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Determinants

Multiple Choice Questions

351.

If the system of linear equations
x + ky + 3z = 0
3x + ky – 2z = 0
2x + 4y – 3z = 0

has a non-zero solution (x,y,z), then xz/y² is equal to

352.

If $|\begin{matrix} x - 4 & 2 x & 2 x \\ 2 x & x - 4 & 2 x \\ 2 x & 2 x & x - 4 \end{matrix}| = (A + B x) (x - A)^{2}$ , then the ordered pair (A,B) is equal to

(4,5)
(-4,-5)
(-4,3)
(-4,5)

353.

Let A = $(\begin{matrix} x + 2 & 3 x \\ 3 & x + 2 \end{matrix})$ , B = $(\begin{matrix} x & 0 \\ 5 & x + 2 \end{matrix})$ , Then all solutions of equation det(AB) = 0 is

1, - 1, 0, 2
1, 4, 0, - 2
1, - 1, 4, 3
- 1, 4, 0, 3

354.

The value of det A, where

A = $A = (\begin{matrix} 1 & \cos (θ) & 0 \\ - \cos (θ) & 1 & \cos (θ) \\ - 1 & - \cos (θ) & 1 \end{matrix})$ , lies

in the closed interval [1, 2]
in the closed interval [0, 1]
in the open interval (0, 1)
in the open interval (1, 2)

355.

The points (- a, - b), (a, b), (0, 0) and (a, ab), $a \neq 0, b \neq 0$ are always

collinear
vertices of a parallelogram
vertices of a rectangle
lie on a circle

356.

The number of distinct real roots of

$|\begin{matrix} \sin (x) & \cos (x) & \cos (x) \\ \cos (x) & \sin (x) & \cos (x) \\ \cos (x) & \cos (x) & \sin (x) \end{matrix}| = 0 in the interval - \frac{π}{4} \leq x \leq \frac{π}{4}$ is

357.

$If f : [0, π / 2] \to R is defined as f (θ) = |\begin{matrix} 1 & \tan (θ) & 1 \\ - \tan (θ) & 1 & \tan (θ) \\ - 1 & - \tan (θ) & 1 \end{matrix}| Then, the range of f is$

$(2, \infty)$
$(- \infty, - 2]$
$[2, \infty)$
$(- \infty, 2]$

358.

If a is an imaginary cube root of unity, then the value of the determinant

$|\begin{matrix} 1 + w & w^{2} & - w \\ 1 + w^{2} & w & - w^{2} \\ w + w^{2} & w & - w^{2} \end{matrix}|$ is

- 2w
- 3w²
- 1
0

359.
Let $n \geq 2$ be an integer,
$A = [\begin{matrix} \cos (2 π / 3) & \sin (2 π / n) & 0 \\ - \sin (2 π / n) & \cos (2 π / n) & 0 \\ 0 & 0 & 1 \end{matrix}]$ and I is the identity matrix of order 3. Then,

Aⁿ = I and A^{n - 1} $\neq$ I

A^m $\neq$ I for any positive integer m

A is not invertible

A^m = 0 for a positive integer m

Aⁿ = I and A^{n - 1} $\neq$ I

$Given, A = [\begin{matrix} \cos (2 π / n) & \sin (2 π / n) & 0 \\ - \sin (2 π / n) & \cos (2 π / n) & 0 \\ 0 & 0 & 1 \end{matrix}] Now, A \times A = [\begin{matrix} \cos (2 π / n) & \sin (2 π / n) & 0 \\ - \sin (2 π / n) & \cos (2 π / n) & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} \cos (2 π / 3) & \sin (2 π / n) & 0 \\ - \sin (2 π / n) & \cos (2 π / n) & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \cos^{2} (2 π / n) - \sin^{2} (2 π / n) & \cos (2 π / n) \sin (2 π / n) + \cos (2 π / n) \sin (2 π / n) & 0 \\ - \sin (2 π / n) \cos (2 π / n) - \sin (2 π / n) \cos (2 π / n) & - \sin^{2} (2 π / n) + \cos^{2} (2 π / n) & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \cos (2 \times 2 π / n) & \sin (2 \times 2 π / n) & 0 \\ \sin (2 \times 2 π / n) & \cos (2 \times 2 π / n) & 0 \\ 0 & 0 & 1 \end{matrix}]$

Similarly,

$A^{n} = [\begin{matrix} \cos (2^{n - 1} \times \frac{2 π}{n}) & \sin (2^{n - 1} \times \frac{2 π}{n 0}) & 0 \\ - \sin (2^{n - 1} \times \frac{2 π}{n}) & \cos (2^{n - 1} \times \frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}]$

$As [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ = I

and A^{n - 1}= $[\begin{matrix} \cos (2^{n - 2} \times \frac{2 π}{n}) & \sin (2^{n - 2} \times \frac{2 π}{n 0}) & 0 \\ - \sin (2^{n - 2} \times \frac{2 π}{n}) & \cos (2^{n - 2} \times \frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}]$