Question

pls tell me..
if the value of the determinant is 26 root 6,, then find x
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Answer 1

$\[\begin {bmatrix} </p><p>3\sqrt{6} & -4\sqrt{6} \\</p><p>5\sqrt{3} & x</p><p>\end { bmatrix} </p><p>\]$

$= 26\sqrt{6}$

$\implies 3\sqrt{6}\times x - 5\sqrt{3} \times (-4\sqrt{6}) = 26 \sqrt{6}$

$\implies 3\sqrt{6}x + 20\sqrt{6} \times \sqrt{3} = 26 \sqrt{6}$

/* Divide each term by √6, we get */

$\implies 3x + 20\sqrt{3} = 26$

$\implies 3x = - 20\sqrt{3} + 26$

$\implies 3x = - 20\times 1.732+ 26$

$\implies 3x = - 34.64 + 26$

$\implies 3x = - 8.64$

$\implies x = \frac{- 8.64}{3}$

$\implies x = - 2.88$

Therefore.,

$\red {Value \:of \:x } \green {= -2.88}$

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