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.

Let Q = $(\begin{matrix} \cos (\frac{π}{4}) & - \sin (\frac{π}{4}) \\ \sin (\frac{π}{4}) & \cos (\frac{π}{4}) \end{matrix}) and x = (\begin{matrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{matrix})$ , then Q³x is equal to

$(\begin{matrix} 0 \\ 1 \end{matrix})$
$(\begin{matrix} - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{matrix})$
$(\begin{matrix} - 1 \\ 0 \end{matrix})$
$(\begin{matrix} - \frac{1}{\sqrt{2}} \\ - \frac{1}{\sqrt{2}} \end{matrix})$

22.

If the matricx A = $[\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2 \end{matrix}]$ , then Aⁿ = $[\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a \end{matrix}], n \in N$ where

a = 2n, b = 2ⁿ
a = 2ⁿ, b = 2n
a = 2ⁿ, b = n2^{n - 1}
a = 2ⁿ, b = n2ⁿ

23.

If $f (x) = |\begin{matrix} 1 & x & x + 1 \\ 2 x & x (x - 1) & (x + 1) x \\ 3 x (x - 1) & x (x - 1) (x - 2) & (x + 1) x (x - 1) \end{matrix}|$

Then, f(100) is equal to

24.
For a matrix $A = (\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{matrix})$ , if U₁, U₂, and U₃. are $3 \times 1$ column matrices satisfying ${AU}_{1} = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), {AU}_{2} = (\begin{matrix} 2 \\ 3 \\ 0 \end{matrix}), {AU}_{3} = (\begin{matrix} 2 \\ 3 \\ 1 \end{matrix})$ and U is $3 \times 3$ matrix whose columns are U₁, U₂, and U₃. Then, sum of the elements of U^{- 1}

6

0

1

2/3

Let

$U_{i} = (\begin{matrix} a_{i} \\ b_{i} \\ c_{i} \end{matrix}), where i = 1, 2, 3 ∴ {AU}_{1} = (\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{matrix}) (\begin{matrix} a_{1} \\ b_{1} \\ c_{1} \end{matrix}) \Rightarrow (\begin{matrix} a_{1} \\ 2 a_{1} + b_{1} \\ 3 a_{1} + 2 b_{1} + c_{1} \end{matrix}) = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) [∵ {AU}_{1} = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})]$

$∴ a_{1} = 1, 2 a_{1} + b_{1} = 0 \Rightarrow 2 + b_{1} = 0 \Rightarrow b_{1} = - 2 and 3 a_{1} + 2 b_{1} + c_{1} = 0 \Rightarrow 3 + 2 (- 2) + c_{1} = 0 \Rightarrow 3 - 4 + c_{1} = 0 \Rightarrow c_{1} = 1$

Similarly, AU₂ = $(\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{matrix}) (\begin{matrix} a_{2} \\ b_{2} \\ c_{2} \end{matrix})$

$\Rightarrow (\begin{matrix} a_{2} \\ 2 a_{2} + b_{2} \\ 3 a_{2} + 2 b_{2} + c_{2} \end{matrix}) = (\begin{matrix} 2 \\ 3 \\ 0 \end{matrix}) [∵ {AU}_{2} = (\begin{matrix} 2 \\ 3 \\ 0 \end{matrix})]$

$∴ a_{2} = 2 and 2 a_{2} + b_{2} = 3 \Rightarrow 2 \times 2 + b_{2} = 3 \Rightarrow 4 + b_{2} = 3 \Rightarrow b_{2} = - 1 and 3 a_{2} + 2 b_{2} + c_{2} = 0 \Rightarrow 3 \times 2 + 2 (- 1) + c_{2} = 0 \Rightarrow 6 - 2 + c_{2} = 0 \Rightarrow c_{2} = - 4$

Now, AU₃ = $(\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{matrix}) (\begin{matrix} a_{3} \\ b_{3} \\ c_{3} \end{matrix})$

$\Rightarrow (\begin{matrix} a_{3} \\ 2 a_{3} + b_{3} \\ 3 a_{3} + 2 b_{3} + c_{3} \end{matrix}) = (\begin{matrix} 2 \\ 3 \\ 0 \end{matrix}) [∵ {AU}_{3} = (\begin{matrix} 2 \\ 3 \\ 0 \end{matrix})]$

$∴ a_{3} = 2 and 2 a_{3} + b_{3} = 3 \Rightarrow 2 \times 2 + b_{3} = 3 \Rightarrow 4 + b_{3} = 3 \Rightarrow b_{3} = - 1 and 3 a_{3} + 2 b_{3} + c_{3} = 1 \Rightarrow 3 \times 2 + 2 (- 1) + c_{3} = 1 \Rightarrow 6 - 2 + c_{3} = 1 \Rightarrow c_{2} = - 3$

$∴ U = (\begin{matrix} 1 & 2 & 2 \\ - 2 & - 1 & - 1 \\ 1 & - 4 & - 3 \end{matrix}) \Rightarrow |U| = 1 (3 - 4) - 2 (6 + 1) + 2 (8 + 1) = - 1 - 14 + 18 = 3 Also, adj (U) = {[\begin{matrix} - 1 & - 7 & 9 \\ - 2 & - 5 & 6 \\ 0 & - 3 & 3 \end{matrix}]}^{T} = [\begin{matrix} - 1 & - 2 & 0 \\ - 7 & - 5 & - 3 \\ 9 & 6 & 3 \end{matrix}] ∴ U^{- 1} = \frac{1}{|U|} adj (U)$

$\Rightarrow U^{- 1} = [\begin{matrix} - \frac{1}{3} & - \frac{2}{3} & 0 \\ - \frac{7}{3} & - \frac{5}{3} & - 1 \\ 3 & 2 & 1 \end{matrix}]$

Hence, the sum of all elements of u^- ¹ is 0

25.

Let I denote the 3 x 3 identity matrix and P be a matrix obtained by rearranging the columns of I. Then,

there are six distinct choices for P and det (P) = 1
there are six distinct choices for P and det (P) = $\pm 1$
there are more than one choices for P and some of them are not invertible
there are more than one choices for P and P^{- 1} = I in each choice

26.

If I = $(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) and P = (\begin{matrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 2 \end{matrix})$ . Then, the matrix P³ + 2P² is equal to