# 9.8 Solving systems with cramer's rule  (Page 2/11)

 Page 2 / 11

We eliminate one variable using row operations and solve for the other. Say that we wish to solve for $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If equation (2) is multiplied by the opposite of the coefficient of $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ in equation (1), equation (1) is multiplied by the coefficient of $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ in equation (2), and we add the two equations, the variable $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ will be eliminated.

Now, solve for $\text{\hspace{0.17em}}x.$

Similarly, to solve for $\text{\hspace{0.17em}}y,$ we will eliminate $\text{\hspace{0.17em}}x.$

Solving for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ gives

Notice that the denominator for both $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is the determinant of the coefficient matrix.

We can use these formulas to solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y,\text{\hspace{0.17em}}$ but Cramer’s Rule also introduces new notation:

• $\text{\hspace{0.17em}}\text{\hspace{0.17em}}D:$ determinant of the coefficient matrix
• ${D}_{x}:$ determinant of the numerator in the solution of $x$
$x=\frac{{D}_{x}}{D}$
• ${D}_{y}:$ determinant of the numerator in the solution of $\text{\hspace{0.17em}}y$
$y=\frac{{D}_{y}}{D}$

The key to Cramer’s Rule is replacing the variable column of interest with the constant column and calculating the determinants. We can then express $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a quotient of two determinants.

## Cramer’s rule for 2×2 systems

Cramer’s Rule    is a method that uses determinants to solve systems of equations that have the same number of equations as variables.

Consider a system of two linear equations in two variables.

$\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}$

The solution using Cramer’s Rule is given as

$x=\frac{{D}_{x}}{D}=\frac{|\begin{array}{cc}{c}_{1}& {b}_{1}\\ {c}_{2}& {b}_{2}\end{array}|}{|\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}|},\text{\hspace{0.17em}}\text{\hspace{0.17em}}D\ne 0;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{​}\text{​}\text{\hspace{0.17em}}y=\frac{{D}_{y}}{D}=\frac{|\begin{array}{cc}{a}_{1}& {c}_{1}\\ {a}_{2}& {c}_{2}\end{array}|}{|\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}|},\text{\hspace{0.17em}}\text{\hspace{0.17em}}D\ne 0.$

If we are solving for $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ column is replaced with the constant column. If we are solving for $\text{\hspace{0.17em}}y,\text{\hspace{0.17em}}$ the $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ column is replaced with the constant column.

## Using cramer’s rule to solve a 2 × 2 system

Solve the following system using Cramer’s Rule.

Solve for $\text{\hspace{0.17em}}x.$

$x=\frac{{D}_{x}}{D}=\frac{|\begin{array}{rr}\hfill 15& \hfill 3\\ \hfill 13& \hfill -3\end{array}|}{|\begin{array}{rr}\hfill 12& \hfill 3\\ \hfill 2& \hfill -3\end{array}|}=\frac{-45-39}{-36-6}=\frac{-84}{-42}=2$

Solve for $\text{\hspace{0.17em}}y.$

$y=\frac{{D}_{y}}{D}=\frac{|\begin{array}{rr}\hfill 12& \hfill 15\\ \hfill 2& \hfill 13\end{array}|}{|\begin{array}{rr}\hfill 12& \hfill 3\\ \hfill 2& \hfill -3\end{array}|}=\frac{156-30}{-36-6}=-\frac{126}{42}=-3$

The solution is $\text{\hspace{0.17em}}\left(2,-3\right).$

Use Cramer’s Rule to solve the 2 × 2 system of equations.

$\left(3,-7\right)$

## Evaluating the determinant of a 3 × 3 matrix

Finding the determinant of a 2×2 matrix is straightforward, but finding the determinant of a 3×3 matrix is more complicated. One method is to augment the 3×3 matrix with a repetition of the first two columns, giving a 3×5 matrix. Then we calculate the sum of the products of entries down each of the three diagonals (upper left to lower right), and subtract the products of entries up each of the three diagonals (lower left to upper right). This is more easily understood with a visual and an example.

Find the determinant    of the 3×3 matrix.

$A=\left[\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}\end{array}\right]$
1. Augment $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ with the first two columns.
$\mathrm{det}\left(A\right)=|\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}{a}_{1}\\ {a}_{2}\\ {a}_{3}\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}|$
2. From upper left to lower right: Multiply the entries down the first diagonal. Add the result to the product of entries down the second diagonal. Add this result to the product of the entries down the third diagonal.
3. From lower left to upper right: Subtract the product of entries up the first diagonal. From this result subtract the product of entries up the second diagonal. From this result, subtract the product of entries up the third diagonal.

can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
Where do the rays point?
Spiro
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas