Write the system of linear equations represented by the augmen...

Question

Write the system of linear equations represented by the augmented matrix. Use x, y, and z, or, if necessary, w, x, y, and z, for the variables.

$\begin{bmatrix}1 & 1 & 4 & 1 & \vert & 3 \-1 & 1 & -1 & 0 & \vert & 7 \2 & 0 & 0 & 5 & \vert & 11 \0 & 0 & 12 & 4 & \vert & 5\end{bmatrix}$

Accepted Answer

Hey everyone here we are asked to determine the system of linear equations associated to the given augmented matrix. So first our given matrix has four rows and the first row has the values of 00 negative 35 and six. Our second row is 104, negative five. And to our third row has negative 6708 and 10. And finally our last row has 2009 and four. So with an augmented matrix we need to recall that it has this vertical line here which represents an equal sign in the equation. So the column to the right of this line are all constant values and then to the left, all of these terms are coefficients of the variables. So we are told to use the variables W, X, Y and Z. So from left to right we have coefficients of W, then X, Y and Z. So now with this information we can start writing out our four equations. So starting with the first row we have zero W plus zero x minus three, Y plus five, Z is equal to six. And now we can move on to the second row where we will get our second equation. So we have one W plus zero X plus four, Y minus five, Z is equal to two. And now looking at our third world for our third equation, we get negative six, W plus seven, X plus zero Y plus eight, Z is equal to 10. And for our final equation from the fourth row we have two, W plus zero X plus zero, Y plus nine, Z is equal to four. And now all that is left for us to do is simplify these equations. We can get get rid of our terms with zero coefficients because those are simply equal to zero. And we can also simplify the terms with a coefficient of one because a one is already implied when the variable is on its own. So again, starting with our first equation, we have the simplified form of negative three, Y plus five, Z is equal to six because we get rid of the first two terms the W and X. Next four. Our second equation here, we can get rid of this one coefficient for W. And we can also get rid of this zero X. So we are left with W plus four, Y minus five, Z is equal to two. And then moving on to our third equation, we just have to get rid of this zero Y. And we are left with negative at six. W plus seven X plus eight Z is equal to 10. And finally moving on to our final equation, we have to just get rid of the zero X. And zero Y. So we are left with two, W plus nine, Z is equal to four. So now we are left with our system of equations here. We have four different equations. Leaving us with answer choice. A Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Key Concepts

Augmented Matrix Representation

System of Linear Equations

Variables and Their Correspondence

Watch next