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Matrices Inter First Year Important

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The key takeaways are about different types of matrices like symmetric, skew-symmetric matrices and how to perform operations like finding the trace, inverse, adjoint and determinant of matrices. It also discusses how to solve systems of linear equations using matrix inversion and Cramer's rule.

The different types of matrices discussed are symmetric, skew-symmetric and identity matrices. A symmetric matrix is equal to its transpose while a skew-symmetric matrix is equal to the negative of its transpose.

To find the inverse of a matrix, you calculate the adjoint and divide it by the determinant. The adjoint of a matrix is the transpose of the cofactors. You can also use matrix inversion to solve systems of linear equations.

Matrices

1.

VSAQ

Sol:

Define trace of the matrix and find trace of

A+A=

A=
=

AA=

2.

Sol: sum of elements in the principle diagonal of a


square matrix is called trace of A.
=1+ (-1) +1=1.
Define a symmetric matrix, If
A=

=
7.

Sol:
2X=B-A
2X=

Sol: A square matrix A is said to be symmetric matrix if


A=A.

3.

If
A=

, then find x.

x=6.

X=
8.

Define skew symmetric,

Find the cofactor of the elements 2, -5 in the


matrix

If
Sol: the cofactor of the elements 2 is =

Matrix, then find x.


Sol: A square matrix A is said to be skew symmetric
matrix

the cofactor of the elements -5 is =


=-(2-5)
=-(-3)
=3.

if A=-A
x=-4

4.

9.

If A=

Sol:

Sol:
1 (3x+24) =45
3x=45-24
3x=21 x=7

5.

Find the det of

=-31-4(-44) +9(-17)
=-31+176-153
=176-184
=-8.
10. If A

If
Find the value of x, y, z and a.

Sol: A

Sol:

=0

Equating the corresponding elements


x=5+3|| 2y=2+8 || z=-2-2 ||a=6+4
x=8, y=5, z=-4 and a=10
6.

If

-2-k=0k=-2.
11. Find the ad joint and inverse of

Matrices
Sol:
ifA=
A=

16. If A=
=

=
A=

12. If
. (1)
A+B=

(2)

From (1) & (2)

=0.

SAQ

13. If A=
Sol: A=

1.

If A=

Sol: A=

14. Find the rank of the matrix


Sol:

=
Rank of A=3(number of non -zero rows)

15. Construct a matrix A=


.
Let A=
.

2.

If A=
Sol:

Matrices
=

5.

If I=
Sol:

=
=
=

=
or

=
=

3.

If A=
Sol:

6.

If A=
Sol: let p (n) =

= -4+1+4
=1
A

Step-1: put n=1

A=

Step-2: let us assume that p(k) is true for n=k


Put n=k
Step-3: put n=k+1

=
4.

Find the inverse of diag [a b c].


Sol: let A=

7.

IfA=
(h/w)
Sol: s(n) is true for n=k+1.

Matrices
8.

Show that

Sol: L.H.S

Sol:
=

=
=

=
=
12. If

=
=

9.

Show that

Sol:

, then show that

Sol:

(h/w)
=cos

10. Show that

L.H.S

=
Sol: L.H.S
=
=
=
=

=
e

11. Show that

=-sin

Matrices

LAQ (2

1.
=
Sol: L.H.S
=

(2)

3.

If

Sol:

R.H.S

2.

L.H.S

Matrices
=
4.

Expanding along

Sol:

6.

S.T

(a-b) (b-c) (c-a) (ab+bc+ca).

Sol: L.H.S

=
=

= (a-b) (b-c)

Expanding along
= (a-b) (b-c)

=
=
b(c-a) +(c-a

5.

(c-a

= (a-b) (b-c)
Sol:
= (a-b) (b-c)
Expanding along
= (a-b) (b-c)
=

= (a-b) (b-c)
= (a-b) (b-c) (c-a) (ab+bc+ca).

Matrices
7.

S.T

(a-b) (b-c) (c-a)


Sol: A=

Sol:

Now A. adjA=

=
=
=
=

Expanding along
=
=
(a-b) (b-c) (c-a)

A.

8.

x=4.
9.

If A=

=I

Similarly we can prove that

= detA.I

=I

Matrices
10. Solve the following equations by using Cramers rule
and matrix inversion method.
a)
b)

x=3, y=1 and z=1.


(b).
Sol: let A=

, X=

and B=

=2
=2(1+1) +1(1-1) +3(-1-1)
=2(2) +1(0) +3(-2)
=4-6 = -2
=6
(a).
Sol: let A=

, X=

and B=

=9(1+1) +1(6-2) +3(-6-2)


=9(2) +1(4) +3(-8)
=18+4-24=22-24=-2
=2

=3

-4

+5

=3(-7+16) -4(14-40) +5(-4+5)


=3(9)-4(-26) +5(1)
=27+10+5
=136

=18

-4

+5

=18(-7+16) -4(91-160) +5(-26+20)


=18(9)-4(-69) +5(-6)
=162+276-30
=408

=2(6-2) -9(1-1) +3(2-6)


=2(4) +1(0) +3(-24)
=8-12=-4
=2
=2(2+6) +1(2-6) +9(-1-1)
=2(8) +1(-4) +9(-2)
=16-4-18
=16-22=-6
x=

=1, y=

, z=

, X=

and B=

x=1, y=2 and z=3.


(c). A=
=1

=3

-18

+5

=3(91-160) -18(14-40) +5(40-65)


=3(-69)-18(-26) +5(-25)
=-207+468 -125
=136

=3

-4

+18

=3(-20+26) -4(40-65) +5(-4+5)


=3(6)-4(-25)+5(1)
=18+100+18
=136
x=

, y=

, z=

=1(-5-7) -1(-2-14) +1(2-10)


=1(-12) -1(-16) +1(-8)
=-12+16-8=- 4
=9
=9(-5-7) -1(-52-0) +1(52-0)
=9(-12) -1(-52) +1(52)
=-108+52+52
=-108+104=-4
=

=1

=1(-52-0) -9(-2-14) +1(0-104)


=1(-52) -9(-16) +1(-104)

Matrices
=-52+144-104
=-156+144=-12

AdjA=
=1

=1(0-52) -1(0-104) +9(2-10)


=1(-52) -1(-104) +9(-8)
=-52+104-72
=-124+104=-20
x=

=1, y=

, z=

x=1, y=3 and z=5.


(d). A=

, X=

and B=

AdjA =

=2
DetA=
2(9)-4(-26) +5(1)
=136

=2(-8-1) +1(4-3) +3(-1-6)


=2(-9) +1(1) +3(-7)
=--18+1-21
=-38

A-1=
X=A-1.B

=8

=8(-8-1) +1(-16-0) +3(4-0)


=8(-9) +1(-16) +3(4)
=-72-16+12
=-76

=2
==

=2(-16-0) -8(4-3) +3(0-12)


=2(-16) -8(1) +3(-12)
=-32-8-36
=-76

(b). Sol: let A=

And A-1=

=2(0-4+1(0-12+8(-1-6)
=2(-4) +1(-12) +8(-7)
=-8-12-56
=76
=2, y=

=2

, z=

x=2, y=2 and z=2.


Matrix inversion method:
(a). A=
AX=BX=A-1.B
And A-1=

, X=

AX=BX=A-1.B

=2

x=

x=3, y=1 and z=1.

, X=

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


=2(2) +1(0) +3(-2)
=4-6 = -2
AdjA=

and B=
=

and B=

Matrices
=
=
==

x=1, y=3 and z=5.

AdjA =
(d).let A=

, X=

and B=

A-1=
=2

X=A-1.B

=2(-8-1) +1(4-3) +3(-1-6)


=2(-9) +1(1) +3(-7)
=--18+1-21
=-38

AdjA=
==
(c). A=

x=1, y=2 and z=3


, X=
=1

=1(-5-7) -1(-2-14) +1(2-10)


=1(-12) -1(-16) +1(-8)
=-12+16-8=- 4
AdjA=

and B=

AdjA =
A-1=
X=A-1.B

AdjA =

==

A-1=
X=A-1.B
=

x=2, y=2 and z=2.

Matrices

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