For values of a and b for which the system is consistent, find its general solution and express your answer in the usual column form.

Question

Consider the system of equations with the following augmented matrix, in which a and b are real numbers.

(1 0 2 a^2 3| b)

(0 a b 0 1| a)

(0 0 0 a 0| b^2-1)

(b) For values of a and b for which the system is consistent, find its general solution and express your answer in the usual column form. Note that a and b are arbitrary here. Either of them could be zero, positive, or negative, and the system will likely have different solutions depending on their values. As a result, it’s best to consider cases in which the values of a and b are in different ranges (for example, when a = 0 or a ̸= 0, and perhaps different possibilities for b within those). Describe your cases carefully and organize your solution neatly.

Answer 1

In this problem there are two different cases that we must examine: firstly Case 1: $a \neq 0$ and then Case 2: a = 0

We start with the augmented matrix below

\[

\left[

\begin{array}{ccccc|c}

1 & 0 & 2 & a^2 & 3 & b \\

0 & a & b & 0 & 1 & a \\

0 & 0 & 0 & a & 0 & b^2 - 1

\end{array}

\right]

\]

Case 1: a $\neq$

Here we end up with the x_4 = \frac{b^2 - 1}{a} from the third row. Knowing this we can sub this into the first and second rows and obtain the following equations

1st row:

x_1 +2x_3 + a^2 (\frac{b^2-1}[a}) + 3x_5 = b ->

x_1 = b - 2x_3 - 3x_5 - a (b^2 - 1)

2nd row:

a x_2 + b x_3 + x_5 = a ->

x_2 = 1 - \frac{b}{a} x_3 - \frac{1}{a} x_5

The ultimate result is a column vector where

(x_1, x_2, x_3, x_4, x_5) = (b - a(b^2 - 1), 1, 0, \frac{b^2 - 1}{a}, 0)

We now move on to the second case

Case 2: a = 0

As we get the following matrix with a = 0

\[

\left[

\begin{array}{ccccc|c}

1 & 0 & 2 & 0 & 3 & b \\

0 & 0 & b & 0 & 1 & 0 \\

0 & 0 & 0 & 0 & 0 & b^2 - 1

\end{array}

\right]

\]

We get the two following conditions that must be met for the solutions

b^2 - 1 = 0

bx_3 + x_5 = 0

We note that b = \pm 1

We now evaluate the two subases

Subcase 2a: a = 0 and b\neq \pm 1

In this subcase we end up with no solution

Subcase 2b: a = 0 b = \pm 1

We end up with the following as the case

(x_1, x_2, x_3, x_4, x_5) = (b, 0, 0, 0, 0)