# Thread: Matrix(Stucked at part iii)

1. ## Matrix(Stucked at part iii)

Question

i)Find the inverse of $\left[ \begin{array}{cc}
2 & k-2 \\
3 & k \end{array} \right]$

ii) Hence find the solutions of the simultaneous equations
$2x+(k-2)y=4$
$3x+ky=5$

iii) State the value of k for which there are no solutions to the simultaneous equations.

Attempted solutions

i) $
\left[ \begin{array}{cc}2 & {k-2} \\3 & k \end{array} \right]^{-1} = \frac{1}{2k-3(k-2)}\left[ \begin{array}{cc}k & -(k-2) \\-3 & 2 \end{array} \right]
$

$
= \frac{1}{2k-(3k-6)}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right]
$

$
= \frac{1}{-k+6}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right]
$

$
= \left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \right]
$

ii) $\left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \left[ \begin{array}{cc}
4 \\
5 \end{array} \right]\right]$

$
=\left[ \begin{array}{cc}\frac{4k}{-k+6} + \frac{5(-k+2)}{-k+6} \\ \frac{-12}{-k+6} + \frac{10}{-k+6} \end{array} \right]
$

$
= \left[ \begin{array}{cc}\frac{-k+10}{-k+6} \\\frac{-2}{-k+6} \end{array} \right]
$

$x=\frac{-k+10}{-k+6}$ $y=\frac{-2}{-k+6}$

Stucked @ iii

2. Originally Posted by Punch
Question

i)Find the inverse of $\left[ \begin{array}{cc}
2 & k-2 \\
3 & k \end{array} \right]$

ii) Hence find the solutions of the simultaneous equations
$2x+(k-2)y=4$
$3x+ky=5$

iii) State the value of k for which there are no solutions to the simultaneous equations.

Attempted solutions

i) $
\left[ \begin{array}{cc}2 & {k-2} \\3 & k \end{array} \right]^{-1} = \frac{1}{2k-3(k-2)}\left[ \begin{array}{cc}k & -(k-2) \\-3 & 2 \end{array} \right]
$

$
= \frac{1}{2k-(3k-6)}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right]
$

$
= \frac{1}{-k+6}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right]
$

$
= \left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \right]
$

ii) $\left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \left[ \begin{array}{cc}
4 \\
5 \end{array} \right]\right]$

$
=\left[ \begin{array}{cc}\frac{4k}{-k+6} + \frac{5(-k+2)}{-k+6} \\ \frac{-12}{-k+6} + \frac{10}{-k+6} \end{array} \right]
$

$
= \left[ \begin{array}{cc}\frac{-k+10}{-k+6} \\\frac{-2}{-k+6} \end{array} \right]
$

$x=\frac{-k+10}{-k+6}$ $y=\frac{-2}{-k+6}$

Stucked @ iii
by gausssian elimination, we have (in augmented form)

$\left[ \begin{array}{ccc}
2 & k-2 & 4\\
0 & \frac{6-k}{2} & 1\end{array} \right]$

this system of equations will have no solution if $\frac{6-k}{2} = 0$, that is for $k = 6$

3. Originally Posted by dedust
by gausssian elimination, we have (in augmented form)

$\left[ \begin{array}{ccc}
2 & k-2 & 4\\
0 & \frac{3-k}{2} & 1\end{array} \right]$

this system of equations will have no solution if $\frac{3-k}{2} = 0$, that is for $k = 3$
after going through a few examples in the book, maybe you are wrong and answer to part iii should be k=6

this is why
$x=\frac{-k+10}{-k+6}$ $y=\frac{-2}{-k+6}$
since denomnator = -k+6, if k=6, -6+6=0

4. Originally Posted by Punch
Question

i)Find the inverse of $\left[ \begin{array}{cc}
2 & k-2 \\
3 & k \end{array} \right]$

ii) Hence find the solutions of the simultaneous equations
$2x+(k-2)y=4$
$3x+ky=5$

iii) State the value of k for which there are no solutions to the simultaneous equations.

Attempted solutions

i) $
\left[ \begin{array}{cc}2 & {k-2} \\3 & k \end{array} \right]^{-1} = \frac{1}{2k-3(k-2)}\left[ \begin{array}{cc}k & -(k-2) \\-3 & 2 \end{array} \right]
$

$
= \frac{1}{2k-(3k-6)}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right]
$

$
= \frac{1}{-k+6}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right]
$

$
= \left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \right]
$

ii) $\left[ \begin{array}{cc}\frac{k}{-k+6} & \frac{-k+2}{-k+6} \\\frac{-3}{-k+6} & \frac{2}{-k+6} \end{array} \left[ \begin{array}{cc}
4 \\
5 \end{array} \right]\right]$

$
=\left[ \begin{array}{cc}\frac{4k}{-k+6} + \frac{5(-k+2)}{-k+6} \\ \frac{-12}{-k+6} + \frac{10}{-k+6} \end{array} \right]
$

$
= \left[ \begin{array}{cc}\frac{-k+10}{-k+6} \\\frac{-2}{-k+6} \end{array} \right]
$

$x=\frac{-k+10}{-k+6}$ $y=\frac{-2}{-k+6}$

Stucked @ iii
If $A\mathbf{x} = \mathbf{b}$

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

This will only happen if $A^{-1}$ exists.

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

$|A| = - k + 6$.

What value of $k$ will make $|A| = 0$?

5. Originally Posted by Punch
after going through a few examples in the book, maybe you are wrong and answer to part iii should be k=6

this is why
$x=\frac{-k+10}{-k+6}$ $y=\frac{-2}{-k+6}$
since denomnator = -k+6, if k=6, -6+6=0
oops,..wrong in gaussian elimination,..it should be $\left[ \begin{array}{ccc}
2 & k-2 & 4\\
0 & \frac{6-k}{2} & 1\end{array} \right]$

6. Originally Posted by dedust
oops,..wrong in gaussian elimination,..it should be $\left[ \begin{array}{ccc}
2 & k-2 & 4\\
0 & \frac{6-k}{2} & 1\end{array} \right]$
gaussian elimination is alien to me, the simplest method is probably to turn denominator equals to zero

7. Originally Posted by Punch
gaussian elimination is alien to me, the simplest method is probably to turn denominator equals to zero
The denominator you are referring to is called the "determinant".

The determinant of A is $|A|$.

See my post above...

8. Originally Posted by Prove It
The denominator you are referring to is called the "determinant".

The determinant of A is $|A|$.

See my post above...
Yay I got it now, if ad-bc=0 it means that equation has no solution !