coefficients of z.
    .
    . .
    Evaluate the determinants. .
    Simplify. .
    Simplify. .
    Simplify. .
    Find x, y, and z. .
    Substitute in the values. .
    Simplify. .
    Write the solution as an ordered triple. .
    Check that the ordered triple is a solution
    to all three original equations.
    We leave the check to you.
      The solution is \((2,−3,−4)\).
    Example \(\PageIndex{17}\)

    Solve the system of equations using Cramer’s Rule: \(\left\{\begin{array} {l} 3x+8y+2z=−5\\2x+5y−3z=0\\x+2y−2z=−1 \end{array} \right.\)

    Answer

    \((−9,3,−1)\)

    Example \(\PageIndex{18}\)

    Solve the system of equations using Cramer’s Rule: \(\left\{\begin{array} {l} 3x+y−6z=−3\\2x+6y+3z=0\\3x+2y−3z=−6 \end{array} \right.\)

    Answer

    \((−6,3,−2)\)

    Cramer’s rule does not work when the value of the D determinant is 0, as this would mean we would be dividing by 0. But when \(D=0\), the system is either inconsistent or dependent.

    When the value of \(D=0\) and \(D_x,\space D_y\) and D are all zero, the system is consistent and dependent and there are infinitely many solutions.

    When the value of \(D=0\) and \(D_x,\space D_y\) and \(D_z\) are not all zero, the system is inconsistent and there is no solution.

    DEPENDENT AND INCONSISTENT SYSTEMS OF EQUATIONS

    For any system of equations, where the value of the determinant \(D=0\),

    \[ \begin{array} {lll} \textbf{Value of determinants} &\textbf{Type of system} &\textbf{Solution} \\ {D=0\text{ and }D_x,\space D_y\text{ and }D_z\text{ are all zero}} &\text{consistent and dependent} &\text{infinitely many solutions} \\ {D=0\text{ and }D_x,\space D_y\text{ and }D_z\text{ are not all zero}} &\text{inconsistent} &\text{no solution} \end{array} \nonumber\]

    In the next example, we will use the values of the determinants to find the solution of the system.

    Example \(\PageIndex{19}\)

    Solve the system of equations using Cramer’s rule : \(\left\{\begin{array} {l} x+3y=4\\−2x−6y=3 \end{array} \right.\)

    Answer

    \(\begin{array} {ll} {} &{\left\{\begin{array} {l} x+3y=4\\−2x−6y=3 \end{array} \right.} \\ {\begin{array} {l} \text{Evaluate the determinantD,using the} \\ \text{coefficients of the variables.} \end{array}} &{D=\left|\begin{matrix} 1&3\\−2&−6\end{matrix}\right|} \\ {} &{D=−6−(−6)} \\ {} &{D=0} \end{array} \)

    We cannot use Cramer’s Rule to solve this system. But by looking at the value of the determinants \(D_x\) and \(D_y\), we can determine whether the system is dependent or inconsistent.

    \(\begin{array} {ll} {\text{Evaluate the determinant }D_x.} &{D_x=\left|\begin{matrix} 4&3\\3&−6\end{matrix}\right|} \\ {} &{D_x=−24−9} \\ {} &{D_x=15} \end{array} \)

    Since all the determinants are not zero, the system is inconsistent. There is no solution.

    Example \(\PageIndex{20}\)

    Solve the system of equations using Cramer’s rule: \(\left\{\begin{array} {l} 4x−3y=8\\8x−6y=14 \end{array} \right.\)

    Answer

    no solution

    Example \(\PageIndex{21}\)

    Solve the system of equations using Cramer’s rule: \(\left\{\begin{array} {l} x=−3y+4\\2x+6y=8 \end{array} \right.\)

    Answer

    infinite solutions

    Solve Applications using Determinants

    An interesting application of determinants allows us to test if points are collinear. Three points \((x_1,y_1)\), \((x_2,y_2)\) and \((x_3,y_3)\) are collinear if and only if the determinant below is zero.

    \[\left|\begin{matrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{matrix}\right|=0\nonumber\]

    TEST FOR COLLINEAR POINTS

    Three points \((x_1,y_1)\), \((x_2,y_2)\) and \((x_3,y_3)\) are collinear if and only if

    \[\left|\begin{matrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{matrix}\right|=0\nonumber\]

    We will use this property in the next example.

    Example \(\PageIndex{22}\)

    Determine whether the points \((5,−5)\), \((4,−3)\), and \((3,−1)\) are collinear.

    Answer
      .
    Substitute the values into the determinant.
    \((5,−5)\), \((4,−3)\), and \((3,−1)\)
    .
    Evaluate the determinant by expanding
    by minors using column 3.
    .
    Evaluate the determinants. .
    Simplify. .
    Simplify. .
      The value of the determinate is 0, so the
    points are collinear.
    Example \(\PageIndex{23}\)

    Determine whether the points \((3,−2)\), \((5,−3)\), and \((1,−1)\) are collinear.

    Answer

    yes

    Example \(\PageIndex{24}\)

    Determine whether the points \((−4,−1)\), \((−6,2)\), and \((−2,−4)\) are collinear.

    Answer

    yes

    Access these online resources for additional instruction and practice with solving systems of linear inequalities by graphing.

    • Solving Systems of Linear Inequalities by Graphing
    • Systems of Linear Inequalities

    Key Concepts

    • Determinant: The determinant of any square matrix \(\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\), where a, b, c, and d are real numbers, is

      \[\left|\begin{matrix}a&b\\c&d\end{matrix}\right|=ad−bc\nonumber\]

    • Expanding by Minors along the First Row to Evaluate a 3 × 3 Determinant: To evaluate a \(3×3\) determinant by expanding by minors along the first row, the following pattern:
      A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.
    • Sign Pattern: When expanding by minors using a row or column, the sign of the terms in the expansion follow the following pattern.

      \[\left|\begin{matrix}+&−&+\\−&+&−\\+&−&+\end{matrix}\right|\nonumber\]

    • Cramer’s Rule: For the system of equations \(\left\{\begin{array} {l} a_1x+b_1y=k_1\\a_2x+b_2y=k_2\end{array}\right.\), the solution \((x,y)\) can be determined by
      x is Dx upon D and y is Dy upon D where D is determinant with row 1: a1, b1 and row 2 a2, b2, use coefficients of the variables; Dx is determinant with row 1: k1, b1 and row 2: k2, b2, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1 and row 2: a2, k2, replace the y coefficients with constants.
      Notice that to form the determinant D, we use take the coefficients of the variables.
    • How to solve a system of two equations using Cramer’s rule.
      1. Evaluate the determinant D, using the coefficients of the variables.
      2. Evaluate the determinant \(D_x\). Use the constants in place of the x coefficients.
      3. Evaluate the determinant \(D_y\). Use the constants in place of the y coefficients.
      4. Find x and y. \(x=\frac{D_x}{D}\), \(y=\frac{D_y}{D}\).
      5. Write the solution as an ordered pair.
      6. Check that the ordered pair is a solution to both original equations.
      7. Dependent and Inconsistent Systems of Equations: For any system of equations, where the value of the determinant \(D=0\),\[ \begin{array} {lll} \textbf{Value of determinants} &\textbf{Type of system} &\textbf{Solution} \\ {D=0\text{ and }D_x,\space D_y\text{ and }D_z\text{ are all zero}} &\text{consistent and dependent} &\text{infinitely many solutions} \\ {D=0\text{ and }D_x,\space D_y\text{ and }D_z\text{ are not all zero}} &\text{inconsistent} &\text{no solution} \end{array} \nonumber\]
      8. Test for Collinear Points: Three points \((x_1,y_1)\), \((x_2,y_2)\), and \((x_3,y_3)\) are collinear if and only if

        \[\left|\begin{matrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{matrix}\right|=0\nonumber\]

    Glossary

    determinant
    Each square matrix has a real number associated with it called its determinant.
    minor of an entry in a 3×33×3 determinant
    The minor of an entry in a 3×33×3 determinant is the 2×22×2 determinant found by eliminating the row and column in the 3×33×3determinant that contains the entry.
    square matrix
    A square matrix is a matrix with the same number of rows and columns.

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