# Use Cramer's rule to solve the given system: x-z=4 2y-z=11 3x+4y=19

## Question:

Use Cramer's rule to solve the given system:

{eq}x-z=4 \\ 2y-z=11 \\ 3x+4y=19 {/eq}

## Cramer's Rule:

The Cramer rule is a technique used to solve systems of equations. It is based on the calculation of determinants for both the coefficient matrix and the replacement of each column by the vector of independent terms. The method allows to obtain the system solution in a quick way.

## Answer and Explanation:

The solution to the system using the Cramer's is:

{eq}\displaystyle x=\frac{\Delta_x}{\Delta}\\ \displaystyle y=\frac{\Delta_y}{\Delta}\\ \displaystyle z=\frac{\Delta_z}{\Delta}\\ {/eq}

Calculus of {eq}\Delta {/eq}, {eq}\Delta_x {/eq}, {eq}\Delta_y {/eq} and {eq}\Delta_z {/eq}.

{eq}\displaystyle \Delta=\begin{bmatrix}{1}&{0}&{-1}\\{0}&{2}&{-1}\\{3}&{4}&{0}\end{bmatrix}=0+0+0-[-6+(-4)+0]\\ \displaystyle \Delta=6+4\\ \displaystyle \Delta=10\\ \displaystyle \Delta_x=\begin{bmatrix}{4}&{0}&{-1}\\{11}&{2}&{-1}\\{19}&{4}&{0}\end{bmatrix}=0+0+(-44)-[-38+(-16)+0]\\ \displaystyle \Delta_x=-44+38+16\\ \displaystyle \Delta_x=-44+54\\ \displaystyle \Delta_x=10\\ \displaystyle \Delta_y=\begin{bmatrix}{1}&{4}&{-1}\\{0}&{11}&{-1}\\{3}&{19}&{0}\end{bmatrix}=0+(-12)+0+-[-33+(-19)+0]\\ \displaystyle \Delta_y=-12+33+19\\ \displaystyle \Delta_y=-12+52\\ \displaystyle \Delta_y=40\\ \displaystyle \Delta_z=\begin{bmatrix}{1}&{0}&{4}\\{0}&{2}&{11}\\{3}&{4}&{19}\end{bmatrix}=38+0+0-[24+44+0]\\ \displaystyle \Delta_z=38-24-44\\ \displaystyle \Delta_z=38-68\\ \displaystyle \Delta_z=-30\\ {/eq}

Calculating the values of the variables:

{eq}\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{10}{10}=1\\ \displaystyle y=\frac{\Delta_y}{\Delta}=\frac{40}{10}=4\\ \displaystyle z=\frac{\Delta_z}{\Delta}=\frac{-30}{10}=-3\\ {/eq}

The solution to the system is:

{eq}\displaystyle x=1\\ \displaystyle y=4\\ \displaystyle z=-3 {/eq}

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from Algebra II Textbook

Chapter 10 / Lesson 12