In each part find a linear system in the unknowns x1, x2,...

In each part find a linear system in the unknowns $x_{1}, x_{2}, x_{3}, \ldots,$ that corresponds to the given augmented matrix. $$\text { (a) }\left[\begin{array}{ccccc} 0 & 3 & -1 & -1 & -1 \\ 5 & 2 & 0 & -3 & -6 \end{array}\right]$$ $$\text { (b) }\left[\begin{array}{rrrrr} 3 & 0 & 1 & -4 & 3 \\ -4 & 0 & 4 & 1 & -3 \\ -1 & 3 & 0 & -2 & -9 \\ 0 & 0 & 0 & -1 & -2 \end{array}\right]$$

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00:01 In each part we have two pots a and b.

00:06 We want to find a linear system in the unknowns x1, x2, etc., that corresponds to the given augmented matrix.

00:16 So we have two augmented matrices.

00:19 We are going to write the corresponding linear system using the variables x1, x2, 3, etc.

00:26 So let's see the first one here and we know that the last column corresponds to the right -hand side of the linear system.

00:37 So, we must have 1, 2, 3, and 4 variables from x1 to x4.

00:44 The linear system will be in a matrix vector form like this.

00:49 0, 3, negative 1, negative 1, 5, 2, 0, 0 ,0, 0 ,000, that is the same augmentic matrix without the last column, which is the right -hand side vector.

01:06 And now we have 1, 2, 3, and 4 variables.

01:10 So we get x1, x2, x3, x4.

01:17 And that got to be equal to the right -hand side vector, negative 1, negative 6, the last column of the augmented matrix.

01:26 And that now corresponds to the linear system with two equations.

01:34 0 times x1, we get 0, then 3.

01:38 3 times x2 minus x3 minus x4.

01:44 That is, we do the inner product of the first row of the matrix and the vector, the unknown vector.

01:53 We get this.

01:55 And that got to be equal to the first component of the vector on the right -hand side, negative 1...

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