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Matrices

Multiple Choice Questions

271.

If A = $[\begin{matrix} 2 & 3 & 1 \\ 0 & - 1 & 5 \end{matrix}]$ and B = $[\begin{matrix} 1 & 2 & - 6 \\ 0 & - 1 & 3 \end{matrix}]$ , then 3A - 4B is equal to

$[\begin{matrix} 2 & 1 & 27 \\ 0 & 1 & 3 \end{matrix}]$
$[\begin{matrix} 2 & 1 & 27 \\ 0 & 1 & 3 \end{matrix}]$
$[\begin{matrix} 3 & 5 & - 5 \\ 0 & - 2 & 8 \end{matrix}]$
None of these

272.

If x is a complex root of the equation $|\begin{matrix} 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{matrix}| + |\begin{matrix} 1 - x & 1 & 1 \\ 1 & 1 - x & 1 \\ 1 & 1 & 1 - x \end{matrix}| = 0$ , then $x^{2005} + \frac{1}{x^{2005}}$ is equal to

273.

If A² - A + I = 0, then matrix A^-1 will be equal to

A + I
I - A
A
A - I

274.

For how many values of x in the interval [- 4, - 1] the matrix $[\begin{matrix} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{matrix}]$ is singular ?

275.

If A = $[\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix}]$ , then $A$ is a

singular
non-singular
symmetric
unit matrix

276.

Let $f (x) = |\begin{matrix} \sin (3 x) & 1 & 2 {(\cos (\frac{3 x}{2}) + \sin (\frac{3 x}{2}))}^{2} \\ \cos (3 x) & - 1 & 2 \cos^{2} (\frac{3 x}{2}) - \sin^{2} (\frac{3 x}{2}) \\ \tan (3 x) & 4 & 1 + 2 \tan (3 x) \end{matrix}|$ . Then, the value of f'(x) at x = (2n + 1) $π$ , n $\in I$ (the set of integers) is equal to

(- 1)ⁿ
(- 1)ⁿ^{+ 1}
3
9

277.

The value of $|\begin{matrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{matrix}|$ is

0
a + b + c
abc
1

278.

If 2x - $[\begin{matrix} 1 & 2 \\ 7 & 4 \end{matrix}] = [\begin{matrix} 3 & 2 \\ 0 & - 2 \end{matrix}]$ , then the value of x will be

$[\begin{matrix} 2 & 2 \\ 7 & 4 \end{matrix}]$
$[\begin{matrix} 1 & 2 \\ 7 / 2 & 2 \end{matrix}]$
$[\begin{matrix} 2 & 2 \\ 7 / 2 & 1 \end{matrix}]$
None of the above

279.
If $[\begin{matrix} 0 & 2 β & γ \\ α & β & - γ \\ α & - β & γ \end{matrix}]$ is orthagonal, then the values of $α, β and γ$ will be

$α = \pm \frac{1}{\sqrt{5}}, β = \pm \frac{1}{\sqrt{3}}, γ = \pm \frac{1}{\sqrt{2}}$

$α = \pm \frac{1}{\sqrt{6}}, β = \pm \frac{1}{\sqrt{7}}, γ = \pm \frac{1}{\sqrt{3}}$

$α = \pm \frac{1}{\sqrt{3}}, β = \pm \frac{1}{\sqrt{3}}, γ = \pm \frac{1}{\sqrt{3}}$

$α = \pm \frac{1}{\sqrt{2}}, β = \pm \frac{1}{\sqrt{6}}, γ = \pm \frac{1}{\sqrt{3}}$

$α = \pm \frac{1}{\sqrt{2}}, β = \pm \frac{1}{\sqrt{6}}, γ = \pm \frac{1}{\sqrt{3}}$

$Let A = [\begin{matrix} 0 & 2 β & γ \\ α & β & - γ \\ α & - β & γ \end{matrix}] Since, A is an orthogonal matrix ∴ AA' = I \Rightarrow [\begin{matrix} 0 & 2 β & γ \\ α & β & - γ \\ α & - β & γ \end{matrix}] [\begin{matrix} 0 & α & α \\ 2 α & β & - β \\ γ & - γ & γ \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 4 β^{2} + γ^{2} & 2 β^{2} - γ^{2} & - 2 β^{2} + γ^{2} \\ 2 β^{2} - γ^{2} & α^{2} + β^{2} + γ^{2} & α^{2} - β^{2} - γ^{2} \\ - 2 β^{2} + γ^{2} & α^{2} - β^{2} - γ^{2} & α^{2} + β^{2} + γ^{2} \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] ∴ 4 β^{2} + γ^{2} = 1, 2 β^{2} - γ^{2} = 0, α^{2} + β^{2} + γ^{2} = 1, α^{2} - β^{2} - γ^{2} = 0 [∵ comparing the corresponding entries] On solving these equations, we get α = \pm \frac{1}{\sqrt{2}}, β = \pm \frac{1}{\sqrt{6}}, γ = \pm \frac{1}{\sqrt{3}}$