Solve each system of equations using matrices. Use Gaussian...

Solve each system of equations using matrices.
Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
$$\left\{\begin{array}{cc}
3 a-b-4 c= & 3 \\
2 a-b+2 c= & -8 \\
a+2 b-3 c= & 9
\end{array}\right.$$

Key Concepts

Back-Substitution

A technique applied after Gaussian elimination in which the equations, now in triangular form, are solved backwards from the last equation to the first. Each solved variable is substituted into the previous equations to help find the remaining values.

Gauss-Jordan Elimination

An extension of Gaussian elimination where further row operations are performed to convert the augmented matrix into reduced row echelon form. This fully simplified form directly reveals the solution for each variable without requiring back-substitution.

Gaussian Elimination

A method that uses elementary row operations to transform an augmented matrix into an upper triangular or echelon form. This simplified form makes it easier to solve for the variables through sequential elimination.

Matrix Representation

A tool for organizing the coefficients of a system of linear equations into a structured format, facilitating the application of systematic methods to solve the system.

System of Linear Equations

A collection of two or more linear equations involving the same set of variables. Solving these systems means finding values for the variables that satisfy all equations simultaneously.

Augmented Matrix

A form of matrix that includes both the coefficients of the variables and the constants from the right-hand side of the equations. It combines all the information of the system into a single matrix, allowing for streamlined operations.

Transcript

00:01 Given the system, 3a minus b minus 4c equals 3, 2a minus b plus 2c equals negative 8, and a plus 2b minus 3c equals 9, now we want to get this to an augmented matrix.

00:16 First with the coefficients, 3, negative 1, negative 4, 2, negative 1, 2, 2, negative 3, negative 3, negative 3, 9.

00:26 We can do this right away because it's already organized, a's, b, c's equals constants.

00:32 I'm going to go about this with the gaussian elimination with back substitution.

00:39 So i'm going to get a diagonal of ones.

00:41 One way to do this is we already have a 1 in the first position.

00:47 We can interchange rows 1 in 3.

00:52 So row 2 will stay the same.

00:55 And there are many ways to do this.

00:57 This is not the only way.

00:59 Our new row 1 will be 1 -2 -9, and our new row 3 -9 now we need zeros below this one.

01:09 We could take 2 times row 1 and subtract row 2 and put that in row 2.

01:16 So row 1 and 3 unchanged.

01:23 And our new row 2 would be 2 times 1 is 2 minus 2 is 0.

01:30 2 times 2 is 4 minus negative 1 is 5.

01:34 2 times negative 3 is negative 6 minus 2 is negative 8 and 2 times 9 is 18.

01:40 Minus negative 8 is 26.

01:45 Now we need to 0 in row 3 column 1.

01:51 So we could take row 3 and subtract 3 times row 1 for a new row 3...

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