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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Determinants
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Let $a,b,c$ be the real numbers. Then the following system of equations in $x,y$ and $z$: $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$, $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$, $\large\frac{-x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ has

$\begin{array}{1 1}(a)\;No \;solution&(b)\;unique\;solution\\(c)\;infinitely\;many\;solutions&(d)\;finitely\;many\;solutions\end{array}$

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Let $\large\frac{x^2}{a^2}=$$X$
$\large\frac{y^2}{b^2}$$=Y$
$\large\frac{z^2}{c^2}$$=Z$
Then the given system of equations becomes
$X+Y-Z=1$
$X-Y+Z=1$
$-X+Y+Z=1$
This is the new system of equations.
For New system we have,
$D=\begin{vmatrix}1&1&-1\\1&-1&1\\-1&1&1\end{vmatrix}$
$\;\;=1(-1-1)-1(1+1)-1(1-1)$
$\;\;\;=-4\neq 0$
New system of equations has unique solution.
$D_1=\begin{vmatrix}1 &1&-1\\1&-1&1\\1&1&1\end{vmatrix}$
$\;\;\;=1(-1-1)-1(1-1)-1(1+1)$
$\;\;\;=-4$
$D_2=\begin{vmatrix}1 &1&-1\\1&-1&1\\1&1&1\end{vmatrix}$
$\;\;\;=1(-1-1)-1(1-1)-1(1+1)$
$\;\;\;=-4$
$D_3=\begin{vmatrix}1 &1&-1\\1&-1&1\\1&1&1\end{vmatrix}$
$\;\;\;=1(-1-1)-1(1+1)-1(1-1)$
$\;\;\;=-4$
Now $X=\large\frac{D_1}{D}=\frac{-4}{-4}=$$1$
$Y=\large\frac{D_2}{D}=\frac{-4}{-4}=$$1$
$Z=\large\frac{D_3}{D}=\frac{-4}{-4}=$$1$
$\Rightarrow x=\pm a,y=\pm b,z=\pm c$
Hence (b) is the correct answer.
answered Nov 21, 2013 by sreemathi.v
edited Mar 19, 2014 by sharmaaparna1
 

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