how to find if the same matrix has one solution, infinitely many solutions and no solution

Question

The matrix is as follows: $$ \left[ \begin{array}{ccc|cc} a&0&b&2\\ a&a&4&4\\ 0&a&2&b \end{array} \right] $$

I need to find when this matrix will have one, infinitely many, and no solutions by expressing condition on a, b and c

I did put the matrix in reduced row echelon form as follow:

$$ \left[ \begin{array}{ccc|cc} 1&0&0&\frac{2-b}{a}\\ 0&1&0&\frac{b-2}{a}\\ 0&0&1&\frac{2-b}{a} \end{array} \right] $$

but now I'm stuck about how to express the conditions on a, b, and c

Answer 1

Reduction should have gotten you here:

$$ \left[ \begin{array}{ccc|cc} a&0&b&2\\ 0&a&4-b&2\\ 0&0&b-2&b-2 \end{array} \right] $$

This is far enough to analyze the different cases:

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• Unique:

We must have a leading term for each column. Therefore, $b \neq 2$ and $a \neq 0$.

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• No Solutions:

For this to occur we must have a row where the left side is all zeros, but the right side is not. This can occur if:

1. $a = b = 0$
2. $a = 0$ and $b= 4$

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• Infinitely Many Solutions:

This occurs when we have a free variable. This is achieved by getting a whole row (including the right side) or column to be zeros:

1. $a = 0$
2. $b = 2$