Solve using the Gaussian elimination method 4x&-3Y&=1 \\ 2x&+2Y&=4


Solve using the Gaussian elimination method

{eq}\begin{Bmatrix} 4x&-3Y&=1 \\ 2x&+2Y&=4 \end{Bmatrix} {/eq}

Gaussian Elimination

When solving a system of equations in augmented matrix form, we do so by performing row operations in a method called Gaussian elimination. The three types of row operations that we can perform are: swapping two rows, multiplying a row by a constant, and adding multiples of rows to other rows. The goal is to construct the identity matrix on the left side of this augmented matrix. The solutions to this system are then in the right column.

Answer and Explanation:

Let's perform row operations in order to put this matrix into row reduced echelon form. We want to have a square identity matrix on the left, and then we want to have the solutions to our system in the right. As each column represents a variable, we can remove the x and y, as well as the equals sign, from this matrix.

{eq}\begin{align*} \begin{bmatrix} 4&-3&1 \\ 2&+&4 \end{bmatrix} & R_2 = \frac{1}{2}R_2\\ \begin{bmatrix} 4&-3&1 \\ 1&1&2 \end{bmatrix} & R_1 \leftrightarrow R_2\\ \begin{bmatrix} 1&1&2\\ 4&-3&1 \end{bmatrix} &R_2 = R_2 - 4R_1\\ \begin{bmatrix} 1&1&2\\ 0&-7&-7 \end{bmatrix} &R_2 = -\frac{1}{7}R_2\\ \begin{bmatrix} 1&1&2\\ 0&1&1 \end{bmatrix} &R_1 = R_1 - R_2\\ \begin{bmatrix} 1&0&1\\ 0&1&1 \end{bmatrix} \end{align*} {/eq}

Thus, the solution to this system is that both x and y equal 1.

Learn more about this topic:

How to Solve Linear Systems Using Gaussian Elimination

from Algebra II Textbook

Chapter 10 / Lesson 6

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