Let a, b, c be positive real numbers. The fol | vedclass

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(B) Let $\frac{x^2}{a^2} = X, \frac{y^2}{b^2} = Y$ and $\frac{z^2}{c^2} = Z$.The system of equations becomes:$X + Y - Z = 1$$X - Y + Z = 1$$-X + Y + Z

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(B) Let $\frac{x^2}{a^2} = X, \frac{y^2}{b^2} = Y$ and $\frac{z^2}{c^2} = Z$.The system of equations becomes:$X + Y - Z = 1$$X - Y + Z = 1$$-X + Y + Z = 1$The coefficient matrix $A$ is given by:$A = \begin{bmatrix} 1 & 1 & -1 \ 1 & -1 & 1 \ -1 & 1 & 1 \end{bmatrix}$Calculating the determinant $|A|$:$|A| = 1((-1)(1) - (1)(1)) - 1((1)(1) - (1)(-1)) + (-1)((1)(1) - (-1)(-1))$$|A| = 1(-2) - 1(2) - 1(0) = -2 - 2 = -4$Since $|A| = -4 
eq 0$,the system has a unique solution for $(X, Y, Z)$.Solving the system:Adding the equations: $(X+Y-Z) + (X-Y+Z) = 1+1 \implies 2X = 2 \implies X = 1$.Similarly,$Y = 1$ and $Z = 1$.Thus,$\frac{x^2}{a^2} = 1, \frac{y^2}{b^2} = 1, \frac{z^2}{c^2} = 1$,which gives $x = \pm a, y = \pm b, z = \pm c$. Since there are finite values for $x, y, z$,the system has a unique solution for $(X, Y, Z)$ and finitely many solutions for $(x, y, z)$. However,in the context of the matrix system for $(X, Y, Z)$,it is a unique solution.

Let $a, b, c$ be positive real numbers. The following system of equations in $x, y, z$:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$
$\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
$-\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
has:

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Let $a, b, c$ be positive real numbers. The following system of equations in $x, y, z$:$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$$\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$-\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$has: