# Determinants - Class 12 - Mathematics

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## Extra Questions - Determinants | NCERT | Mathematics | Class 12

The value of

$$ \begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} + \begin{vmatrix} 0 & 3 \\ 4 & -2 \end{vmatrix} $$

is

A) -9

B) 8

C) -10

D) 5

To find the value of the given expression, we need to calculate the determinants of each of the 2x2 matrices and then sum them.

The determinant of a 2x2 matrix

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix}$$

is calculated by the formula \( ad - bc \).

1. For the first matrix:

$$ \begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} = (2)(1) - (0)(-1) = 2 - 0 = 2 $$

2. For the second matrix:

$$ \begin{vmatrix} 0 & 3 \\ 4 & -2 \end{vmatrix} = (0)(-2) - (4)(3) = 0 - 12 = -12 $$

Now, summing these results:

$$ 2 + (-12) = -10 $$

Therefore, the value of the given expression is \(-10\), and the correct answer is:

C) -10

If $a, \beta, \gamma$ are real numbers, then $\Delta=\left|\begin{array}{ccc}1 & \cos (\beta-\alpha) & \cos (\gamma-\alpha) \ \cos (\alpha-\beta) & 1 & \cos (\gamma-\beta) \ \cos (\alpha-\gamma) & \cos (\beta-\gamma) & 1\end{array}\right|$

A) -1

B) $\cos \alpha \cos \beta \cos \gamma$

C) $\cos \alpha + \cos \beta + \cos \gamma$

D) None of these

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How many combinations of two-digit numbers with 8 can be made from the following numbers?

$$ 8, 5, 2, 1, 7, 6 $$

A) 9

B) 10

C) 11

D) 12

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Find the values of $x$ for which the inequality $\frac{8x^{2} + 16x - 51}{(2x - 3)(x + 4)} > 3$ holds.

The correct answer is:

A. $x \geq 4$

B. $ -4 \leq x \leq -3$

C $\frac{3}{2} < x < \frac{5}{2}$ or $x < \frac{5}{2}$ or $-3 < x < \frac{3}{2}$

D. $ x < -4$ $x > \frac{5}{2}$ or $ x < \frac{3}{2}$