Matrices

More Topic from Mathematics

Question 1

If A and B are square matrices of size n × n such that A2 − B2 = (A − B) (A + B), then which of the following will be always true?

  • A = B

  • AB = BA

  • either of A or B is a zero matrix

  • either of A or B is an identity matrix

Solution

B.

AB = BA

A2 − B2 = (A − B) (A + B)
A2 − B2 = A2 + AB − BA − B2
⇒ AB = BA

Question 2

If A2 – A + I = 0, then the inverse of A is

  • A + I

  • A

  • A – I

  • I – A

Solution

D.

I – A

Given A2 – A + I = 0
A–1A2 – A–1A + A–1 – I = A–1⋅0 (Multiplying A–1 on both sides)
⇒ A - I + A-1 = 0 or A–1 = I – A.

Question 3

If a2 + b2 + c2 = -2 and straight f left parenthesis straight x right parenthesis space equals space open vertical bar table row cell 1 plus straight a squared straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row cell left parenthesis 1 plus straight a squared right parenthesis straight x end cell cell 1 plus straight b squared straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row cell left parenthesis 1 plus straight a squared right parenthesis straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell 1 plus straight c squared straight x end cell end table close vertical bar then f(x) is a polynomial of degree

  • 1

  • 0

  • 2

  • 3

Solution

C.

2

straight f left parenthesis straight x right parenthesis space equals space open vertical bar table row cell 1 space plus left parenthesis straight a squared plus straight b squared plus straight c squared plus 2 right parenthesis straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis straight x end cell row cell 1 plus left parenthesis straight a squared plus straight b squared space plus straight c squared space plus 2 right parenthesis straight x end cell cell 1 plus straight b squared straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis straight x end cell row cell 1 plus space left parenthesis straight a squared plus straight b squared space plus straight c squared plus 2 right parenthesis straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell 1 plus straight c squared straight x end cell end table close vertical bar
Applying space straight C subscript 1 space rightwards arrow with space on top space straight C subscript 1 space plus straight C subscript 2 plus space straight C subscript 3
space equals space open vertical bar table row 1 cell left parenthesis 1 plus straight b squared right parenthesis space straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row 1 cell 1 plus straight b squared space straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row 1 cell left parenthesis 1 plus straight b squared right parenthesis space straight x end cell cell 1 plus straight c squared space straight x end cell end table close vertical bar space therefore space straight a squared space plus straight b squared space plus straight c squared space plus 2 space equals space 0
straight f left parenthesis straight x right parenthesis space equals space space open vertical bar table row 0 cell straight x minus 1 end cell 0 row 0 cell 1 minus straight x end cell cell straight x minus 1 end cell row 1 cell left parenthesis 1 plus straight b squared right parenthesis space straight x end cell cell 1 plus straight c squared space straight x end cell end table close vertical bar space space semicolon space Applying space straight R subscript 1 space rightwards arrow space straight R subscript 1 minus straight R subscript 2
straight f left parenthesis straight x right parenthesis space equals space left parenthesis straight x minus 1 right parenthesis squared
Hence space degree space equals space 2
Question 4

If P = open square brackets table row 1 straight alpha 3 row 1 3 3 row 2 4 4 end table close square brackets is the adjoint of a 3 x3 matrix A and |A| = 4, then α is equal to 

  • 4

  • 11

  • 5

  • 0

Solution

B.

11

Given, 
straight P space equals space open square brackets table row 1 straight alpha 3 row 1 3 3 row 2 4 4 end table close square brackets
|P| = 1(12-12)-α (4-6) +3(4-6)
 = 2α -6.
∵ P =adj(A)
∴ |P| = |adj A | = |A|3-1 = |A|2 = 16
[∵ |adj A| = |A|n-1 order is 3 x3
∴ 2α -6 = 16
2α = 22
α = 11

Question 5

If S is the set of distinct values of 'b' for which the following system of linear equations
x + y + z = 1
x + ay + z = 1
ax + by + z = 0
has no solution, then S is

  • a singleton

  • an empty set

  • an infinite set

  • a finite set containing two or more elements

Solution

A.

a singleton

straight D space equals space open vertical bar table row 1 1 1 row 1 straight a 1 row straight a straight b 1 end table close vertical bar space equals space 0
rightwards double arrow straight a space equals space 1
and space at space straight a space equals space 1
straight D subscript 1 space equals straight D subscript 2 space equals space straight D subscript 3 space equals space 0

but at a = 1 and b =1
First two equations are x +y+ z =1
and third equations is x + y +z = 0
⇒ There is no solution
therefore, b = {1}
⇒ it is a singleton set