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

121.

If $|A| \neq 0$ and A is of order n, then adj(adj A) is equal to

${|A|}^{n}$
${|A|}^{2}$
${|A|}^{n - 1} . I$
${|A|}^{n - 2} . A$

122.

If A is any non-zero square matrix of order n, then A (adj A) is equal to

I
$|A|$ I
${|A|}^{n}$ I
None of these

123.
If A = $[\begin{matrix} 1 & \tan (x) \\ - \tan (x) & 1 \end{matrix}]$ , then the value of $|A^{T} A^{- 1}|$ is

$\cos (4 x)$

$\sec^{2} (x)$

$- \cos (4 x)$

1

$\sec^{2} (x)$

$We have, A = [\begin{matrix} 1 & \tan (x) \\ - \tan (x) & 1 \end{matrix}] Now, A^{T} = [\begin{matrix} 1 & - \tan (x) \\ \tan (x) & 1 \end{matrix}] adj (A) = [\begin{matrix} 1 & \tan (x) \\ - \tan (x) & 1 \end{matrix}] and A^{- 1} = \frac{1}{|A|} (adj A) = \frac{1}{\sec^{2} (x)} [\begin{matrix} 1 & - \tan (x) \\ \tan (x) & 1 \end{matrix}] Now, |A^{T} A^{- 1}| = |\frac{1}{\sec^{2} (x)} [\begin{matrix} 1 & - \tan (x) \\ \tan (x) & 1 \end{matrix}] [\begin{matrix} 1 & - \tan (x) \\ \tan (x) & 1 \end{matrix}]| = |\frac{1}{\sec^{2} (x)} [\begin{matrix} 1 - \tan^{2} (x) & - \tan (x) - \tan (x) \\ \tan (x) + \tan (x) & 1 - \tan^{2} (x) \end{matrix}]| = \frac{1}{\sec^{2} (x)} [\begin{matrix} 1 - \tan^{2} (x) & - 2 \tan (x) \\ 2 \tan (x) & 1 - \tan^{2} (x) \end{matrix}] = \frac{{(1 - \tan^{2} (x))}^{2} + 4 \tan^{2} (x)}{\sec^{2} (x)} = \frac{{(1 + \tan^{2} (x))}^{2}}{\sec^{2} (x)} = \sec^{2} (x)$

124.

If A = $[\begin{matrix} 4 & 11 \\ 2 & 6 \end{matrix}]$ , then A^-1 is equal to

$[\begin{matrix} - 1 & \frac{- 11}{2} \\ 3 & 2 \end{matrix}]$
$[\begin{matrix} 3 & \frac{- 11}{2} \\ - 1 & 2 \end{matrix}]$
$[\begin{matrix} 3 & 2 \\ - 1 & \frac{- 11}{2} \end{matrix}]$
None of these

125.

The area of the triangle formed by the points (a, b + c), (b, c + a) and (c, a + b) will be

abc
a² + b² + c²
ab + bc + ca
0

126.

$If A = [\begin{matrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{matrix}], then \det A is equal to$

127.

$If x^{2} + y^{2} + z^{2} \neq 0, x = cy + bz, y = az + cx and z = bx + ay, then a^{2} + b^{2} + c^{2} + 2 abc is equal to$

1
2
a + b + c
ab + bc + ca

128.

$adj [\begin{matrix} 1 & 0 & 2 \\ - 1 & 1 & - 2 \\ 0 & 2 & 1 \end{matrix}] = [\begin{matrix} 5 & a & - 2 \\ 1 & 1 & 0 \\ - 2 & - 2 & b \end{matrix}], then [a b] is equal to$