Find the value of x in 2 4 5 1 - 2x 4 6 x-

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Byju's Answer

Standard XII

Mathematics

Cross Product of Two Vectors

Find the valu...

Question

Find the value of $x$ in $∣ ∣ ∣ \begin{matrix} 2 & 4 5 & 1 \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} 2 x & 4 6 & x \end{matrix} ∣ ∣ ∣$ .

\pm \sqrt{3}

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Solution

The correct option is A $\pm \sqrt{3}$
Since both determinants are equal, thus,
$2 x^{2} - 24 = 2 - 20$
$2 x^{2} = 2 - 20 + 24$
$2 x^{2} = 6$
$x^{2} = 3$
$x = \pm \sqrt{3}$

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Similar questions

Q. Find the values of

x

, if
(i)

∣ ∣ ∣ \begin{matrix} 2 & 4 5 & 1 \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} 2 x & 4 6 & x \end{matrix} ∣ ∣ ∣

(ii)

∣ ∣ ∣ \begin{matrix} 2 & 3 4 & 5 \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} x & 3 2 x & 5 \end{matrix} ∣ ∣ ∣

Find values of x, if

(i) $|\begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix}| = |\begin{matrix} 2 x & 4 \\ 6 & x \end{matrix}|$

(ii) $|\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}| = |\begin{matrix} x & 3 \\ 2 x & 5 \end{matrix}|$

Q. Find the values of x, if

(i)

|\begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix}| = |\begin{matrix} 2 x & 4 \\ 6 & x \end{matrix}|

(ii)

|\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}| = |\begin{matrix} x & 3 \\ 2 x & 5 \end{matrix}|

(iii)

|\begin{matrix} 3 & x \\ x & 1 \end{matrix}| = |\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}|

(iv) If

|\begin{matrix} 3 x & 7 \\ 2 & 4 \end{matrix}| = 10

, find the value of x.

Q. Find values of x , if (i) 2 4 5 1 = 2 x 4 6 x (ii) 2 3 4 5 = x 3 2 x 5

Q. Find the values of x, if

(i)

|\begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix}| = |\begin{matrix} 2 x & 4 \\ 6 & x \end{matrix}|

(ii)

|\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}| = |\begin{matrix} x & 3 \\ 2 x & 5 \end{matrix}|

(iii)

|\begin{matrix} 3 & x \\ x & 1 \end{matrix}| = |\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}|

(iv) If

|\begin{matrix} 3 x & 7 \\ 2 & 4 \end{matrix}| = 10

, find the value of x.

(v)

|\begin{matrix} x + 1 & x - 1 \\ x - 3 & x + 2 \end{matrix}| = |\begin{matrix} 4 & - 1 \\ 1 & 3 \end{matrix}|

(vi)

|\begin{matrix} 2 x & 5 \\ 8 & x \end{matrix}| = |\begin{matrix} 6 & 5 \\ 8 & 3 \end{matrix}|

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Find the value of x in ∣∣∣2451∣∣∣=∣∣∣2x46x∣∣∣.

The correct option is A ±√3Since both determinants are equal, thus,2x2−24=2−202x2=2−20+242x2=6x2=3x=±√3

Find the value of $x$ in $∣ ∣ ∣ \begin{matrix} 2 & 4 5 & 1 \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} 2 x & 4 6 & x \end{matrix} ∣ ∣ ∣$ .

The correct option is A $\pm \sqrt{3}$
Since both determinants are equal, thus,
$2 x^{2} - 24 = 2 - 20$
$2 x^{2} = 2 - 20 + 24$
$2 x^{2} = 6$
$x^{2} = 3$
$x = \pm \sqrt{3}$