# Thread: Matrix, find variable

1. ## Matrix, find variable

For some reason im just not getting this, if anyone could help me solve this problem, explaining exactly how they did it, i would realy appreciate it.

7x + 6y - 3z = -7
7x - 7y + kz = 4
7x - 20y - 3z = 10

For the system to be consistent k must not = ?

2. Originally Posted by Juggalomike
For some reason im just not getting this, if anyone could help me solve this problem, explaining exactly how they did it, i would realy appreciate it.

7x + 6y - 3z = -7
7x - 7y + kz = 4
7x - 20y - 3z = 10

For the system to be consistent k must not = ?
Write the system in matrix form...

$\left[\begin{matrix} 7 & 6 & -3 \\ 7 & -7 & k \\ 7 & -20 & -3\end{matrix}\right] \left[\begin{matrix} x \\ y \\ z \end{matrix}\right] = \left[\begin{matrix} -7 \\ 4 \\ 10 \end{matrix}\right]$.

I'm going to rewrite this equation as

$A\mathbf{x} = \mathbf{b}$.

Using some matrix algebra

$A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}$

$I\mathbf{x} = A^{-1}\mathbf{b}$

$\mathbf{x} = A^{-1}\mathbf{b}$.

So does it make sense that to have a solution, you need to be able to premultiply both sides by the inverse matrix of $A$. Therefore this inverse matrix has to exist.

$A^{-1}$ exists when $|A| \neq 0$.

So evaluate $|A|$, and solve for $k$ when $|A| = 0$

This will tell you the value(s) of $k$ that you can not have.

3. I tried solving the sum ..Can i know if k should not be = -3 is correct..

4. Let's see.

For a $3 \times 3$ matrix $A = \left[\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right]$

then $|A| = aei + bfg + cdh - ceg - afh - bdi$.

So in your case

$|A| = 7(-7)(-3) + 6(k)(7) + (-3)(7)(-20) - (-3)(-7)(7) - 7(k)(-20) - 6(7)(-3)$

$= 147 + 42k + 420 - 147 + 140k + 126$

$= 546 + 182k$

Solving $|A| = 0$ for $k$ gives

$546 + 182k = 0$

$182k = -546$

$k = -3$.

So yes, you can not let $k = -3$