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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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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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