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.

Matrices

Long Answer Type

201.

Using elementary row operations, find the inverse of the following matrix:

$[\begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix}]$

Short Answer Type

202.

For a 2 x 2 matrix, A = [ a_ij ] whose elements are given by a_ij = $\frac{i}{j}$ , write the value of _a12 .

203.

For what value of x, the matrix $[\begin{matrix} 5 - x & x + 1 \\ 2 & 4 \end{matrix}]$ is singular?

204.

Write A^-1 for A = $[\begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix}]$

Long Answer Type

205.

Using elementary transformations, find the inverse of the matrx

$(\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0 \end{matrix})$

Short Answer Type

206.

If $(\begin{matrix} 2 & 3 \\ 5 & 7 \end{matrix}) (\begin{matrix} 1 & - 3 \\ - 2 & 4 \end{matrix}) = (\begin{matrix} - 4 & 6 \\ - 9 & x \end{matrix})$ , write the value of x.

207.

Simplify: $\cos θ [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] + \sin θ [\begin{matrix} \sin θ & - \cos θ \\ \cos θ & \sin θ \end{matrix}]$

Long Answer Type

208.

Using elementary operations, find the inverse of the following matrix:

$(\begin{matrix} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix})$

Multiple Choice Questions

209.
If P = $open square brackets table row 1 straight alpha 3 row 1 3 3 row 2 4 4 end table close square brackets$ is the adjoint of a 3 x3 matrix A and |A| = 4, then α is equal to

4

11

5

5

Given,
$straight P space equals space open square brackets table row 1 straight alpha 3 row 1 3 3 row 2 4 4 end table close square brackets$
|P| = 1(12-12)-α (4-6) +3(4-6)
= 2α -6.
∵ P =adj(A)
∴ |P| = |adj A | = |A|3-1 = |A|2 = 16
[∵ |adj A| = |A|n-1 order is 3 x3
∴ 2α -6 = 16
2α = 22
α = 11

547 Views

210.

Let A = $open square brackets table row 1 0 0 row 2 1 0 row 3 2 1 end table close square brackets$ . If u₁ and u₂ are column matrices such that Au1 = $open square brackets table row 1 row 0 row 0 end table close square brackets$ and Au2 = $open square brackets table row 0 row 1 row 0 end table close square brackets$ , then u₁ +u₂ is equal to

$open square brackets table row cell negative 1 end cell row 1 row 0 end table close square brackets$
$open square brackets table row cell negative 1 end cell row cell space space space 1 end cell row cell negative 1 end cell end table close square brackets$
$open square brackets table row cell negative 1 end cell row cell negative 1 end cell row 0 end table close square brackets$
$open square brackets table row cell negative 1 end cell row cell negative 1 end cell row 0 end table close square brackets$