# 11.8 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.

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