Determinants NCERT Solutions
NCERT Exercise 4.5 Q:9-18
Question:9 . `[(2, 1, 3), (4, -1, 0), (-7, 2, 1)]`
Solution:
Let A=`[(2, 1, 3), (4, -1, 0), (-7, 2, 1)]`
|A|=`[(2, 1, 3), (4, -1, 0), (-7, 2, 1)]`
=`2(-1-0)-(4-0)` `+3(8-7)`
`=-2-4+3=-3`
`A_11=(-1-0)`
`=-A_12=(-4-0)`
`=-4,A_13=(8-7)=1`
`A_21=-(1-6)=5,A_22`
`=(2+21)=23,A_23=`
`=-(4+7)=-11`
`A_31=(0+3)=3,A_32`
`=-(0-12)=12,A_33=`
`=(-2-4)=-6`
`A_21=-(1-6)`
`=5,A_22=(2+21)`
`=23,A_23`
`=-(4+7)=-11`
`A_31= (0+3)`
`=3,A_32=-(0-12)`
`=12, A_33`
`=(-2-4)=-6`
`adj A= [(-1, -4, 1), (5, 23, -11), (3, 12, -6)]`
`= [(-1, 5, 3), (-4, 23, 12), (1, -11, -6)]`
`A^-1=1/|A| adj A`
`=1/-3 [(-1, 5, 3), (-4, 23, 12), (1, -11, -6)]`
Question:10 .
`[(1, -1, 2),(0, 2, -3), (3, -2, 4)]`
Solution:
Let A =`[(1, -1, 2),(0, 2, -3), (3, -2, 4)]`
`|A|=[(1, -1, 2),(0, 2, -3), (3, -2, 4)]`
`=1(8-6)-(-1)(0+9)` `+2(0-6)`
`=2+9-12=-1`
`A_11=(8-6)`
`=2,A_12=`
`=-(0+9)=-9,A_13`
`=(0-6)=-6`
`A_21=(-4+4)`
`=0,A_22`
`=(4-6)=-2,A_23`
`=(-2+3)=-1`
`A_31=(3-4)`
`=1,A_32=`
`=-(-3-0)=3,A_33`
`=(2-0)=2`
`adj A =[(2, -9, -6),(0, -2, -1),(-1, 3, 2)]`
` =[(-2, 0, 1),(9, 2, -3),(6, 1, -2)]`
Question:11 .
`[(1, 0, 0),(0, cosalpha, sin alpha), (0, sin alpha, -cos alpha)]`
Solution:
Let A=`[(1, 0, 0),(0, cosalpha, sin alpha), (0, sin alpha, -cos alpha)]`
|A|`[(1, 0, 0),(0, cosalpha, sin alpha), (0, sin alpha, -cos alpha)]`
`=1(-cos^2alpha-sin^2alpha)`
`=-(cos^2alpha+sin^2alpha)=-1`
`A_11=(-cos^2alpha-sin^2alpha)`
`=-(-cos^2alpha+sin^2alpha)`
`-1,A_12 = -(0-0)`
`=0,A_13=(0-0)=0`
`A_21=-(0-0)=0,A_22`
`=(-cosalpha-0)=`
`=-cosalpha,A_23`
`-(sinalpha-0)=-sinalpha`
`A_31=(0-0)=0,A_32`
`=-(sinalpha-0)`
`=-sinalpha,A_33`
`=(cosalpha-0)=cos alpha`
`adj A[(-1, 0, 0), (0, -cosalpha, -sinalpha), (0, -sinalpha, cosalpha)]^1`
`=[(-1, 0, 0), (0, -cosalpha, -sinalpha), (0, -sinalpha, cosalpha)]`
`A^-1=1/|A| adj A`
`1/-1[(-1, 0, 0), (0, -cosalpha, -sinalpha), (0, -sinalpha, cosalpha)]`
`=[(1, 0, 0), (0, cosalpha, sinalpha), (0, sinalpha, -cosalpha)]`
Question:12 .
Let A=`[(3, 7), (2, 5)]` and B =`[(6, 8), (7, 9)]`,verify that `(AB)^-1=B^-1A^-1`.
Solution:
Here A=`[(3, 7), (2, 5)]`
`|A|=[(3, 7), (2, 5)]`
`15 -4=1`
`A_11=5,A_12`
`=-2, A_21=`
`=-7,A_22=3`
`adj A=[(5, -2), (-7, 3)]`
`=[(5, -7), (-2, 3)]`
`A^-=1/|A|adj A`
`1/1[(5, -7), (-2, 3)]` `=[(5, -7), (-2, 3)]`
`B=[(6, 8), (7, 9)]`
`|B|=[(6, 8), (7, 9)]`
`=54-56=-2`
`_11=9,B_12=-7,A_21`
`=-8,B_22 =6`
`adj B =[(9, -7),(-8, 6)]` `=[(9, -8), (-7, 6)]`
`B^-1=1/|B| adj B`
`=1/-2[(9, -8),(-7, 6)]` `=1/2[(-9, 8), (7, -6)]`
Now`AB=[(3, 7), (2, 5)][(6, 8), (7, 9)]`
`[(18+49, 24+63), (12+35, 16+45)]`
`=[(67, 87), (47, 61)]`
`|AB|=|(67, 87), (47, 61)|`
`=4087-4089=-2`
`AB_11=61,AB_12`
`-47,AB_21=`
`-87,AB_22=67`
`:. adj |AB|`
`=[(61, -47), (-87, 67)]'` `=[(61, -87), (-47, 67)]`
`(AB)^-1 =1/|AB| adj |AB|`
`1/-2 [(61, -87), (-47, 67)]`
`1/2[(-61, -87), (47, -67)]`
Now` B^-1A^-1=`
`1/2[(-9, 8), (7, -6)][(5, -7), (-2, 3)]`
`=1/2[(-45-16, 63+24),(35+12, -49-18)]`
`=1/2[(-61, 87), (47, -67)]`
Thus `(AB)^-1=B^_1A^-1`
Question:13 .
If A`=[(3, 1), (-1, 2) ]` show that `A^2-5A+71=0` Hence find `A^-1`.
Solution:
Here A`=[(3, 1), (-1, 2) ]`
`:. A^2=A.A`
`=[(3, 1), (-1, 2) ][(3, 1), (-1, 2) ]`
`=[(9-1, 3+2), (-3-2, -1+4)]` `[(8, 5), (-5, 3)]`
`:.A^2-5A+71`
`=[(8, 5), (-5, 3)]-5[(3, 1)(-1, 2)]` `+7[(1, 0)(0, 1)]`
`=[(8, 5), (-5, 3)][(15, 5)(-5, 10)]` `+[(7, 0)(0, 7)]`
`=[(8-15+7, 5-5+0), (-5+5+0, 3-10+7)]` `=[(0, 0),(0, 0)]`
Thus `A^2-5A+71=0`
pre -multiplying by `A^-1`on both side we get
`A^-1(A^2-5A+71)=A^-1.`0
`=>A^-1A^2-5A^-1A` `+7A^-1 1=0`
`=> A-51+7A^-1=0`
`=> A^-1 =1/7 (51-A)`
`=1/7(5[(1, 0), (0, 1)]-[(3, 1), (-1, 2)])`
`=1/7([(5, 0), (0, 5)]-[(3, 1), (-1, 2)])`
`=1/7[(2, -1), (1, 3)]`
Thus `A^-1=1/7[(2, -1), (1, 3)]`
Question:14 .
Find the matrix A =`[(3, 2), (2, 1)]` find the numbers a and b suchthat `A^2-aA+bI=0`.
Solution:
Here A =`[(3, 2), (2, 1)]`
`:. A^2=A.A=A `
`=[(3, 2), (2, 1)][(3, 2), (2, 1)]`
`=[(9+2, 6+2), (3+1, 2+1)]=[(11, 8), (4, 3)]`
`:.A^2-aA+bI=0`
`=>[(11, 8), (4, 3)]+a[(3, 2), (1, 1)]` `+b[(1,0), (0, 1)]`=0
`=>[(11, 8), (4, 3)]+[(3a, 2a), (a, a)]` `+[(b, 0), (0, b)]=0`
`=>[(11+3a+b, 8+2a+0), (4+a+0, 3+a+b)]=0`
`=>4+a=0 and 3+a+b=0`
`=>a=-4and 3-4+b=0`
`=>b=0`
Thus a= -4 and b=1
Question:15 .Find the matrix `A=[(1, 1, 1), (1, 2, -3), (2, -1, 3)]`show that `A^3-6A^2+5A+11I=0`
Hence find `A^-1`.
Solution:
Here A=`[(1, 1, 1), (1, 2, -3), (2, -1, 3)]`
`A^2=A.A=`
`[(1, 1, 1), (1, 2, -3), (2, -1, 3)][(1, 1, 1), (1, 2, -3), (2, -1, 3)]`
`=[(1+1+2, 1+2-1, 1-3+3), (1+2-6, 1+4+3, 1-6-9),(2-1+6, 2-2-6, 2+3+9)]`
`=[(4, 2, 1), (-3, 8, -14), (7, -3, 14)]`
`A^3=A^2.A=`
`[(4, 2, 1), (-3, 8, -14), (7, -3, 14)]` `[(1, 1, 1), (1, 2, -3), (2, -1, 3)]`
`=[(4+2+2, 4+4+-1, 4-6+3), (-3+8-28, -3+16+14, -3-24-42), (7-3+28, 7-6-14, 7+9+42)]`
`= [(8, 7, 1), (-23, 27, -69), (32, -13, 58)]`
Now `A^3-6A62+5A+11I`
`=[(8, 7, 1), (-23, 27, -69), (32, -13, 58)]` `6[(24, 12, 6), (-3, 8, -14), (7, -3, 14)]+5` `[(1, 1, 1), (1, 2, -3), (2, -1, 3)]+11[(1, 0, 0), (0, 1, 0), (0, 0, 1)]`
`=[(8, 7, 1), (-23, 27, -69), (32, -13, 58)]` `-[(24, 12, 6), (-18, 48, -14), (42, -18, 84)]`
`+[(5, 5, 5), (5, 10, -15), (10, -5, 15)]` `+[(11, 0, 0), (0, 11, 0), (0, 0, 11)]`
`=[(8-24+5+11, 7-12+5+0, 1-6+5+0), (-2323+18+5+0, 27-448+10+11, -69+84-15+0), (32-42+10+0, -13+18-5+0, 58-84+155+11)]`
`=[(0, 0, 0), (0, 0, 0), (0, 0, 0)]`=0
Thus`A^3-6A^2+5A+11I=0`
Pre-multiplying both sides by `A^-1`, we get
`A^-1(A^3-6A^2+5A+11I)` `=A^-1.0`
`=>A^-1A^3-6A^-1A^2+` `5A^-1A+11A^-I=0`
`A^2-6A+51+11A^-1=0`
`=>A^-1=1/11(6A-A^2-5I)`
`1/11(6[(1, 1, 1), (1, 2, -3), (2, -1, 3)]-[(4, 2, 1), (-3, 8, -14), (7, -3, 14)]` `-5[(1, 0, 0), (0, 1, 0), (0, 0, 1)])`
`1/11([(6, 6, 6), (6, 12, -18), (12, -6, 18)]-[(4, 2, 1), (-3, 8, -14), (7, -3, 14)]` `-[(5, 0, 0), (0, 5, 0), (0, 0, 5)])`
`1/11 [(6-4-5, 6-2-0, 6-1-0),(6+3-0, 12-8-5, -18+14-0), (12-7-0, -6+3-0, 18-14-5)]`
`1/11[(-3, 4, 5), (9, -1, -4), (5, -3, -1)]`
Thus `A^-1=1/11[(-3, 4, 5), (9, -1, -4), (5, -3, -1)]`
Question:16 .
If A`=[(2, -1, 1), (-1, 2, -1), (1, -1, 2)]`, verify that `A^3-6A^2+9A-4I=0 ` and hence find `A^-1.
Solution:
Here A`=[(2, -1, 1), (-1, 2, -1), (1, -1, 2)]`
`A^2=A.A=`
`[(2, -1, 1), (-1, 2, -1), (1, -1, 2)]A`=[(2, -1, 1), (-1, 2, -1), (1, -1, 2)]`
`=[(4+1+1, -2-2-1, 2+1+2), (-2-2-1, 1+4+1, -1-2-2), (2+1+2, -1-2-2, 1+1+4)]`
`=[(6, -5, 5), (-5, 6, -5), (5, -5, 6)]`
`A^3=A^2.A=`
`=[(6, -5, 5), (-5, 6, -5), (5, -5, 6)][(2, -1, 1), (-1, 2, -1), (1, -1, 2)]`
`=[(12+5+5, -6-10-5, 6+5+10), (-10-6-5, 5+12+5, -5-6-10), (10+5+6, -5-10-6, 5+5+12)]`
`=[(22, -22, 21), (-21, 22, -21), (21, -21, 22)]`
Now `A^3-6A^2+9A-4I`
`=[(22, -22, 21), (-21, 22, -21), (21, -21, 22)]-[(36, -30, 30), (-30, 36, -30), (30, -30, 36)]` `+[(18, -9, 9), (-9, 18, -9), (9, -9, 18)]` `-[(4, 0, 0), (0, 4, 0), (0, 0, 4)]`
`=[(22-36+18-4, -21+30-9-0, 21-30+9-0), (-21+30-9-0, 22-36+8-4, -21+30-9-), (21-30+9-0, -21+30-9-0, 22-36+18-4)]`
`=[(0, 0, 0), (0, 0, 0), (0, 0, 0)]=0`
Thus `A^3-6A^2+9A-4I=0`
Pre-multiplying by `A^-1` on both sides, we get
`A^-1(A^3-6A^2+9A-4I)=A^-1.0`
`=>A^-1A^3-6A^-1A^2` `+9A^-1A-4A^-1 I=0`
`A^2-6A+9I-4A^-1=0`
`=>A^-1 =1/4 (A^2-6A+9I)`
`=1/4([(6, -5, 5), (-5, 6, -5), (5, -5, 6)]-6[(2, -1, 1), (-1, 2, -1), (1, -1, 2)]` `+9[(1, 0, 0), (0, 1, 0), (0, 0, 1)])`
`=1/4([(6, -5, 5), (-5, 6, -5), (5, -5, 6)]-[(12, -6, 6), (-6, 12, -6), (6, -6, 12)]` `+[(9, 0, 0), (0, 9, 0), (0, 0, 9)])`
`1/4[(6-12+9, -5+6+0, 5-6+0), (-5+6+0, 6-12+9, -5+6+0), (5-6+0, -5+6+0, 6-12+9)]`
=`1/4[(3, 1, -1), (1, 3, 1), (-1, 1, 3)]`
Thus `A^-1=1/4[(3, 1, -1), (1, 3, 1), (-1, 1, 3)]`
Question:17 .
Let A be a non singular square matrix of order `3xx3` Then |adj A| is equal to
(A)|A|
(B)`|A|^2`
(C)`|A|^3`
(D)`3|A|`
Solution:
We know that (adj A)A=|A|I
`=|A|[(1, 0, 0), (0, 1, 0), (0, 0, 1)]`
Now `|(adj A).A|=[(|A|, 0, 0), (0, |A|, 0), (0, 0, |A|)]`
`|(adj A)||A|`
`=|A|^3[(1, 0, 0), (0, 1, 0), (0, 0, 1)]=|A|^3.I `
` :.|adj A| |A|^2`Thus answer is (B).
Question:18.
If A is an invertible matrix of order 2 then det (A^-1)is equal to
(A)det(A)
(B)`1/(det(A))`
(C)1
(D)0
Solution:
we know that `A A^-1 =I`
`:.|AA^-1|=|I|`
`=>|A||A^-1=1`
`=>|A^-1|=1/|A|`
Thus answer is (B).
Reference: