Answer to Question #288311 in Linear Algebra for Sabelo Xulu

Question

Q1.

Consider the sets

A = {x $\in$ Z : x = 2 (y - 2) for some

y $\in$ Z} and B = {x $\in$ Z : x = 2z for some z $\in$ Z}. Then

(i) A and B are equal,

(ii) A {\displaystyle \subsetneq } B,

(iii) B {\displaystyle \subsetneq } A

(1) (i) only

(2) (ii) only.

(3) (iii) only

(4) None of the given answers is true.

Q2.

Consider a function f: R → R^+ defined as f(x) =e^x: Then the image of S ={x ∈ R: 0 ≤ x^2 - 9} is

(i) (-∞; e^-3]U[e^3;∞);

(ii) [e^-3; e3^];

(iii)[e^3;∞)

(4) None of the given answers is true.

Q3.

The two distinct square roots of i is

(1) √2 + √2i and -√2 - √2i

(2) 2√2 + 2√2i and - 2√2 - 2√2i

(3)√2/2 + √2/2(I) and - √2/2 - √2/2(I)

(4) None of the given answers is true.

Answer 1

$Q1.\\ A=\{x\in \Z:x=2(y-2), y\in \Z\}\\ B=\{x\in \Z: x=2y, y\in Z\}\\ \text{If, $k\in A$ then $k=2(k'-2)$ for some $k'\in \Z$}\\ k'\in \Z\Rightarrow k''=k'-2\in \Z\\ \text{Now, } k=2k'', k''\in \Z\Rightarrow k\in B\Rightarrow A\subseteq B\\ \text{Simillarly,}\\ \text{If, $l\in B$ then $l=2l'$ for some $l'\in \Z$}\\ l'\in \Z\Rightarrow l''=l'+2\in \Z\\ \text{Now, } l=2(l''-2), l''\in \Z\Rightarrow l\in A\Rightarrow B\subseteq A\\ \text{Hence, } A=B$

So option (1) is correct.

$Q2.\\ f: R → R^+ \ni f(x)=e^x\\$

Now $S =\{x ∈ R: 0 ≤ x^2 - 9\}$

$\therefore x^2-9\geq 0\Rightarrow (x+3)(x-3)\geq 0\\ \Rightarrow x \in (-\infin, -3] \cup [3,\infin)\\ \therefore S=\{x| x \in (-\infin, -3] \cup [3,\infin)\}$

So, Image of S is $(0,\frac{1}{e^3}]\cup [e^3, \infin)$

Therefore option (1) is correct option.

$Q3.\\ |i|=1\\ \therefore \text{We can write $i$ as }e^\frac{i\pi}{2}. \\\text{Square roots of $e^\frac{i\pi}{2}$ are $\pm e^\frac{i\pi}{4}$}\\ \text{Now, } \\e^\frac{i\pi}{4}=cos(\pi/4)+i~sin(\pi/4)\\=\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2}=\dfrac{\sqrt2}{2}+\dfrac{i\sqrt2}{2}$

So square roots of i are $\dfrac{\sqrt2}{2}+\dfrac{i\sqrt2}{2}, -\dfrac{\sqrt2}{2}-\dfrac{i\sqrt2}{2}$

$\therefore$ Option (3) is the correct option.