\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Precalculus - Cramer's Rule

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This page covers how to solve linear systems of equations using determinants.

We can use matrices and determinants to solve systems of linear equations using a technique called Cramers Rule. We will show how to do this with 2 equations and 2 unknowns. However, the concept can be extended to higher order systems.

Problem

We are given a system of linear equations of the form
\( \begin{array}{ccccc} ax & + & by & = & z_0 \\ cx & + & dy & = & z_1 \end{array} \)
where \( a, b, c, d, z_0, z_1 \) are all real constants and the variables are \( x\) and \(y \).

We need to find what values of \(x\) and \(y\) solve this equation. There are three possible cases.
1. \(x\) and \(y\) are real, unique and not equal.
2. \(x\) and \(y\) are real and equal.
3. \(x\) and \(y\) are complex.

The key to determining which case holds is to look at the determinant of the coefficient matrix, i.e.
\( \begin{vmatrix} a & b \\ c & d \end{vmatrix} \)
If this determinant is zero, then we cannot use this technique and either case 2 or 3 hold. If this case is non-zero, then case 1 holds and we can solve this problem. Let's call the coefficient matrix \(A\) and so
\( \abs{A} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} \)

Solution

As long as the determinant of the coefficient matrix is NOT zero, we can solve this system of equations and the values are given by
\(\displaystyle{ x = \frac{ \begin{vmatrix} z_0 & b \\ z_1 & d \end{vmatrix} }{\abs{A}} }\)       and       \(\displaystyle{ y = \frac{ \begin{vmatrix} a & z_0 \\ c & z_1 \end{vmatrix} }{\abs{A}} }\)

In each case above, notice that we have replaced the column of matrix \(A\) corresponding to the variable we are calculating with the \(z\) constants.

Here are a couple of videos with examples.

PatrickJMT - video 1

video by PatrickJMT

PatrickJMT - video 2

video by PatrickJMT

The best way to learn Cramer's Rule is by watching someone work specific problems and then working plenty on your own. To get started, here is a video showing, in general, how to use Cramer's Rule, then he does a specific example. After this video, you should be able to work problems on your own.

Thinkwell - Using Cramer's Rule [7min-33secs]

video by Thinkwell

Here is a video showing a proof or justification for Cramer's Rule.

PatrickJMT - Cramer's Rule : A Proof / Justification for a System of 2 Linear Equations, 2 Unknowns [10min-12secs]

video by PatrickJMT

Important!

Although most of the solutions below do not show the work, it is important to check your work when using Cramer's Rule. As you saw from the videos and you will see in the solutions, it is very, very easy to make mistakes that propagate and give you all wrong answers. So it is important to plug your answers back into all of the equations to make sure you have the correct answers. Your work may be considered incomplete if you do not check your answers (depending on what your instructor expects). So just get used to always check your answers. It could be the difference between a whole letter grade and it takes only a few seconds.

Okay, now you are ready for some practice problems.

Practice

Solve these linear systems using Cramer's Rule.

Let's start with some 2x2 systems

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x+4y=-14; \) \( -2x-3y=11 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x+4y=-14; \) \( -2x-3y=11 \)

Final Answer

\(x=2, y=-5\)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x+4y=-14; \) \( -2x-3y=11 \)

Solution

1884 video

video by Krista King Math

Final Answer

\(x=2, y=-5\)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 5x+7y=-1; \) \( 6x+8y=1 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 5x+7y=-1; \) \( 6x+8y=1 \)

Final Answer

\(x=15/2, y=11/2\)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 5x+7y=-1; \) \( 6x+8y=1 \)

Solution

1885 video

video by Thinkwell

Final Answer

\(x=15/2, y=11/2\)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 4x-2y=10; \) \( 3x-5y=11 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 4x-2y=10; \) \( 3x-5y=11 \)

Final Answer

\( x=2, y=-1 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 4x-2y=10; \) \( 3x-5y=11 \)

Solution

1894 video

video by mattemath

Final Answer

\( x=2, y=-1 \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x-y=-3; \) \( x+4y=17 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x-y=-3; \) \( x+4y=17 \)

Final Answer

\( x=1, y=4 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x-y=-3; \) \( x+4y=17 \)

Solution

1895 video

video by MIP4U

Final Answer

\( x=1, y=4 \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x-3y=16;\) \( x+2y=1 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x-3y=16;\) \( x+2y=1 \)

Final Answer

\( x=5, y=-2 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x-3y=16;\) \( x+2y=1 \)

Solution

1897 video

video by MIP4U

Final Answer

\( x=5, y=-2 \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 5y = 26 \)
\( 5x - 4y = -1 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 5y = 26 \)
\( 5x - 4y = -1 \)

Final Answer

\( (3, 4) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 5y = 26 \)
\( 5x - 4y = -1 \)

Solution

2855 video

Final Answer

\( (3, 4) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x - 2y = -4 \)
\( 4x - y = 3 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x - 2y = -4 \)
\( 4x - y = 3 \)

Final Answer

\( (2, 5) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x - 2y = -4 \)
\( 4x - y = 3 \)

Solution

2856 video

Final Answer

\( (2, 5) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 3y = 13 \)
\( 3x - 5y = -9 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 3y = 13 \)
\( 3x - 5y = -9 \)

Final Answer

\( (2, 3) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 3y = 13 \)
\( 3x - 5y = -9 \)

Solution

2857 video

Final Answer

\( (2, 3) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x - 7y = 1 \)
\( 3x + y = 13 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x - 7y = 1 \)
\( 3x + y = 13 \)

Final Answer

\( (4, 1) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x - 7y = 1 \)
\( 3x + y = 13 \)

Solution

2858 video

Final Answer

\( (4, 1) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x - 2y = -3 \)
\( 3x + y = 5 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x - 2y = -3 \)
\( 3x + y = 5 \)

Final Answer

\( (1, 2) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x - 2y = -3 \)
\( 3x + y = 5 \)

Solution

2859 video

Final Answer

\( (1, 2) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x - y = 4 \)
\( 2x + y = 2 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x - y = 4 \)
\( 2x + y = 2 \)

Final Answer

\( (2, -2) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x - y = 4 \)
\( 2x + y = 2 \)

Solution

2860 video

Final Answer

\( (2, -2) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 4x - 2y = 10 \)
\( 3x - 5y = 11 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 4x - 2y = 10 \)
\( 3x - 5y = 11 \)

Final Answer

\( (2, -1) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 4x - 2y = 10 \)
\( 3x - 5y = 11 \)

Solution

2861 video

video by mattemath

Final Answer

\( (2, -1) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x - 3y = 1 \)
\( x + 2y = 11 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x - 3y = 1 \)
\( x + 2y = 11 \)

Final Answer

\( (5, 3) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x - 3y = 1 \)
\( x + 2y = 11 \)

Solution

2862 video

Final Answer

\( (5, 3) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x = 2y + 5 \)
\( 4y = 6x - 8 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x = 2y + 5 \)
\( 4y = 6x - 8 \)

Final Answer

no solution

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x = 2y + 5 \)
\( 4y = 6x - 8 \)

Solution

2863 video

Final Answer

no solution

close solution

Log in to rate this practice problem and to see it's current rating.

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x + 2y = -3 \)
\( -3x - 6y = 9 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x + 2y = -3 \)
\( -3x - 6y = 9 \)

Final Answer

infinite number of solutions

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x + 2y = -3 \)
\( -3x - 6y = 9 \)

Solution

2864 video

Final Answer

infinite number of solutions

close solution

Log in to rate this practice problem and to see it's current rating.

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x + 7y = 5 \)
\( -2x + y = -9 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x + 7y = 5 \)
\( -2x + y = -9 \)

Final Answer

\( (4, -1) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x + 7y = 5 \)
\( -2x + y = -9 \)

Solution

2865 video

Final Answer

\( (4, -1) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 3y = -4 \)
\( -x + y = 7 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 3y = -4 \)
\( -x + y = 7 \)

Final Answer

\( (-5, 2) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 3y = -4 \)
\( -x + y = 7 \)

Solution

2866 video

Final Answer

\( (-5, 2) \)

close solution

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Now you are ready for some 3x3 systems.

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x+3y+z=2; \) \( -x+2y+3z=-1; \) \( -3x-3y+z=0 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x+3y+z=2; \) \( -x+2y+3z=-1; \) \( -3x-3y+z=0 \)

Final Answer

\( x=4, y=-3, z=3 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x+3y+z=2; \) \( -x+2y+3z=-1; \) \( -3x-3y+z=0 \)

Solution

1896 video

video by MIP4U

Final Answer

\( x=4, y=-3, z=3 \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x-y-z=2; \) \( 4x+y-z=-5; \) \( 6x-2y+2z=17 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x-y-z=2; \) \( 4x+y-z=-5; \) \( 6x-2y+2z=17 \)

Final Answer

\( x=1/2, y=-4, z=3 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x-y-z=2; \) \( 4x+y-z=-5; \) \( 6x-2y+2z=17 \)

Solution

1902 video

video by MIP4U

Final Answer

\( x=1/2, y=-4, z=3 \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x+2z=9; \) \( 2y+z=8; \) \( 4x-3y=-2 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x+2z=9; \) \( 2y+z=8; \) \( 4x-3y=-2 \)

Final Answer

\( x=1, y=2, z=4 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x+2z=9; \) \( 2y+z=8; \) \( 4x-3y=-2 \)

Solution

1903 video

video by PatrickJMT

Final Answer

\( x=1, y=2, z=4 \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x-y+z=4; \) \( 2x+y+z=7; \) \( -x-2y+2z=-1 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x-y+z=4; \) \( 2x+y+z=7; \) \( -x-2y+2z=-1 \)

Final Answer

\( x=3, y=0, z=1 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x-y+z=4; \) \( 2x+y+z=7; \) \( -x-2y+2z=-1 \)

Solution

1904 video

video by PatrickJMT

Final Answer

\( x=3, y=0, z=1 \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( -x+2y-3z=1; \) \( 2x+z=0; \) \( 3x-4y+4z=2 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( -x+2y-3z=1; \) \( 2x+z=0; \) \( 3x-4y+4z=2 \)

Final Answer

\( x=-4/5, y=-3/2, z=-8/5 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( -x+2y-3z=1; \) \( 2x+z=0; \) \( 3x-4y+4z=2 \)

Solution

1905 video

video by mattemath

Final Answer

\( x=-4/5, y=-3/2, z=-8/5 \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x - 3y + 3z = -6 \)
\( -2x + 4y + z = 3 \)
\( 3x - 5y +4z = -9 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x - 3y + 3z = -6 \)
\( -2x + 4y + z = 3 \)
\( 3x - 5y +4z = -9 \)

Final Answer

\( (0, 1, -1) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x - 3y + 3z = -6 \)
\( -2x + 4y + z = 3 \)
\( 3x - 5y +4z = -9 \)

Solution

This problem is solved in 2 consecutive videos.

2867 video

2867 video

Final Answer

\( (0, 1, -1) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x - 2y + z = 2 \)
\( 4x + 3y - 2z = 4 \)
\( 5x - 3y + 3z = 8 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x - 2y + z = 2 \)
\( 4x + 3y - 2z = 4 \)
\( 5x - 3y + 3z = 8 \)

Final Answer

\( (1, 2, 3) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x - 2y + z = 2 \)
\( 4x + 3y - 2z = 4 \)
\( 5x - 3y + 3z = 8 \)

Solution

2868 video

Final Answer

\( (1, 2, 3) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x + y - z = -2 \)
\( 2x - y + z = 0 \)
\( x - 2y + 3z = 1 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x + y - z = -2 \)
\( 2x - y + z = 0 \)
\( x - 2y + 3z = 1 \)

Final Answer

\( (-2/3, -7/3, -1) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( x + y - z = -2 \)
\( 2x - y + z = 0 \)
\( x - 2y + 3z = 1 \)

Solution

2870 video

Final Answer

\( (-2/3, -7/3, -1) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + y - z = 1 \)
\( 3x + 2y + 2z = 13 \)
\( 4x - 2y + 3z = 9 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + y - z = 1 \)
\( 3x + 2y + 2z = 13 \)
\( 4x - 2y + 3z = 9 \)

Final Answer

\( (1, 2, 3) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + y - z = 1 \)
\( 3x + 2y + 2z = 13 \)
\( 4x - 2y + 3z = 9 \)

Solution

2871 video

Final Answer

\( (1, 2, 3) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 3y - 5z = 1 \)
\( x + y - z = 2 \)
\( 2y + z = 8 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 3y - 5z = 1 \)
\( x + y - z = 2 \)
\( 2y + z = 8 \)

Final Answer

\( (1, 3, 2) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 2x + 3y - 5z = 1 \)
\( x + y - z = 2 \)
\( 2y + z = 8 \)

Solution

2872 video

Final Answer

\( (1, 3, 2) \)

close solution

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Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x + y - z = 0 \)
\( -2x + 5y + 4z = -1 \)
\( 3x + 2y + z = 1 \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x + y - z = 0 \)
\( -2x + 5y + 4z = -1 \)
\( 3x + 2y + z = 1 \)

Final Answer

\( (1/2, -2/3, 5/6) \)

Problem Statement

Unless otherwise instructed, solve this linear system using Cramer's Rule.
\( 3x + y - z = 0 \)
\( -2x + 5y + 4z = -1 \)
\( 3x + 2y + z = 1 \)

Solution

2873 video

Final Answer

\( (1/2, -2/3, 5/6) \)

close solution

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Really UNDERSTAND Precalculus

Topics You Need To Understand For This Page

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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