Math501 assignment no 1

Student name=mubeen habibullah

Student Id=bc240410868

Question 1: Marks=05 Find a parametric equation of the line through and , where

Solution •

P1=⎣⎢⎢⎢⎡−514⎦⎥⎥⎥⎤ P2=⎣⎢⎢⎢⎡−3−17⎦⎥⎥⎥⎤

r(t)=P1+t(P2-P1)

Where t is the parameters.

Find the direction vector (d)

d= P2-P1

=⎣⎢⎢⎢⎡−3−17⎦⎥⎥⎥⎤-⎣⎢⎢⎢⎡−51−4⎦⎥⎥⎥⎤

=⎣⎢⎢⎢⎡2−211⎦⎥⎥⎥⎤Ans.

1.write the parametric equation.

r(t)=⎣⎢⎢⎢⎡−51−4⎦⎥⎥⎥⎤+t⎣⎢⎢⎢⎡2−211⎦⎥⎥⎥⎤

Components form.

x=-5+2t

y=1-2t

z=-4+11t

Question 2: Consider a linear transformation T from R2 into R2 defined by:T⎝⎛xy⎠⎞=⎝⎛−5x+3yx−7y⎠⎞ . Show that T is invertible, and also find T^-1⎝⎛57⎠⎞

Solution.

1.write the transformation T in matrix form

The transformation T maps [x,y]

to [-5x + 3y x -7y]

T([x,y])=[-5. 3. 1. 7][x,y]

Invertibility.

Compute the determinant of A;

det(A)=(-5)(-7)-(3)(1)= 35-3=32

Find the inverse matrix T^-1

Using formula;

A^-1=det(A)1[d −b −c a]

A=[-5. 3. -1. -7]

A^-1=32[−7−3−1−5]1

=[−327 −32332−1−325]

Verify that T^-1 correctly reverse the transformation T

To verify

T^-1(T([x,y]))

=A^-1A:

[−327−323−321−325][-5. 3. 1. -7]

=[1. 0. 0. 1]

T^-1([st])=[−327s−323s−321s−325t] Answer.

Question 3: Consider a linear transformation T from R2 into R2 formed by first performing a reflection across x-axis, and then performing a vertical shear mapping that transforms e1 into e1+2e2 (leaving e2 unchanged). Determine the standard matrix for the linear transformation T.

^Solution

^{Reflection across the axis}

^{(x,y) Into (x,-y)}

R=⎣⎡100−1⎦⎤

Vertical shear maping

e1=(1,0) into e1+2e₂=(1,2)

While S=⎣⎡1 021⎦⎤

Combine the transformation

T=S.R

T=⎣⎡1021⎦⎤.⎣⎡1001⎦⎤=⎣⎡10 2−1⎦⎤Answer

Question 4: Consider a matrix A ⎣⎡b −2−1 a⎦⎤and its inverse A^-1=⎣⎡1 21 3⎦⎤ . Find the values of a and b by verifying your result using the property .

Solution.

Multiply A and A^-1

A×A^-1=⎣⎡b −2−1 a⎦⎤×⎣⎡1 21 3⎦⎤

Multiply row by column

A×A^-1=b(1)+(-2)(1)=b-2

b(2)+(-2)(3)=2b-6

(-1)(1)+(a)(1)=-1+a

(-1)(2)+(a)(3)=-2+3a

Equate A×A^-1to I

⎣⎡b−2 2b−6−1+a −2+3a ⎦⎤=⎣⎡1 00 1⎦⎤b−2=12b−6=0−1+a=0−2+3a=1b−2=1b=3verify2(3)−6=06−6=00=0b=3which is,true thatfrom−1+a=0a=1from−2+3a=1−2+3(1)=1−2+3=11=1a=1,b=3 answer