$∣ ∣ ∣ ∣ \begin{matrix} 24 & 25 & 26 25 & 26 & 27 26 & 27 & 27 \end{matrix} ∣ ∣ ∣ ∣ =$

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Solution

The correct option is C 1
$∣ ∣ ∣ ∣ \begin{matrix} 24 & 25 & 26 25 & 26 & 27 26 & 27 & 27 \end{matrix} ∣ ∣ ∣ ∣$

Apply operations $R_{3} \to R_{3} - R_{2}, R_{2} \to R_{2} - R_{1}$

$= ∣ ∣ ∣ ∣ \begin{matrix} 24 & 25 & 26 1 & 1 & 1 1 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣$

Expand along $R_{3}$

$= 1 (25 - 26) - 1 (24 - 26) = - 1 + 2 = 1$

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Similar questions

Q. Solve:

x^{2} - 22 x + 112 = 0

A two digit number is such that the product of its digits is 28. When 112 is subtracted from the number, the digits interchange their places. Find the second degree equation in x from which we can find tens digit.

Q. I.

\frac{4}{\sqrt{x}} + \frac{7}{\sqrt{x}} = \sqrt{x}

II.

y^{2} - \frac{(11)^{\frac{5}{2}}}{\sqrt{y}} = 0

{sin}^{- 1} \frac{5}{x} + {sin}^{- 1} \frac{12}{x} = \frac{π}{2}, (x > 0)

then

x =

Simplify (√3+√11)²+(√3-√11)²

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∣∣ ∣∣242526252627262727∣∣ ∣∣=

The correct option is C 1∣∣ ∣∣242526252627262727∣∣ ∣∣ Apply operations R3→R3−R2,R2→R2−R1=∣∣ ∣∣242526111110∣∣ ∣∣Expand along R3=1(25−26)−1(24−26)=−1+2=1

$∣ ∣ ∣ ∣ \begin{matrix} 24 & 25 & 26 25 & 26 & 27 26 & 27 & 27 \end{matrix} ∣ ∣ ∣ ∣ =$