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Determinants

Multiple Choice Questions

431.

The value of $|\begin{matrix} 1 & 1 & 1 \\ bc & ca & ab \\ b + c & c + a & a + b \end{matrix}|$ is

1
0
(a - b)(b - c)(c - a)
(a + b)(b + c)(c + a)

432.

The value of $|\begin{matrix} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{matrix}|$ is :

$441 \times 446 \times 4510$
0
- 1
1

433.

Inverse of the matrix $[\begin{matrix} 1 & - 2 \\ 3 & 4 \end{matrix}]$ is

$\frac{1}{10} [\begin{matrix} 1 & - 2 \\ 3 & 4 \end{matrix}]$
$\frac{1}{10} [\begin{matrix} 4 & 2 \\ - 3 & 1 \end{matrix}]$
$[\begin{matrix} 4 & 2 \\ - 3 & 1 \end{matrix}]$
$\frac{1}{10} [\begin{matrix} 4 & - 2 \\ - 3 & 1 \end{matrix}]$

434.

$|\begin{matrix} b^{2} c^{2} & bc & b + c \\ c^{2} a^{2} & ca & c + a \\ a^{2} b^{2} & a b & a + b \end{matrix}|$ is equal to

$\frac{1}{abc} (ab + bc + ca)$
ab + bc + ca
0
a + b + c

435.

If w = $- \frac{1}{2} + \frac{\sqrt{3} i}{2}$ , the value of the determinant $|\begin{matrix} 1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w \end{matrix}|$ is

zero
3
- 1
1

436.

If f(x) = $|\begin{matrix} \sin (x) & \cos (x) & \tan (x) \\ x^{3} & x^{2} & x \\ 2 x & 1 & 1 \end{matrix}|$ , then $\lim_{x \to 0} \frac{f (x)}{x^{2}}$ is

- 1
3
1
zero

437.
If for a triangle ABC, $|\begin{matrix} 1 & a & b \\ 1 & b & a \\ 1 & c & c \end{matrix}| = 0$ , then the value of $\sin^{2} (A) + \sin^{2} (B) + \sin^{2} (C)$ is

$\frac{4}{9}$

$\frac{9}{4}$

1

$\frac{3 \sqrt{3}}{2}$

$\frac{9}{4}$

$Given, |\begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{matrix}| = 0 1 (c^{2} - ab) - a (c - a) + b (b - c) = 0 c^{2} - ab - ca + a^{2} + b^{2} - bc = 0 2 a^{2} + 2 b^{2} + 2 c^{2} - 2 ab - 2 bc - 2 ca = 0 \Rightarrow a^{2} + b^{2} - 2 ab + b^{2} + c^{2} - 2 bc + c^{2} + a^{2} - 2 ca = 0 \Rightarrow {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} = 0 \Rightarrow a = b = c \Rightarrow A = B = C = 60 ° ∴ \sin^{2} (A) + \sin^{2} (B) + \sin^{2} (C) = {(\frac{\sqrt{3}}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2} = \frac{9}{4}$

438.

The inverse of the matrix $[\begin{matrix} 5 & - 2 \\ 3 & 1 \end{matrix}]$ is

$\frac{1}{11} [\begin{matrix} 1 & 2 \\ - 3 & 5 \end{matrix}]$
$[\begin{matrix} 1 & 2 \\ - 3 & 5 \end{matrix}]$
$\frac{1}{13} [\begin{matrix} - 2 & 5 \\ 1 & 3 \end{matrix}]$
$[\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$

439.

If A = $[\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 1 & 1 & 1 \end{matrix}], 10 B = [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & α \\ 1 & - 2 & 3 \end{matrix}]$ and B is the inverse of A, then the value of $α$ is