Determinants NCERT Solutions
NCERT Exercise 4.6 Q:11-16
Question:11.`2x+y+z=1`
`x-2-z=3/2`
`3y-5z=9`
Solution:
The system of equations can be written in the form AX =B,
Where `A=[(2, 1, 1), (2, -4, -2), (0, 3, -5)],X`
`=[(x), (y), (z)]and B= [(1), (3), (9)]`
Now `|A| =|(2, 1, 1), (2, -4, -2), (0, 3, -5)|`
`-2(20+6)-1(-10-0)` `+1(6-0)`
`=52+10+6=68`
`A_11=(20+6)`
`=26,A_12=-(-10-0)=`
`=10,A_13=(6-0)=6`
`A_21=-(-5-3)`
`=8,A_22=(-10-0)`
`=-10,A_23=-(6-0)=-6`
`A_31=(-2+4)`
`=2,A_32=-(-4-2)`
`=6,A_33=-(-8-2)=-10`
`adj A =[(26, 10, 6), (8, -10, -6), (2, 6, -10)]`
`=[(26, 8, 2), (10, -10, -6), (6, -6, -10)]`
`:.A^-1=1/|A| adj A`
`=1/68[(26, 8, 2), (10, -10, 6), (6, -6, -10)]`
`:.X=A^-1B`
`=[(x), (y), (z)]`
`=1/68[(26, 8, 2), (10, -10, 6), (6, -6, -10)][(1), (3), (9)]`
`=1/68[(26+24+18), (10-30+54), (6-18-90)]`
`=1/68[(68), (34), (-102)]=[(1), (1/2), (-3/2)]`
`x=1,y=1/2 and z=-3/2`
Question:12 .
`x-y+z=4`
`2x+y-3z=0`
`x+y+z=2`
Solution:
The system of equations can be written in the form AX =B,
Where `A =[(1, -1, 1), (2, 1, -3), (1, 1, 1)],X`
`=[(x), (y), (z)]and B= [(4), (0), (2)]`
Now`|A| =[(1, -1, 1), (2, 1, -3), (1, 1, 1)]`
`=1(1+3)-(-1)` `(2+3)+1(2-1)`
`=4+5+1=10`
`A_11=(1+3)`
`=4,A_12=-(2+3)`
`=-5,A_13=(2-1)=1`
`A_21=-(-1-1)`
`=2,A_22=(1-1)`
`=0,A_23=-(1+1)=-2`
`A_31=(3-1)`
`=2,A_32=-(-3-2)`
`=5,A_33=(1+2)=3`
`adj A= [(4, -5, 1), (2, 0, -2), (2, 5, 3)]`
`=[(4, 2, 2), (-5, 0, 5), (1, -2, 3)]`
`:.A^-1=1/|A| adj A`
`=1/10[(4, 2, 2), (-5, 0, 5), (1, -2, 3)]`
`:.X=A^-1B`
`[(x), (y), (z)]`
`=1/10[(4, 2, 2), (-5, 0, 5), (1, -2, 3)][(4), (0), (2)]`
`=1/10[(16+0+4), (-20+0+10), (4-0+6)]`
`=1/10[(20), (-10), (10)]=[(2), (-1), (1)]`
`:. x=2,y=-1 and z=1.`
Question:13 .
`2x+3y+3z=5`
`x-2y+z=-4`
`3x-y-2z=3`
Solution:
The system of equations can be written in the form AX =B,
Where `A=[(2, 3, 3),(1, -2, 1),(3, -1, -2)]`,X
`=[(x), (y), (z)]and B=[(5),(-4),(3)]`
`|A|=[(2, 3, 3),(1, -2, 1),(3, -1, -2)]`
`=2(4+1)-3(-2-3)` `+3(-1+6)
`=10+15+15=40`
`A_11=(4+1)`
`=5,A_12=(-2-3)`
`=5,A_13=(-1+6)=5`
`=A_21=-(-6+3)`
`=3,A_22=(-4+9)`
`=-13,A_23=-(-2-9)=11`
`=A_31=-(3+6)`
`=9,A_32=-(2-3)`
`=1,A_33=(-4-3)=-7`
adj `A=[(5,5,5), (3, -13, 11), (9, 1, -7)]`
`=[(5,5,5), (3, -13, 11), (9, 1, -7)]`
`:.A^-1=1/|A|adj A`
``1/40[(5, 3, 9), (5, -13, 1), (5,11,-7)]`
`:.X=A^-1B`
`[(x), (y), (z)]`
`=1/40 [(5, 3, 9), (5, -13, 1), (5, 11, -7)][(5), (-4), (3)]`
`=1/40 [(25-12+27), (25+52+3), (25-44-21)]`
`=1/40[(40), (80), (-40)] =[(1), (2), (-1)]`
`:.x=1,y=2and z=-1`
Question:14 .
`x-y+2z=7`
`3x+4y-5z=-5`
`2x-y+3z=12`
Solution:
The system of equations can bewritten in the form AX=B,
Where `A=[(1, -1, 2), (3, 4, -5), (2, -1, 3)],X`
`=[(x), (y), (z)]and B=[(7), (-5), (12)]`
Now `|A|=|(1, -1, 2), (3, 4, -5), (2, -1, 3)|`
`=1(12-5)-(-1)` `(9+10)+2(-3-8)`
`=7+19-22=4`
`A_11=12-5`
`=7,A_12=-(9+10)`
`=-19,A_13=(-3-8)=-11`
`=A_21=-(-3+2)`
`=1,A_22=(3-4)`
`=-1,A_23=-(-1+2)=-1`
`A_31 =(5-8)`
`=-3,A_32`
`=-(-5-6)=11,A_33` `=(4+3)=7`
`adj A= [(7, -19, -11), (1, -1, -1), (-3, 11, 7)]`
`[(7, 1, -3), (-19, -1, 11), (-11, -1, 7)]`
`:.A^-1=1/|A| adj A`
`=1/4[(7, 1, -3), (-19, -1, 11), (-11, -1, 7)]`
`:.X+A^-1 B`
`[(x), (y), (z)]`
`1/4[(7, 1, -3), (-19, -1, 11), (-11, -1, 7)][(7), (-5), (12)]`
`1/4[(49-5-36), (-113+5-132), (-77+5+84)]` `=1/4[(8), (24), (12)]=[(2), (6), (3)]`
`:. x=2,y=6 and z=3`
Question:15 .
If `A= [(2, -3, 5), (3, 2, -4), (1, 1, -2)]`, find `A^_1`, using `A^-1` solve the system of equations.
`2x-3y+5z=11`
`3x+2y-4z=-5`
`x+y-2z=-3`
Solution:
Here `A= [(2, -3, 5), (3, 2, -4), (1, 1, -2)]`
`|A|= [(2, -3, 5), (3, 2, -4), (1, 1, -2)]`
`2(-4+4)-(-3)(-6+4)` `+5(3-2)`
`=0-6+5=-1`
`A_11=(-4+4)`
`=0,A_12=-(-6+4)`
`=2,A_13=(3-2)-1`
`A_21=-(6-5)`
`=-1,A_22=-(-4-5)`
`=-9,A_23=-(2+3)=-5`
`A_31=(12-10)`
`=2,A_32=-(-8-15)`
`=23,A_33=4+9=13`
`adj A=[(0, 2, 1), (-1, -9, -5), (2, 23, 13)]`
`=[(0, -1, 2), (2, -9, 23), (1, -5, 13)]`
`:.A^-1=1/|A| adj A`
`=1/-1[(0, -1, 2), (2, -9, 23), (1, -5, 13)]`
`[(0, 1, -2), (-2, 9, -23), (-1, 5, -13)]`
Now `X=A^-1B`
`[(x), (y), (z)]`
`[(0, 1, -2), (-2, 9, -23), (-1, 5, -13)][(11), (-5), (-3)]`
`=[(0-5+6), (-22-45+69), (-11-25+39)]=[(1), (2), (3)]`
`x=1, y = 2 and z = 3`
Question:16. The cost of a4kg onion, 3 kg wheat and 2 kg rice is Rs.60. Teb cost of 2kg onion, 4kg wheat and 6 kg rice is Rs. 90 .The cost of 6kg onion, 2kgwheat and 3kg rice is Rs.70. Find cost of each item per kg by matrix method.
Solution:
Let cost of 1kg onion=Rs.x
cost of 1kg rice =Rs.z
`:. 4x+3y+2z=60`
`2x+4y+6z=90`
`6x +2y+3z=70`
The system of equations can be written in the form AX=B, where
`A=[(4, 3, 2), (2, 4, 6), (6, 2, 3)],X=`
`[(x), (y), (z)]and B=[(60), (90), (70)]`
Now`|A|=[(4, 3, 2), (2, 4, 6), (6, 2, 3)],X=`
`=4(12-12)-3(6-36)` `+2(4-24)`
`=0+90-0=90`
`A_11=(12-12)`
`0,A_12=-(6-36)`
`=30,A_13=(4-24)=-20`
`A_21=-(9-4)`
`=-5,A_22=(12-12)`
`=0,A_23=-(8-18)=10`
`A_31=(18-8)`
`10,A_32=-(24-4)`
`=-20,A_33=(16-6)=10`
`adj A[(0, 30, -20), (-5, 0, 10), (10, -20, 10)]`
`[(0, -5, 10), (30, 0, -20), (-20, 10, 10)]`
`:. A^1=1/|A| adj A`
`=1/50[(0, -5, 10), (30, 0, -20), (-20, 10, 10)]`
`:. X=A^-1 B`
`[(x), (y), (z)]`
`=1/50[(0, -5, 10), (30, 0, -20), (-20, 10, 10)][(60), (90), (70)]`
`=1/20[(0-450+700), (1800+0-1400), (-1200+900+700)]`
`=1/50 [(250), (400), (400)]=[(5), (8), (8)]`
`:.x=5, y=8and z=8`
Thus cost of 1kg onion =Rs.5`
cost of 1kg wheat =Rs. 8
cost of 1 kg rice = Rs.8.
Reference: