Answer to Question #98546 in Algebra for Shaheli Chattopadhyay

Question

3. (a) If y = a + 13 is a solution of the cubic equation 3
y3= 4y + B = 0 such that 34 = — A, find the
quadratic equation having a 3and 133as
roots.
(b) Without Expanding the determinant 2
1 2 3
4 5 6
7 8 9
find its value.
4. (a) Prove that A u (A n B) = A for any two
subsets A and B of a set U.
(b) Check if the following system of equations
can be solved by cramer's rule. If it can be
solved, then apply the rule for solving the
system ; otherwise solve it by the Gaussian
elimination method.
x+y+z=6
x + 2y + 3z = 10
x -F 2y + 3z = 6.
5. (a) Show that 22n — 3n —1 is divisible by 9 using
the principle of mathematical induction.
(b) Find the greatest value of (3 — x)5(2 + x)6,
for — 2 < x < 3.

Answer 1

3.(a) Find the quadratic equation having $a=3$ and $a=133$ as roots.  

Given $a_1=3, a_2=133.$ Complete the quadratic equation

$$C(a-3)(a-133)=0, C\not=0$$

$$Ca^2 -136Ca+399C=0, C\not=0$$

(b)

$$\begin{pmatrix} 1& 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$$

Subtract row1 multiplied by 4 from row2 

$$\begin{pmatrix} 1& 2 & 3 \\ 0 & -3 & -6 \\ 7 & 8 & 9 \end{pmatrix}$$

Subtract row1 multiplied by 7 from row3 

$$\begin{pmatrix} 1& 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{pmatrix}$$

Subtract row2 multiplied by 2 from row3

$$\begin{pmatrix} 1& 2 & 3 \\ 0 & -3 & -6 \\ 0 & 0 & 0 \end{pmatrix}$$

The determinant

$$\begin{vmatrix} 1& 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}=\begin{vmatrix} 1& 2 & 3 \\ 0 & -3 & -6 \\ 0 & 0 & 0 \end{vmatrix}=0$$

4. (a)

$$A\cap B=\{{x:x\in A\ and\ x\in B}\}$$

$$A\cup(A\cap B)=\{{x:x\in A\ or ( \ x\in A\ and\ x\in B)}\}$$

Hence

$$A\cup(A\cap B)=\{{x:x\in A}\}=A$$

(b)

$$x+y+z=6$$ $$x+2y+3z=10$$ $$x-2y+3z=6$$

$$\varDelta=\begin{vmatrix} 1& 1 & 1 \\ 1 & 2& 3 \\ 1 & -2& 3 \end{vmatrix}=$$ $$=1\begin{vmatrix} 2 & 3 \\ -2 & 3 \end{vmatrix}-1\begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix}+1\begin{vmatrix} 1 & 2 \\ 1 & -2 \end{vmatrix}=$$ $$=(2(3)-3(-2))-(1(3)-3(1))+(1(-2)-2(1))=$$ $$=12-0-4=8\not=0$$

$$\varDelta_x=\begin{vmatrix} 6 & 1 & 1 \\ 10 & 2& 3 \\ 6 & -2 & 3 \end{vmatrix}=$$ $$=6\begin{vmatrix} 2 & 3 \\ -2 & 3 \end{vmatrix}-1\begin{vmatrix} 10 & 3 \\ 6 & 3 \end{vmatrix}+1\begin{vmatrix} 10 & 2 \\ 6 & -2 \end{vmatrix}=$$ $$=6(2(3)-3(-2))-(10(3)-3(6))+(10(-2)-2(6))=$$ $$=72-12-32=28$$

$$\varDelta_y=\begin{vmatrix} 1& 6 & 1 \\ 1 & 10 & 3 \\ 1 & 6 & 3 \end{vmatrix}=$$ $$=1\begin{vmatrix} 10 & 3 \\ 6 & 3 \end{vmatrix}-6\begin{vmatrix} 1 & 3 \\ 3 & 3 \end{vmatrix}+1\begin{vmatrix} 1 & 10 \\ 1 & 6 \end{vmatrix}=$$ $$=(10(3)-3(6))-6(1(3)-3(1))+(1(6)-10(1))=$$ $$=12-0-4=8$$

$$\varDelta_z=\begin{vmatrix} 1& 1 & 6 \\ 1 & 2& 10 \\ 1 & -2 & 6 \end{vmatrix}=$$ $$=1\begin{vmatrix} 2 & 10 \\ -2 & 6 \end{vmatrix}-1\begin{vmatrix} 1 & 10 \\ 1 & 6 \end{vmatrix}+6\begin{vmatrix} 1 & 2 \\ 1 & -2 \end{vmatrix}=$$ $$=(2(6)-10(-2))-(1(6)-10(1))+6(1(-2)-2(1))=$$ $$=32+4-24=12$$

$$x={\varDelta_x \over \varDelta}={28 \over 8}={7 \over 2}$$

$$y={\varDelta_y \over \varDelta}={8 \over 8}=1$$

$$z={\varDelta_z \over \varDelta}={12 \over 8}={3 \over 2}$$

$$({7 \over 2},1, {3 \over 2})$$

5. (a)   

 Let the statement $P(n)$ given as

$$P(n):2^{2n}-3n-1\ \text{is divisible by 9,}n=1, 2, ....$$

Base case:

$n=1: P(1)=2^{2(1)}-3(1)-1=0,$ is divisible by $9.$

Assume that $P(n)$ is true for some natural number $k,$ i.e.,

$$P(k):2^{2k}-3k-1\ \text{is divisible by 9,}\text{i.e.}$$ $$2^{2k}-3k-1=9q, q\in \N$$

Now, to prove that $P(k+1)$ is true, we have

$$P(k+1)=2^{2(k+1)}-3(k+1)-1=$$ $$=4\cdot2^{2k}-3k-3-1=$$ $$=4\cdot2^{2k}-12k-4+12k+4-3k-3-1=$$ $$=4\cdot(2^{2k}-3k-1)+9k=4(9q)+9=$$ $$=9(4q+1)=9m, m\in \N$$

Thus $P(k+1)$ is true, whenever $P(k)$ is true. Hence, by the Principle of Mathematical Induction $P(n)$ is true for all natural numbers $n.$

(b)

$$f(x)=(3-x)^5(2+x)^6,\ -2<x<3$$

$$f'(x)=-5(3-x)^4(2+x)^6+6(3-x)^5(2+x)^5=$$

$$=(3-x)^4(2+x)^5(-10-5x+18-6x)=$$

$$=(3-x)^4(2+x)^5(8-11x)$$ $$f'(x)=0=>(3-x)^4(2+x)^5(8-11x)=0$$

Find critical number(s)

$$-2, {8 \over 11} , 3$$

First Derivative Test

$$\text{If }-2<x<{8 \over 11}, f'(x)>0, f\text{ increases}$$ $$\text{If }{8 \over 11}<x<3, f'(x)<0, f\text{decreases}$$

The greatest value of $(3-x)^5(2+x)^6$ for $-2<x<3$

$$f({8 \over 11})=(3-{8 \over 11})^5(2+{8 \over 11})^6={25^5\cdot30^6 \over 11^{11}}\approx$$

$$\approx24952$$