Matrices inter first year important
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7/28/2019 Matrices inter first year important
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Matrices
VSAQ
. Define trace of the matrix and find trace of
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, IfA= , then find x.Sol: A square matrix A is said to be symmetric matrix if
A=A.
x=6.
. Define skew symmetric,If
Matrix, then find x.
Sol: A square matrix A is said to be skew symmetric
matrix
if A=-A
x=-4
. If A= Sol: 1 (3x+24) =45 3x=45-24 3x=21 x=7
. If Find the value of x, y, z and a.Sol: 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
. If
Sol: A= A+A= = AA=
=
= 7. IfA= Sol: 2X=B-A2X= X=
8.
Find the cofactor of the elements 2, -5 in the
matrix Sol: the cofactor of the elements 2 is = the cofactor of the elements -5 is =
=-(2-5)
=-(-3)
=3.
9. Find the det of Sol:
=-31-4(-44) +9(-17)
=-31+176-153
=176-184
=-8.
10.If A Sol:A =0 -2-k=0k=-2.
11.Find the ad joint and inverse of ,
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Matrices
Sol:
ifA= =
= = 2.If
=0.
3.If A= Sol:A= =
4.Find the rank of the matrix Sol: = =
Rank of A=3(number of non -zero rows)
5.Construct a matrix A= .Let A= .
A=
16.If A= A= . (1)A+B=
(2)
From (1) & (2)
SAQ
1. If A= .Sol: A=
= =
= = 2. If A=
Sol:
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Matrices
=
= = =
or . If A=
Sol: = -4+1+4
=1
A
. Find the inverse of diag [a b c].Sol: let A= =
5. If I= Sol: = = =
= =.
6. If A= Sol: let p (n) = 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
= 7. IfA= (h/w)
Sol: s(n) is true for n=k+1.
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Matrices
. Show that Sol: = = = = =
. Show that Sol: (h/w)
0. Show that Sol: L.H.S = = = e
1. Show that
Sol: L.H.S
= = = =
12. If , then show that
Sol: =cos =-sin
L.H.S = = =
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Matrices
LAQ (2).
Sol: L.H.S
= = R.H.S
.
L.H.S
=
=
=
(2)
3. If
Sol:
{ }
{
}
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Matrices
. Sol: =
= Expanding along =
.
Sol:
=
= Expanding along =
6. S.T (a-b) (b-c) (c-a) (ab+bc+ca).Sol: L.H.S = =
= (a-b) (b-c)
= (a-b) (b-c)
= b(c-a) +(c-a (c-a = (a-b) (b-c) = (a-b) (b-c)
Expanding along = (a-b) (b-c) = (a-b) (b-c)
= (a-b) (b-c) (c-a) (ab+bc+ca).
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Matrices
7. S.T (a-b) (b-c) (c-a)Sol:
= = =
= Expanding along == (a-b) (b-c) (c-a)
. .
: . .
x=4.. If A=
Sol: A=
Now A. adjA= =
=
= = detA.IA. =ISimilarly we can prove that
= I
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0. Solve the following equations by using Cramers ruleand matrix inversion method.
a) b)
(a).
Sol: let A= , X= and B=
=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
=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=
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 =9(1+1) +1(6-2) +3(-6-2)
=9(2) +1(4) +3(-8)
=18+4-24=22-24=-2
=2 =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=1, y=2 and z=3.(c). A= , X=
and B=
=1 =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)
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Matrices
=-52+144-104
=-156+144=-12
=1 =1(0-52) -1(0-104) +9(2-10)
=1(-52) -1(-104) +9(-8)=-52+104-72
=-124+104=-20x= =1, y= = , z= x=1, y=3 and z=5.(d). A= , X=
and B=
=2 =2(-8-1) +1(4-3) +3(-1-6)
=2(-9) +1(1) +3(-7)
=--18+1-21
=-38
=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
=2 =2(0-4+1(0-12+8(-1-6)
=2(-4) +1(-12) +8(-7)
=-8-12-56=76x= =2, y= = , z= x=2, y=2 and z=2.
Matrix inversion method:
(a). A= , X= and B=
AX=BX=A-1.BAnd A-1=
AdjA=
= = AdjA = DetA=
2(9)-4(-26) +5(1)
=136
A-1= X=A-1.B
=
=
==
x=3, y=1 and z=1.(b). Sol: let A= , X=
and B=
AX=BX=A-1.BAnd A-1=
=2
=2(1+1) +1(1-1) +3(-1-1)
=2(2) +1(0) +3(-2)
=4-6 = -2AdjA=
=
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Matrices
=
AdjA = A-1=
X=A-1.B=
=
== x=1, y=2 and z=3
(c). A= , X= and B=
=1 =1(-5-7) -1(-2-14) +1(2-10)
=1(-12) -1(-16) +1(-8)
=-12+16-8=- 4
AdjA=
= =
AdjA = A-1=
X=A-1.B=
=
==
x=1, y=3 and z=5.
(d).let A= , X= and B= =2 =2(-8-1) +1(4-3) +3(-1-6)
=2(-9) +1(1) +3(-7)
=--18+1-21
=-38 AdjA=
= = AdjA =
A-1= X=A-1.B
=
=
== x=2, y=2 and z=2.
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Matrices