# Matrix(Stucked at part iii)

• Jan 16th 2010, 08:24 PM
Punch
Matrix(Stucked at part iii)
Question

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

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

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

Attempted solutions

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

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

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

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

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

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

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

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

Stucked @ iii
• Jan 16th 2010, 09:07 PM
dedust
Quote:

Originally Posted by Punch
Question

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

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

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

Attempted solutions

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

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

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

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

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

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

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

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

Stucked @ iii

by gausssian elimination, we have (in augmented form)

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

this system of equations will have no solution if $\displaystyle \frac{6-k}{2} = 0$, that is for $\displaystyle k = 6$
• Jan 16th 2010, 09:13 PM
Punch
Quote:

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

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

this system of equations will have no solution if $\displaystyle \frac{3-k}{2} = 0$, that is for $\displaystyle k = 3$

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
$\displaystyle x=\frac{-k+10}{-k+6}$$\displaystyle y=\frac{-2}{-k+6} since denomnator = -k+6, if k=6, -6+6=0 • Jan 16th 2010, 09:29 PM Prove It Quote: Originally Posted by Punch Question i)Find the inverse of \displaystyle \left[ \begin{array}{cc} 2 & k-2 \\ 3 & k \end{array} \right] ii) Hence find the solutions of the simultaneous equations \displaystyle 2x+(k-2)y=4 \displaystyle 3x+ky=5 Giving your answers in terms of K iii) State the value of k for which there are no solutions to the simultaneous equations. Attempted solutions i) \displaystyle \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] \displaystyle = \frac{1}{2k-(3k-6)}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right] \displaystyle = \frac{1}{-k+6}\left[ \begin{array}{cc}k & -k+2 \\-3 & 2 \end{array} \right] \displaystyle = \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) \displaystyle \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] \displaystyle =\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] \displaystyle = \left[ \begin{array}{cc}\frac{-k+10}{-k+6} \\\frac{-2}{-k+6} \end{array} \right] \displaystyle x=\frac{-k+10}{-k+6} \displaystyle y=\frac{-2}{-k+6} Stucked @ iii If \displaystyle A\mathbf{x} = \mathbf{b} Then \displaystyle \mathbf{x} = A^{-1}\mathbf{b}. This will only happen if \displaystyle A^{-1} exists. \displaystyle A^{-1} exists if \displaystyle |A| \neq 0. So in your case \displaystyle |A| = - k + 6. What value of \displaystyle k will make \displaystyle |A| = 0? • Jan 16th 2010, 09:35 PM dedust Quote: 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 \displaystyle x=\frac{-k+10}{-k+6}$$\displaystyle y=\frac{-2}{-k+6}$
since denomnator = -k+6, if k=6, -6+6=0

oops,..wrong in gaussian elimination,..it should be $\displaystyle \left[ \begin{array}{ccc} 2 & k-2 & 4\\ 0 & \frac{6-k}{2} & 1\end{array} \right]$
• Jan 16th 2010, 10:07 PM
Punch
Quote:

Originally Posted by dedust
oops,..wrong in gaussian elimination,..it should be $\displaystyle \left[ \begin{array}{ccc} 2 & k-2 & 4\\ 0 & \frac{6-k}{2} & 1\end{array} \right]$

gaussian elimination is alien to me, the simplest method is probably to turn denominator equals to zero
• Jan 16th 2010, 10:19 PM
Prove It
Quote:

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 $\displaystyle |A|$.

See my post above...
• Jan 16th 2010, 10:24 PM
Punch
Quote:

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

The determinant of A is $\displaystyle |A|$.

See my post above...

Yay I got it now, if ad-bc=0 it means that equation has no solution !