Find x- if -x-5-1- beginbmatrix 180822110 - 2 - 11 -2 - 0 - 3lendbmatrix bmatrix x 1411111endbmatrix -0 -

Here, [x 5 1] [ 1 amp; 0 amp; 2; 0 amp; 2 amp; 1; 2 amp; 0 amp; 3 ][ x; 4; 1; ]=0. ⇒ [x 5 1] ( [ 1 amp; 0 amp; 2; 0 amp; 2 amp; 1 ...

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Standard XII

Mathematics

Properties of Determinants

Find x, if [x...

Question

Find x, if [x -5 -1] $⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x 41 \end{matrix} ⎤ ⎥ ⎦ = 0$ .

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Solution

Here, $[x - 5 - 1] ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x 41 \end{matrix} ⎤ ⎥ ⎦ = 0.$
$\Rightarrow [x - 5 - 1] ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x 41 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠ = 0 \Rightarrow [x - 5 - 1] ⎡ ⎢ ⎣ \begin{matrix} x + 0 + 2 0 + 8 + 1 2 x + 0 + 3 \end{matrix} ⎤ ⎥ ⎦ = 0 \Rightarrow [x (x + 2) + (- 5) (9) + (- 1) (2 x + 3)] = 0 \Rightarrow x^{2} - 48 = 0 \Rightarrow x^{2} = 48 \Rightarrow x = \pm \sqrt{48} = \pm 4 \sqrt{3}$

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Similar questions

Q. If

[\begin{matrix} 2 & 3 & 4 \end{matrix}] ⎡ ⎢ ⎣ \begin{matrix} 1 & x & 3 2 & 4 & 5 3 & 2 & x \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x 20 \end{matrix} ⎤ ⎥ ⎦ = 0

, then

x =

For what values of x, [1 2 1] $⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 0 2 & 0 & 1 1 & 0 & 2 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 02 x \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠ = 0$ ?

If $A = [\begin{matrix} 5 & 4 3 & - 1 \end{matrix}], B = [\begin{matrix} 2 & 1 0 & 4 \end{matrix}]$ and $C = [\begin{matrix} - 3 & 2 1 & 0 \end{matrix}]$ then find A + B - C

Q. If

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[125] {[\begin{matrix} 0 & \frac{1}{2} - 1 & \frac{1}{2} \end{matrix}]}^{5} {[\begin{matrix} 1 & - 1 2 & 0 \end{matrix}]}^{10} {[\begin{matrix} 0 & \frac{1}{2} - 1 & \frac{1}{2} \end{matrix}]}^{5} [\begin{matrix} x^{2} - 5 x + 20 x + 2 \end{matrix}] = [40]

, then

(1 - α) (1 - β) - 50

is equal to ___

$[x 4 - 1] ⎡ ⎢ ⎣ \begin{matrix} 2 & 1 & 0 1 & 0 & 2 0 & 2 & 4 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x 4 - 1 \end{matrix} ⎤ ⎥ ⎦ = 0, t h e n x =$

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Find x, if [x -5 -1] ⎡⎢⎣102021203⎤⎥⎦⎡⎢⎣x41⎤⎥⎦=0.

Here, [x−5−1]⎡⎢⎣102021203⎤⎥⎦⎡⎢⎣x41⎤⎥⎦=0. ⇒[x−5−1]⎛⎜⎝⎡⎢⎣102021203⎤⎥⎦⎡⎢⎣x41⎤⎥⎦⎞⎟⎠=0⇒[x −5 −1]⎡⎢⎣x+0+20+8+12x+0+3⎤⎥⎦=0⇒[x(x+2)+(−5)(9)+(−1)(2x+3)]=0⇒x2−48=0⇒x2=48⇒x=±√48=±4√3

Find x, if [x -5 -1] $⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x 41 \end{matrix} ⎤ ⎥ ⎦ = 0$ .