So im supposed to decide for what h and k this matrix has no solultions, infinite solutions and a unique solution
$$\left[
\begin{array}{cc|c}
1&h&1\\
3&3&k\\
\end{array}
\right]$$
By gaussian elimination i get
$$\left[
\begin{array}{cc|c}
1&h&1\\
0&3-3h&k-3\\
\end{array}
\right] $$
So i should get no solutions when $$3-3h = 0 \: ,k-3 \neq 0, \: and \: 3-3h\neq k-3 $$
And infinitely many solutions when $$3-3h = 0\text{ and }k-3= 0$$
And a unique solution would take the form
$$\left[
\begin{array}{cc|c}
1&0&\frac{3-kh}{3-3h}\\
0&1&\frac{k-3}{3-3h}\\
\end{array}
\right] $$
when $$\: 3-3h\neq 0$$
Are there any cases i have failed to take into consideration?
Best Answer
You'll get a unique solution when the rows of the coefficient matrix are linearly independent, for then the coefficient matrix is invertible. this happens if $h \neq 1.$ If $h=1,$ the rows of the coefficient matrix are dependent, and you will get no or infinitely many solutions depending on the value of $k.$ If $k=3,$ the rows of the augmented matrix are dependent, and you get infinitely many solutions. If $k \neq 3,$ there are no solutions.
Yogi Berra once said "You can see a lot by looking." In this situation, the form of the augmented matrix says a lot. You don't really need Gauss-Jordan to answer your question (modulo a few mental calculations and miscellaneous facts).
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