Determinants NCERT Solutions
NCERT Exercise 4.6
Examine the consistenncy of the system of equations in Exercises 1 to 6
Question:1 .
`x+2y=2`
`2x+3y=3`
Solution
The system of equation canbe written in the form AX=B, where
`A=[(1, 2), (2, 3)],X` ` =[(x), (y)]and B=[(2), (3)]`
Now `|A|=|(1, 2), (2, 3)|`
`=3-4=-1!=0`
So the system of equation is conssistent.
Question:2 .
`2x-y=5`
`x+y=4`
Solution
The system of equations can be written in the form AX + B,where
`A= [(2, -1), (1, 1)]X` ` =[(x), (y)]and B=[(5), (4)]`
Now `|A|= |(2, -1), (1, 1)|`
`=2-(-1)=2+1=3!=0`
Sothe system of equations is consistent.
Question:3 .
`x+3y=5`
`2x+6y=8`
Solution:
The system of equations can be written in the from AX=B, where
`A= [(1, 3), (2, 6)]X` ` =[(x), (y)]and B=[(5), (8)]`
Now `|A|= [(1, 3), (2, 6)]`
`=6-6=0`
So the system of equations is inconsistent.
Question:4 .
`x+y+z=1`
`2x+3y+2z=2`
`ax+ay+2az=4`
Solution:
The system of equations can be written in the form AX=B,
Where `A=[(1, 1, 1 ), (2, 3, 2), (a, a, 2a)],X` `[(x), (y), (z)]and B=[(1), (2), (4)]`
Now `|A|=[(1, 1, 1 ), (2, 3, 2), (a, a, 2a)]`
`=1(6a-2a)-1(4a-2a)` `+1(2a-3a)`
`=4a-2a-a=a!=0`
so the system of equations is consitent.
Question:5 .
`3x-y-2z=2`
`2y-z=-1`
`3x-5y=3`
Solution:
The system of equations can be written in the form AX=B.
Where `A=[(3, -1, -2), (0, 2, -1), (3, -5, 0)],X`
`=[(x), (y), (z)]and B=[(2), (-1), (3)]`
Now `|A|=|(3, -1, -2), (0, 2, -1), (3, -5, 0)|`
`=3(0-5)-(-1)(0+3)+(-2)(0-6)`
`=15+3+12=0`
So the system of equations is inconsitent.
Question:6 .
`5x-y+4z=5`
`2x+3y+5z=2`
`5x-2y+6z=-1`
Solution:
The system of equations can be written in the form AX=B
Where `A=[(5, -1, 4), (2, 3, 5), (5, 2, 6)],X`
`=[(x), (y), (z)]and B=[(5), (2), (1)]`
Now `|A|=[(5, -1, 4), (2, 3, 5), (5, 2, 6)]`
`=3(18+10)-(-1)(12-25)` `+4(-4-15)`
`=140-13-76=51!=0`
So the systemm of equations is consistent.
Solve system of linear equations, using matrix method in Exercises 7 to 14.
Question:7 .
`5x+2y=4`
`7x+3y=5`
Solution:
The system of equations can be written in the form AX=B
Where `A=[(5, 2), (7, 3)],X`
`=[(x), (y)]and B=[(4), (5)]`
Now `|A|=[(5, 2), (7, 3)]`
`=15-14=1`
`:. A_11=3A_12=-7,A_21` `=-2,A_22=5`
`:. adj A=[(3, -7), (-2, 5)]=` `[(3, -2), (-7, 5)]`
`:.A^-1=1/|A|adj A`
`=1/1[(3, -2), (-7, 5)]` `=[(3, -2), (-7, 5)]`
`:.X=A^-1B`
`[(x), (y)]=[(3, -2), (-7, 5)][(4), (5)]`
`=[(12-10), (-28+25)]` `=[(2), (-3)]`
`:. x-2and y=-3`.
Question:8 .
`2x-y=-2`
`3x+4y=3`
Solution:
The system of equations can be written in the form AX= B,
Where `A=[(2, -1), (3, 4)],X`
`=[(x), (y)]and B=[(-2), (3)]`
Now `|A|=[(2, -1), (3, 4)]`
`=18-(-3)=11`
`:.A_11=4,A_12= -3,A_21`
`=-(-1)=1, A_22=2`
`:. adj A =[(4, -3), (1, 2)]`
`:.=[(4, 1), (-3, 2)]`
`:.A^-1=1/|A| adj A`
`=1/11[(4, 1), (-3, 2)]`
`=1/11[(4, 1), (-3, 2)]`
`:.X=A^-1B`
`[(x), (y)]=`
`1/11[(4, 1), (-3, 2)][(-2), (3)]`
`=1/11[(8+3), (6+6)]=[(-5/11), (12/11)]`
`:. x=-5/11 and y=12/11`
Question: 9 .
`4x-3y=3`
`3x-5y=7`
Solution:
The systemm of equations can be written in the form AX = B,
Where `A=[(4, -3), (3, -5)],X`
`=[(x), (y)]and B=[(3), (7)]`
Now `|A|=[(4, -3), (3, -5)]`
`=-20-(-9)` `=-20+9=-11`
`:.A_11=-5,A_12= -3,A_21`
`=-(-3)=3, A_22=4`
`adj A=[(-5, 3), (3, 4)]` `[(-5, 3), (-3, 4)]`
`:.A^-1=1/|A| adj A`
`=1/-11[(-5, 3), (-3, 4)]`
`=1/11[(5, -3), (3, -4)]`
`:.X=A^-1B`
`[(x), (y)]=`
`1/11[(5, -3), (3, -4)][(3), (7)]`
`=1/11[(15-21), (9-28)]`
=`[(-6/11), (-19/11)]`
`:.x=-6/11 and y=19/11`
Question:10 .
`5x+2y=3`
`3x+2y=`
Solution:
The system of equations can bewritten in the form AX = B,
Where `A =[(5, 2), (3, 2)],X=`
`[(x), (y)]and B =[(3), (5)]`
Now `|A| =|(5, 2), (3, 2)|`
`=10 -6=4`
`:.A_11=2,A_12= -3,A_21`
`=-2,A_22=5`
`adj A=[(2, -3), (-2, 5)]` `[(2, -2), (-3, 5)]`
`:.A^-1=1/|A| adj A`
`=1/4[(2, -2), (-3, 5)]`
`1/4 [(2, -2), (-3, 5)]`
`:.X=A^-1B`
`[(x), (y)]=`
`1/4[(2, -2), (-3, 5)][(3), (5)]`
`:.1/4[(6-10),(-9+25)]` `=1/4[(-4), (16)]=[(-1), (4)]`
`:. x=-1 and y =4`
Reference: