Determinants NCERT Solutions
NCERT Exercise 4.4
Write minors and cofactors of the elements of fllowing determinants:
Solution:
(i)Here `Delta=|(2, -4), (0, 3)|`
`:.M_11=M_12=0, M_21` `=-4and M_22=2-c^2-ac`
`A_11=(-1)^1+1 M_11` `=1xx3=3-x`
`A_12 =(-1)^1+2 M_12` `=-1xx0=0-0`
`A_21 =(-1)2+1 M_21` ` =-1xx-4=4`
`A_22=(-1)^2+2 M_22` `=1xx2=2`
Question:1 .(ii)`|(a, c), (b, d)|`
Solution:
(ii)Here` Delta= |(a, c), (b, d)|`
` :. M_11=d,M_12` `=b,M_21=cand M_22= a`
`A_11=(-1)^(1+1) M_11` `=1xxd=d`
`A_12=(-1)^(1+2) M_12` `=-1xxb=-b`
`A_21=(-1)^(2+1) M_21` `=-1xxc=-c`
`A_22=(-1)^(2+2) M_22` `=1xxa=a`
Question:2 .
(i)`|(1, 0, 0), (0, 1, 0), (0, 0,1)|`
Solution:
(i)Here `Delta |(1, 0, 0), (0, 1, 0), (0, 0,1)|`
` M_11 = |(0, 0), (0, 1)|` `=0 - 0 = 0-0 `
` M_12 = |(1, 0), (0, 1)|` `=1 - 0 = 1 `
` M_13 = |(0, 1), (0, 0)|` `=0 - 0 = 0 `
` M_21 = |(0, 0), (0, 1)|` `=0 - 0 = 0 `
` M_22 = |(1, 0), (0, 1)|` `=1 - 0 = 1 `
` M_23 = |(1, 0), (0, 0)|` `=0- 0 = 0 `
`M_31=|(0, 0),(1, 0)|` `= 0 - 0 = 0`
` M_32 = |(1, 0), (0, 0)|` `=0 - 0 = 0 `
` M_33 = |(1, 0), (0, 1)|` `=1 - 0 = 1 `
Also` A_11=(-1)^(1+1)M_11` `=1xx1 =1`
` A_12=(-1)^(1+2)M_12` `=1xx0 =0`
` A_13=(-1)^(1+3)M_13` `=1xx0 =0`
` A_21=(-1)^(2+1)M_21` `=1xx0 =0`
` A_22=(-1)^(2+2)M_22` `=1xx1 =1`
` A_23=(-1)^(2+3)M_23` `=1xx0 =0`
` A_31=(-1)^(3+1)M_31` `=1xx0 =0`
` A_32=(-1)^(3+3)M_33` `=1xx1 =1`
(ii)`|(1, 0, 4), (3, 5, -1), (0, 1, 2)|`
Solution:
(ii)Here `Delta |(1, 0, 4), (3, 5, -1), (0, 1, 2)|`
` M_11 = |(5, -1), (1, 2)|` `=10 -(-1) =11`
` M_12 = |(3, -1), (0, 2)|` `=6 - 0 = 6 `
` M_13 = |(3, 5), (0, 1)|` `=3- 0 = 3 `
` M_21 = |(0, 4), (1, 2)|` `=0 - 4 = -4 `
` M_22 = |(1, 4), (0, 2)|` `=2 - 0 = -2 `
` M_23 = |(1, 0), (0, 1)|` `=1 - 0 = 1 `
` M_31 = |(0, 4), (5, -1)|` `=0 - 20 = -20`
` M_32 = |(1, 4), (3, -1)|` `=-1 - 12= -13`
` M_33 = |(1, 0), (3, 5)|` `=5 - 0 =5 `
Also ` A_11=(-1)^(1+1)M_11` `=1xx11 =11`
` A_12=(-1)^(1+2)M_12` `=-1xx6 =-6`
` A_13=(-1)^(1+3)M_13` `=1xx3 =3`
` A_21=(-1)^(2+1)M_21` `=1xx-4 =-4`
` A_22=(-1)^(2+2)M_22` `=1xx2 =2`
` A_23=(-1)^(2+3)M_23` ` =1xx1 =-1`
` A_31=(-1)^(3+1)M_31` `=1xx-20 =-20`
` A_32=(-1)^(3+2)M_32` `=-1xx13 =13`
` A_33=(-1)^(3+3)M_33` `=1xx5 =5`
Question:3 . Using cofactors of elements of second row, evaluate
`Delta = |(5, 3, 8), (2, 0, 1), (1, 2, 3)|`
Solution:
Here `Delta = |(5, 3, 8), (2, 0, 1), (1, 2, 3)|`
`:.M_21=|(3, 8), (2, 3)|` `=9-16=-7`
`M_22=|(5, 8), (1, 3)|` `=15-8=7`
`M_23=|(5, 3), (1, 2)|` `=10-3=7`
` A_21=(-1)^(2+1)M_21` `=-1xx-7 =7`
` A_22=(-1)^(2+2)M_22` `=1xx7 =7`
` A_23=(-1)^(2+3)M_23` `=1xx7 =-7`
Now expansion of `Delta ` using cofactors of elements of second row
`Delta =a_21A_21` `+a_22A_22+a_23A_23`
`=2xx7+0xx7+1xx-7`
`=14 + 0-7=7`
Question:4 . Using cofactors of elements of third column, evaluate
`Delta = |(1, x, yz), (1, y, zx), (1, z, xy)|`
Solution:
`Delta = |(1, x, yz), (1, y, zx), (1, z, xy)|`
`:.M_13=|(1, y), (1, z)|` `=(z-y)`
`M_23=|(1,x), (1, z)|` `=(z-x)`
`M_33=|(1, x), (1, y)|` `=(y-x)`
` A_13=(-1)^(1+3)M_13` `=1xx(z-y) =(Z-y)`
` A_23=(-1)^(2+3)M_23` `=-1(z-x)=(x-z)`
` A_33=(-1)^(3+3)M_33` `=-1(y-x)=(y-x)`
Now expansion of `Delta `using cofactors of elements of third row
`Delta=a_13A_13+a_23A_23` `+a_33A_33`
`yz(z-y)+ zx(x-z)` `+xy(y-x)`
`= yz^2-y^2z+x^2z-xz^2` `+xy^2-x^2 y`
`=yz(z-y)+ x^2(z-y)` `-(z^2-y^2)`
`=(z-y)[yz+x^2-x(z+y)]`
`=(z-y)[yz=x^2-xz-xy]`
`=(z-y)[z+ (y-x)-x(y-x)]`
`=(z-y)(y-x)(z-x)`
`=(x-y)(y-z)(z-x)`
Question:5 .
If `Delta =|(a_11, a_12, a_13), (a_21, a_22, a_33), (a_31, a_32, a_33)|`
and `A_(ij)` is cofactors of `a_(ij)` then value of `Delta` is given by
(A) `a_11A_31+a_12A_32` `+a_13A_33`
(B) `a_11A_11+a_12A_21` `+a_13A_31`
(C) `a_21A_11+a_22A_12` `+a_23A_13`
(D) `a_11A_11+a_21A_21` `+a_31A_31`
Solution:
Here`Delta =|(a_11, a_12, a_13), (a_21, a_22, a_33), (a_31, a_32, a_33)|`
We know that,
`Delta = a_11 A_11 + a_12A_12 +a_13 A_13` `=a_21A_21+a_22A_22+a_23A_23`
`=a_31A_31+a_32A_32+a_33A_33` `=a_11A_11+a_21A_21+a_31A_31`
`=a_12A_12+a_22A_22+a_32A_32` `=a_13A_13+a_23A_23+a_33A_33`
Reference: