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Solve Systems of Equations Using Determinants

Evaluate the determinant of a

Evaluate the Determinant of a

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 \([\begin{array}{l}a\ b \\ c\ d\end{array}],\) we first write it as \(|\begin{array}{l}a\ b \\ c\ d\end{array}|.\) To get the real number value of the determinant we subtract the products of the diagonals, as shown.

Example

Try it.

Evaluate the determinant of ⓐ \([\begin{array}{l}4\ -2 \\ 3\ -1\end{array}]\) ⓑ \([\begin{array}{llll}-3 & & & -4 \\ -2 & & & 0\end{array}].\)

Solution

ⓐ


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

ⓑ


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

Evaluate the Determinant of a

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

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 \(2\ \times \ 2\) determinant that remains.

Example

Try it.

For the determinant \(|\begin{array}{lllllll}4 & & & -2 & & & 3 \\ 1 & & & 0 & & & -3 \\ -2 & & & -4 & & & 2\end{array}|,\) find and then evaluate the minor of ⓐ \({a}_{1}\) ⓑ \({b}_{3}\) ⓒ \({c}_{2}.\)

Solution

ⓐ


Eliminate the row and column that contains \({a}_{1}.\)
Write the \(2\ \times \ 2\) determinant that remains.
Evaluate.
Simplify.

ⓑ

Eliminate the row and column that contains \({b}_{3}.\)
Write the \(2\ \times \ 2\) determinant that remains.
Evaluate.
Simplify.

ⓒ


Eliminate the row and column that contains \({c}_{2}.\)
Write the \(2\ \times \ 2\) determinant that remains.
Evaluate.
Simplify.

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

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

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

\[|\begin{array}{l}+\ -\ + \\ -\ +\ - \\ +\ -\ +\end{array}|\]

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

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.

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

Notice that to form the determinant \({D}_{x}\) and \({D}_{y},\) we substitute the constants for the coefficients of the variable we are finding.

How to Solve a System of Equations Using Cramer’s Rule

Try it.

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

Solution

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 \(3\ \times \ 3\) which will make our work more interesting!

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

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.

\[|\begin{array}{l}{x}_{1}\ {y}_{1}\ 1 \\ {x}_{2}\ {y}_{2}\ 1 \\ {x}_{3}\ {y}_{3}\ 1\end{array}|=0\]

We will use this property in the next example.

Example

Try it.

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

Solution


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 determinant is 0, so the points are collinear.

Key Concepts

Determinant: The determinant of any square matrix \([\begin{array}{l}a\ b \\ c\ d\end{array}],\) where a, b, c, and d are real numbers, is
\[|\begin{array}{l}a\ b \\ c\ d\end{array}|=ad-bc\]
Expanding by Minors along the First Row to Evaluate a 3 × 3 Determinant: To evaluate a \(3\ \times \ 3\) determinant by expanding by minors along the first row, the following pattern:
Sign Pattern: When expanding by minors using a row or column, the sign of the terms in the expansion follow the following pattern.
\[|\begin{array}{l}+\ -\ + \\ -\ +\ - \\ +\ -\ +\end{array}|\]
Cramer’s Rule: For the system of equations \(\{\begin{array}{l}{a}_{1}x+{b}_{1}y={k}_{1} \\ {a}_{2}x+{b}_{2}y={k}_{2}\end{array},\) the solution \((x,y)\) can be determined by

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 \(D=0,\)
  \(\begin{array}{lllllll}\text{Value of determinants} & & & \text{Type of system} & & & \text{Solution} \\ D=0\ \text{and}\ {D}_{x},{D}_{y}\ \text{and}\ {D}_{z}\ \text{are all zero} & & & \text{consistent and dependent} & & & \text{infinitely many solutions} \\ D=0\ \text{and}\ {D}_{x},{D}_{y}\ \text{and}\ {D}_{z}\ \text{are not all zero} & & & \text{inconsistent} & & & \text{no solution}\end{array}\)
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
  \[|\begin{array}{l}{x}_{1}\ {y}_{1}\ 1 \\ {x}_{2}\ {y}_{2}\ 1 \\ {x}_{3}\ {y}_{3}\ 1\end{array}|=0\]

Solve Systems of Equations Using Determinants

Evaluate the Determinant of a 2 × 2 Matrix

In the following exercises, evaluate the determinant of each square matrix.

Try it.

\([\begin{array}{l}6\ -2 \\ 3\ -1\end{array}]\)

Try it.

\([\begin{array}{l}-4\ 8 \\ -3\ 5\end{array}]\)

Solution

Try it.

\([\begin{array}{llll}-3 & & & 5 \\ 0 & & & -4\end{array}]\)

Try it.

\([\begin{array}{llll}-2 & & & 0 \\ 7 & & & -5\end{array}]\)

Solution

Evaluate the Determinant of a 3 × 3 Matrix

In the following exercises, find and then evaluate the indicated minors.

Try it.

\(|\begin{array}{lllllll}3 & & & -1 & & & 4 \\ -1 & & & 0 & & & -2 \\ -4 & & & 1 & & & 5\end{array}|\)
Find the minor ⓐ \({a}_{1}\) ⓑ \({b}_{2}\) ⓒ \({c}_{3}\)

Try it.

\(|\begin{array}{lllllll}-1 & & & -3 & & & 2 \\ 4 & & & -2 & & & -1 \\ -2 & & & 0 & & & -3\end{array}|\)
Find the minor ⓐ \({a}_{1}\) ⓑ \({b}_{1}\) ⓒ \({c}_{2}\)

Solution

ⓐ 6 ⓑ \(-14\) ⓒ \(-6\)

Try it.

\(|\begin{array}{lllllll}2 & & & -3 & & & -4 \\ -1 & & & 2 & & & -3 \\ 0 & & & -1 & & & -2\end{array}|\)
Find the minor ⓐ \({a}_{2}\) ⓑ \({b}_{2}\) ⓒ \({c}_{2}\)

Try it.

\(|\begin{array}{lllllll}-2 & & & -2 & & & 3 \\ 1 & & & -3 & & & 0 \\ -2 & & & 3 & & & -2\end{array}|\)
Find the minor ⓐ \({a}_{3}\) ⓑ \({b}_{3}\) ⓒ \({c}_{3}\)

Solution

ⓐ 9 ⓑ \(-3\) ⓒ 8

In the following exercises, evaluate each determinant by expanding by minors along the first row.

Try it.

\(|\begin{array}{lllllll}-2 & & & 3 & & & -1 \\ -1 & & & 2 & & & -2 \\ 3 & & & 1 & & & -3\end{array}|\)

Try it.

\(|\begin{array}{lllllll}4 & & & -1 & & & -2 \\ -3 & & & -2 & & & 1 \\ -2 & & & -5 & & & 7\end{array}|\)

Solution

\(-77\)

Try it.

\(|\begin{array}{lllllll}-2 & & & -3 & & & -4 \\ 5 & & & -6 & & & 7 \\ -1 & & & 2 & & & 0\end{array}|\)

Try it.

\(|\begin{array}{lllllll}1 & & & 3 & & & -2 \\ 5 & & & -6 & & & 4 \\ 0 & & & -2 & & & -1\end{array}|\)

Solution

In the following exercises, evaluate each determinant by expanding by minors.

Try it.

\(|\begin{array}{lllllll}-5 & & & -1 & & & -4 \\ 4 & & & 0 & & & -3 \\ 2 & & & -2 & & & 6\end{array}|\)

Try it.

\(|\begin{array}{lllllll}4 & & & -1 & & & 3 \\ 3 & & & -2 & & & 2 \\ -1 & & & 0 & & & 4\end{array}|\)

Solution

\(-24\)

Try it.

\(|\begin{array}{lllllll}3 & & & 5 & & & 4 \\ -1 & & & 3 & & & 0 \\ -2 & & & 6 & & & 1\end{array}|\)

Try it.

\(|\begin{array}{lllllll}2 & & & -4 & & & -3 \\ 5 & & & -1 & & & -4 \\ 3 & & & 2 & & & 0\end{array}|\)

Solution

Use Cramer’s Rule to Solve Systems of Equations

In the following exercises, solve each system of equations using Cramer’s Rule.

Try it.

\(\{\begin{array}{l}-2x+3y=3 \\ x+3y=12\end{array}\)

Try it.

\(\{\begin{array}{l}x-2y=-5 \\ 2x-3y=-4\end{array}\)

Solution

\((7,6)\)

Try it.

\(\{\begin{array}{l}x-3y=-9 \\ 2x+5y=4\end{array}\)

Try it.

\(\{\begin{array}{l}2x+y=-4 \\ 3x-2y=-6\end{array}\)

Solution

\((-2,0)\)

Try it.

\(\{\begin{array}{l}x-2y=-5 \\ 2x-3y=-4\end{array}\)

Try it.

\(\{\begin{array}{l}x-3y=-9 \\ 2x+5y=4\end{array}\)

Solution

\((-3,2)\)

Try it.

\(\{\begin{array}{l}5x-3y=-1 \\ 2x-y=2\end{array}\)

Try it.

\(\{\begin{array}{l}3x+8y=-3 \\ 2x+5y=-3\end{array}\)

Solution

\((-9,3)\)

Try it.

\(\{\begin{array}{l}6x-5y+2z=3 \\ 2x+y-4z=5 \\ 3x-3y+z=-1\end{array}\)

Try it.

\(\{\begin{array}{l}4x-3y+z=7 \\ 2x-5y-4z=3 \\ 3x-2y-2z=-7\end{array}\)

Solution

\((-3,-5,4)\)

Try it.

\(\{\begin{array}{l}2x-5y+3z=8 \\ 3x-y+4z=7 \\ x+3y+2z=-3\end{array}\)

Try it.

\(\{\begin{array}{l}11x+9y+2z=-9 \\ 7x+5y+3z=-7 \\ 4x+3y+z=-3\end{array}\)

Solution

\((2,-3,-2)\)

Try it.

\(\{\begin{array}{l}x+2z=0 \\ 4y+3z=-2 \\ 2x-5y=3\end{array}\)

Try it.

\(\{\begin{array}{l}2x+5y=4 \\ 3y-z=3 \\ 4x+3z=-3\end{array}\)

Solution

\((-3,2,3)\)

Try it.

\(\{\begin{array}{l}2y+3z=-1 \\ 5x+3y=-6 \\ 7x+z=1\end{array}\)

Try it.

\(\{\begin{array}{l}3x-z=-3 \\ 5y+2z=-6 \\ 4x+3y=-8\end{array}\)

Solution

\((-2,0,-3)\)

Try it.

\(\{\begin{array}{l}2x+y=3 \\ 6x+3y=9\end{array}\)

Try it.

\(\{\begin{array}{l}x-4y=-1 \\ -3x+12y=3\end{array}\)

Solution

infinitely many solutions

Try it.

\(\{\begin{array}{l}-3x-y=4 \\ 6x+2y=-16\end{array}\)

Try it.

\(\{\begin{array}{l}4x+3y=2 \\ 20x+15y=5\end{array}\)

Solution

inconsistent

Try it.

\(\{\begin{array}{l}x+y-3z=-1 \\ y-z=0 \\ \text{-}x+2y=1\end{array}\)

Try it.

\(\{\begin{array}{l}2x+3y+z=12 \\ x+y+z=9 \\ 3x+4y+2z=20\end{array}\)

Solution

inconsistent

Try it.

\(\{\begin{array}{l}3x+4y-3z=-2 \\ 2x+3y-z=-12 \\ x+y-2z=6\end{array}\)

Try it.

\(\{\begin{array}{l}x-2y+3z=1 \\ x+y-3z=7 \\ 3x-4y+5z=7\end{array}\)

Solution

infinitely many solutions

Solve Applications Using Determinants

In the following exercises, determine whether the given points are collinear.

Try it.

\((0,1),\) \((2,0),\) and \((-2,2).\)

Try it.

\((0,-5),\) \((-2,-2),\) and \((2,-8).\)

Solution

yes

Try it.

\((4,-3),\) \((6,-4),\) and \((2,-2).\)

Try it.

\((-2,1),\) \((-4,4),\) and \((0,-2).\)

Solution

yes

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Simplify: \(5(-2)-(-4)(1).\)
If you missed this problem, review .

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\(-6\)
Simplify: \(-3(8-10)+(-2)(6-3)-4(-3-(-4)).\)
If you missed this problem, review .

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\(-4\)
Simplify: \(\frac{-12}{-8}.\)
If you missed this problem, review .

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\(\frac{3}{2}\)
Evaluate the determinant of ⓐ \([\begin{array}{l}4\ -2 \\ 3\ -1\end{array}]\) ⓑ \([\begin{array}{llll}-3 & & & -4 \\ -2 & & & 0\end{array}].\)

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ⓐ

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

ⓑ

Write the determinant.
Subtract the products of the diagonals.
Simplify.
Simplify.
Evaluate the determinant of ⓐ \([\begin{array}{l}5\ -3 \\ 2\ -4\end{array}]\) ⓑ \([\begin{array}{llll}-4 & & & -6 \\ 0 & & & 7\end{array}].\)

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ⓐ \(-14;\) ⓑ \(-28\)
Evaluate the determinant of ⓐ \([\begin{array}{l}-1\ 3 \\ -2\ 4\end{array}]\) ⓑ \([\begin{array}{llll}-7 & & & -3 \\ -5 & & & 0\end{array}].\)

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ⓐ 2 ⓑ \(-15\)
For the determinant \(|\begin{array}{lllllll}4 & & & -2 & & & 3 \\ 1 & & & 0 & & & -3 \\ -2 & & & -4 & & & 2\end{array}|,\) find and then evaluate the minor of ⓐ \({a}_{1}\) ⓑ \({b}_{3}\) ⓒ \({c}_{2}.\)

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ⓐ

Eliminate the row and column that contains \({a}_{1}.\)
Write the \(2\ \times \ 2\) determinant that remains.
Evaluate.
Simplify.

ⓑ

Eliminate the row and column that contains \({b}_{3}.\)
Write the \(2\ \times \ 2\) determinant that remains.
Evaluate.
Simplify.

ⓒ

Eliminate the row and column that contains \({c}_{2}.\)
Write the \(2\ \times \ 2\) determinant that remains.
Evaluate.
Simplify.
For the determinant \(|\begin{array}{lllllll}1 & & & -1 & & & 4 \\ 0 & & & 2 & & & -1 \\ -2 & & & -3 & & & 3\end{array}|,\) find and then evaluate the minor of ⓐ \({a}_{1}\) ⓑ \({b}_{2}\) ⓒ \({c}_{3}.\)

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ⓐ 3 ⓑ 11 ⓒ 2
For the determinant \(|\begin{array}{lllllll}-2 & & & -1 & & & 0 \\ 3 & & & 0 & & & -1 \\ -1 & & & -2 & & & 3\end{array}|,\) find and then evaluate the minor of ⓐ \({a}_{2}\) ⓑ \({b}_{3}\) ⓒ \({c}_{2}.\)

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ⓐ \(-3\) ⓑ 2 ⓒ 3
Evaluate the determinant \(|\begin{array}{lllllll}2 & & & -3 & & & -1 \\ 3 & & & 2 & & & 0 \\ -1 & & & -1 & & & -2\end{array}|\) by expanding by minors along the first row.

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Expand by minors along the first row
Evaluate each determinant.
Simplify.
Simplify.
Simplify.
Evaluate the determinant \(|\begin{array}{lllllll}3 & & & -2 & & & 4 \\ 0 & & & -1 & & & -2 \\ 2 & & & 3 & & & -1\end{array}|,\) by expanding by minors along the first row.

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37
Evaluate the determinant \(|\begin{array}{lllllll}3 & & & -2 & & & -2 \\ 2 & & & -1 & & & 4 \\ -1 & & & 0 & & & -3\end{array}|,\) by expanding by minors along the first row.

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7
Evaluate the determinant \(|\begin{array}{lllllll}4 & & & -1 & & & -3 \\ 3 & & & 0 & & & 2 \\ 5 & & & -4 & & & -3\end{array}|\) by expanding by minors.

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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.
Evaluate the determinant \(|\begin{array}{lllllll}2 & & & -1 & & & -3 \\ 0 & & & 3 & & & -4 \\ 3 & & & -4 & & & -3\end{array}|\) by expanding by minors.

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\(-11\)
Evaluate the determinant \(|\begin{array}{lllllll}-2 & & & -1 & & & -3 \\ -1 & & & 2 & & & 2 \\ 4 & & & -4 & & & 0\end{array}|\) by expanding by minors.

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\(-12\)
Solve using Cramer’s Rule: \(\{\begin{array}{l}2x+y=-4 \\ 3x-2y=-6\end{array}.\)
Solve using Cramer’s rule: \(\{\begin{array}{l}3x+y=-3 \\ 2x+3y=6\end{array}.\)

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\((-\frac{15}{7},\frac{24}{7})\)
Solve using Cramer’s rule: \(\{\begin{array}{l}\text{-}x+y=2 \\ 2x+y=-4\end{array}.\)

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\((-2,0)\)

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

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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 \({D}_{y}.\) Use the constants to replace the coefficients of y.

Evaluate the determinants.
Simplify.
Simplify.
Simplify.
Evaluate the determinant \({D}_{z}.\) 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 \((2,-3,-4).\)

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

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\((-9,3,-1)\)
Solve the system of equations using Cramer’s Rule: \(\{\begin{array}{l}3x+y-6z=-3 \\ 2x+6y+3z=0 \\ 3x+2y-3z=-6\end{array}.\)

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

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\(\begin{array}{llllllllllll} & & & & & \{\begin{array}{l}x+3y=4 \\ \text{-}2x-6y=3\end{array} \\ \begin{array}{l}\text{Evaluate the determinant}\ D,\ \text{using the} \\ \text{coefficients of the variables.}\end{array} & & & & & D=|\begin{array}{llll}1 & & & 3 \\ -2 & & & -6\end{array}| \\ & & & & & D=-6-(-6) \\ & & & & & D=0\end{array}\)
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.
\(\begin{array}{llllllllllll}\text{Evaluate the determinant}\ {D}_{x}. & & & & & \ {D}_{x}=|\begin{array}{llll}4 & & & 3 \\ 3 & & & -6\end{array}| \\ & & & & & \ {D}_{x}=-24-9 \\ & & & & & \ {D}_{x}=-33\end{array}\)
Since all the determinants are not zero, the system is inconsistent. There is no solution.
Solve the system of equations using Cramer’s rule: \(\{\begin{array}{l}4x-3y=8 \\ 8x-6y=14\end{array}.\)

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no solution
Solve the system of equations using Cramer’s rule: \(\{\begin{array}{l}x=-3y+4 \\ 2x+6y=8\end{array}.\)

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infinite solutions

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

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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 determinant is 0, so the points are collinear.

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

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

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yes
\([\begin{array}{l}6\ -2 \\ 3\ -1\end{array}]\)
\([\begin{array}{l}-4\ 8 \\ -3\ 5\end{array}]\)

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4
\([\begin{array}{llll}-3 & & & 5 \\ 0 & & & -4\end{array}]\)
\([\begin{array}{llll}-2 & & & 0 \\ 7 & & & -5\end{array}]\)

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10
\(|\begin{array}{lllllll}3 & & & -1 & & & 4 \\ -1 & & & 0 & & & -2 \\ -4 & & & 1 & & & 5\end{array}|\)
Find the minor ⓐ \({a}_{1}\) ⓑ \({b}_{2}\) ⓒ \({c}_{3}\)
\(|\begin{array}{lllllll}-1 & & & -3 & & & 2 \\ 4 & & & -2 & & & -1 \\ -2 & & & 0 & & & -3\end{array}|\)
Find the minor ⓐ \({a}_{1}\) ⓑ \({b}_{1}\) ⓒ \({c}_{2}\)

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ⓐ 6 ⓑ \(-14\) ⓒ \(-6\)
\(|\begin{array}{lllllll}2 & & & -3 & & & -4 \\ -1 & & & 2 & & & -3 \\ 0 & & & -1 & & & -2\end{array}|\)
Find the minor ⓐ \({a}_{2}\) ⓑ \({b}_{2}\) ⓒ \({c}_{2}\)
\(|\begin{array}{lllllll}-2 & & & -2 & & & 3 \\ 1 & & & -3 & & & 0 \\ -2 & & & 3 & & & -2\end{array}|\)
Find the minor ⓐ \({a}_{3}\) ⓑ \({b}_{3}\) ⓒ \({c}_{3}\)

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ⓐ 9 ⓑ \(-3\) ⓒ 8
\(|\begin{array}{lllllll}-2 & & & 3 & & & -1 \\ -1 & & & 2 & & & -2 \\ 3 & & & 1 & & & -3\end{array}|\)
\(|\begin{array}{lllllll}4 & & & -1 & & & -2 \\ -3 & & & -2 & & & 1 \\ -2 & & & -5 & & & 7\end{array}|\)

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\(-77\)
\(|\begin{array}{lllllll}-2 & & & -3 & & & -4 \\ 5 & & & -6 & & & 7 \\ -1 & & & 2 & & & 0\end{array}|\)
\(|\begin{array}{lllllll}1 & & & 3 & & & -2 \\ 5 & & & -6 & & & 4 \\ 0 & & & -2 & & & -1\end{array}|\)

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49
\(|\begin{array}{lllllll}-5 & & & -1 & & & -4 \\ 4 & & & 0 & & & -3 \\ 2 & & & -2 & & & 6\end{array}|\)

Symbols used here

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Systems of Equations Using Determinants

Evaluate the determinant of a
Evaluate the determinant of a
Use Cramer’s Rule to solve systems of equations
Solve applications using determinants
Evaluate the determinant
Evaluate the determinant
Evaluate the determinant
Find

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value


Expand by minors along the first row
Evaluate each determinant.
Simplify.
Simplify.
Simplify.


Expand using the second row.
Be careful of the signs.
Evaluate each determinant.
Simplify.
Simplify.
Add.