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Solve the linear system using the elimination method. $\left\{\begin{aligned}-3 x-2 y+2 z & =-2 \\ -x-3 y+z & =-3 \\ x-2 y+z & =-2\end{aligned}\right.$

   Solve the linear system using the elimination method.
$\left\{\begin{aligned}-3 x-2 y+2 z & =-2 \\ -x-3 y+z & =-3 \\ x-2 y+z & =-2\end{aligned}\right.$
Introduction to Linear Algebra with Applications
Introduction to Linear Algebra with Applications
James DeFranza,… 1st Edition
Chapter 1, Problem 11 ↓
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Solve the linear system using the elimination method. $\left\{\begin{aligned}-3 x-2 y+2 z & =-2 \\ -x-3 y+z & =-3 \\ x-2 y+z & =-2\end{aligned}\right.$
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Transcript

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00:01 For this question, we are asked to solve a system of three equations, the three equations i have written to the left here, using elimination.
00:10 I'm also going to use our knowledge of matrices and my calculator to make solving this a little bit easier.
00:16 The first thing i'm going to do is create a three -by -three matrix based on the values in the equations.
00:23 So in the first equation, i have 1x plus 2y minus 1 z, and the second equation, i have negative 2, x, negative 1, y, and 3z.
00:36 And then the third equation, i have 5x, 3y, and negative 2z.
00:42 So you can see how this 3x -by -3 matrix that i've just written represents those three equations.
00:49 Now, if i multiply this by a matrix of what we want to find, the values of x, y, and c, z, that solve the system.
01:01 Of equations, this will be equal to the values that all three of our equations are equal to 3, negative 5, and 2.
01:16 To make sure this is correct, or to prove that this is correct to you, this is a 3 by 3 matrix, and our unknowns is a 3 by 1 matrix, the number of rows in the first matrix, sorry, the number of columns in the first matrix that we are multiplying, and the number of rows in the second matrix that we are multiplying are equal, so we're able to multiply these two things, and the result will be a 3 by 1 matrix.
01:45 So far, this statement is true.
01:48 Now, if i were to go and find the product that we have here, first i'd be looking for the 3, which is in row 1, column 1.
01:57 So i would look at row 1 of this matrix and column 1 of this matrix.
02:04 I would get 1 times x plus 2 times y minus z equals 3.
02:10 This is exactly the equation that we have in our system of equations that we're trying to solve.
02:16 And if we go through and look for the other two values, it'll be the same thing.
02:19 So this negative 5 would be equal to this second row times this column.
02:25 So you would get negative 2x minus y plus 3z equals negative 5 and so on...
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