If- x - 4-1- bmatrix 2 - 1 - 0 1 - 0 - 2 M0 - 2 - 4lendbmatrix -eginbmatrix x-1 -end bmatrix -0- then - x -A- -2 -10B- -1 -6C- 8 -6D- 3 -6

The correct option is A [ x amp; 4 amp; 1 ][ 2x +4; x 2; 4 ] =0 +4x 8 4 =0 2x^2 +8x 12 =0 x^2 + 4x 6 =0 ⇒ x = 4±√(16+24)/2 = 4 ± 2√(10) ...

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Byju's Answer

Standard XII

Mathematics

Properties of Iota

If[ x ∼ 4-1] ...

Question

If

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

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Solution

The correct option is B

$[\begin{matrix} x & 4 & - 1 \end{matrix}] ⎡ ⎢ ⎣ \begin{matrix} 2 x + 4 x - 2 4 \end{matrix} ⎤ ⎥ ⎦ = 0$

$+ 4 x - 8 - 4 = 0 2 x^{2} + 8 x - 12 = 0 x^{2} + 4 x - 6 = 0 \Rightarrow x = \frac{- 4 \pm \sqrt{16 + 24}}{2} = \frac{- 4 \pm 2 \sqrt{10}}{2} = - 2 \pm \sqrt{10}$

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Similar questions

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If A = [\begin{matrix} 5 & 0 2 & 2 \end{matrix}], then find A \times I_{2 \times 2} .

Q. If

[\begin{matrix} X - Y & 2 & - 2 4 & X & 6 \end{matrix}] + [\begin{matrix} 3 & - 2 & 2 1 & 0 & - 1 \end{matrix}] = [\begin{matrix} 6 & 0 & 0 5 & 2 X + Y & 5 \end{matrix}]

,
then

Q. Find

x

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

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If [x 4−1]⎡⎢⎣210102024⎤⎥⎦⎡⎢⎣x4−1⎤⎥⎦=0,then x=

The correct option is B [x4−1]⎡⎢⎣2x+4x−24⎤⎥⎦=0 +4x−8−4=02x2+8x−12=0x2+4x−6=0⇒x=−4±√16+242=−4±2√102=−2±√10

If

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