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![\left[ \begin{array}{c}<br />
x\\<br />
y\\<br />
z\end{array}\right]=\left[ \begin{array}{ccc}<br />
1 & 2 & m \\<br />
2 & 3 & -2 \\<br />
m & 1 & m^2 \end{array} \right]^{-1} \left[ \begin{array}{c}<br />
0\\<br />
m\\<br />
3\end{array}\right]](http://latex.codecogs.com/png.latex? \left[ \begin{array}{c}<br />
x\\<br />
y\\<br />
z\end{array}\right]=\left[ \begin{array}{ccc}<br />
1 & 2 & m \\<br />
2 & 3 & -2 \\<br />
m & 1 & m^2 \end{array} \right]^{-1} \left[ \begin{array}{c}<br />
0\\<br />
m\\<br />
3\end{array}\right])
If you don't want to use matrices, you can just solve it as you would in basic algebra: Multiplying the first equation by m and subtracting from the third equation elimnates both x and z: (m- m)x+ (1- 2m)y+ (m^2- m^2)z= 3- 0 so (1- 2m)y= 3. IfOriginally Posted by princessamme@hotmail.com
,
. Put that value of y into the first and second equations to get two equations for x and z. if m= 1/2, 1- 2m= 0 so there is no solution.
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