SATHEE: Matrices and Determinants 4 Question 32

Matrices and Determinants 4 Question 32

33. Given, $x = c y + b z, y = a z + c x, z = b x + a y$ , where $x, y$ , $z$ are not all zero, prove that $a^{2} + b^{2} + c^{2} + 2 a b = 1$ .

$(1978, 2 M)$

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Solution:

Given systems of equations can be rewritten as $- x + c y + b y = 0, c x - y + a z = 0$ and $b x + a y - z = 0$

Above system of equations are homogeneous equation. Since, $x, y$ and $z$ are not all zero, so it has non-trivial solution.

Therefore, the coefficient of determinant must be zero.

$∴ | \begin{array}{ccc} - 1 & c & b \\ c & - 1 & a \\ b & a & - 1 \end{array} | = 0$

$\Rightarrow - 1 (1 - a^{2}) - c (- c - a b) + b (c a + b) = 0$

$\Rightarrow a^{2} + b^{2} + c^{2} + 2 a b c - 1 = 0$

$\Rightarrow a^{2} + b^{2} + c^{2} + 2 a b c = 1$

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Matrices and Determinants 4 Question 32

33. Given, x=cy+bz,y=az+cx,z=bx+ay, where x,y, z are not all zero, prove that a2+b2+c2+2ab=1.

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33. Given, $x = c y + b z, y = a z + c x, z = b x + a y$ , where $x, y$ , $z$ are not all zero, prove that $a^{2} + b^{2} + c^{2} + 2 a b = 1$ .