Math Twelve

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.

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