# Finding values such that a matrix has a unique solution

I have a matrix where I'm supposed to find the values of a and b so that the matrix has a unique solution. I have looked through my textbook and there aren't any examples of how to go about this.

$$\left[ \begin{array}{@{}ccc|c@{}} 3&-1&2 & 1 \\ 0& \frac{5}{3}& \frac{14}{3} & \frac{10}{3} \\ 0&0& a-6 & b-4 \\ \end{array} \right]$$

How would I go about solving this problem? Thanks.

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Hint: For a unique solution, we can look for what in the determinant? Regards – Amzoti Feb 8 '13 at 21:31
I'n not sure, we started determinants yesterday. – user1709173 Feb 8 '13 at 21:32
This does not depend on $b$. This is equivalent to the invertibility of the $3\times 3$ matrix on the left. Whose determinant is? – 1015 Feb 8 '13 at 21:32
The determinant must be nonsingular. Can you do it now from the last problem? Regards – Amzoti Feb 8 '13 at 21:32
Consider the case where a=6 and everything else would be my hint. – JB King Feb 8 '13 at 21:38

A system of equations represented by a coefficient matrix $A$ has a unique solution if its determinant is NOT EQUAL to $0$: iff $\det A \neq 0$.

Note that you matrix, below, is an upper triangular matrix. The determinant of a triangular matrix is equal to the product of its diagonal entries:

$$\det A = \det \begin{bmatrix} 3 & -1 &2 \\ 0 & \small\frac53 & \small\frac{14}{3} \\ 0 & 0 & a-6 \\ \end{bmatrix} = 3 \cdot \frac53 \cdot (a - 6)\;=\;5(a-6)$$

$$\det A = 0 \iff a = 6$$

Hence, there exists a unique solution to the system of equations represented by the augmented coefficient matrix if and only if $\;\;\bf{a\neq 0}$:

$$A' = \left[ \begin{array}{@{}ccc|c@{}} 3&-1&2 & 1 \\ 0& \small\frac{5}{3}& \small\frac{14}{3} & \small\frac{10}{3} \\ 0&0& a-6 & b-4 \\ \end{array} \right]$$

Now we can express $x_3$ as a function of $a$ and $b$ for $a\neq 6:$ The solution, in this case, to $x_3$, is given by $x_3 = \large\frac{b-4}{a-6},\;$ provided $a \neq 6$. It turns out that the precise value of $b$ is irrelevant to the uniqueness of a solution ($b$ may take on any value):

So a unique solution exists to the system of equations given by the augmented coefficient matrix if and only if $$a\in (-\infty, 6)\cup (6, \infty)$$

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I'd say determinants are overkill here. I'd start about in the middle of your answer; you get a (unique) value of $x_3$ if and only if $a\ne6$, and then you get unique values of $x_2$ and then of $x_1$. – Gerry Myerson Feb 9 '13 at 5:38