Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

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

$Given, |\begin{matrix} b^{2} c^{2} & bc & b + c \\ c^{2} a^{2} & ca & c + a \\ a^{2} b^{2} & abc & a + b \end{matrix}| Multiplying R_{1}, R_{2}, R_{3} by a, b, c respectively and divide the whole by abc \frac{1}{abc} |\begin{matrix} {ab}^{2} c^{2} & abc & a (b + c) \\ {bc}^{2} a^{2} & bca & b (c + a) \\ {ca}^{2} b^{2} & abc & c (a + b) \end{matrix}| Taking common abc from C_{1} and C_{2}, \frac{(abc) (abc)}{abc} |\begin{matrix} bc & 1 & ab + ac \\ ca & 1 & bc + ab \\ ab & 1 & ca + bc \end{matrix}| Now, C_{1} \Rightarrow C_{1} + C_{3} abc |\begin{matrix} ab + bc + ca & 1 & ab + ac \\ ca + bc + ab & 1 & bc + ab \\ ab + bc + ca & 1 & ca + bc \end{matrix}| Taking common ab + bc + ca from C_{1}, (abc) (ab + bc + ca) |\begin{matrix} 1 & 1 & ab + ac \\ 1 & 1 & bc + ab \\ 1 & 1 & ca + ab \end{matrix}| (abc) (ab + bc + ca) . (0) As C_{1} = C_{2}, so value of △ is zero . ∴ |\begin{matrix} b^{2} c^{2} & bc & b + c \\ c^{2} a^{2} & ca & c + a \\ a^{2} b^{2} & abc & a + b \end{matrix}| = 0$

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

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