Determinants NCERT Solutions
NCERT Miscellaneous Exercise Q9-19
Question:9 .
Evaluate `|(x, y, x+y), (y, x+y, x), (x+y, x, y)|`
Solution:
Let A=`|(x, y, x+y), (y, x+y, x), (x+y, x, y)|`
Operating`R_1-> R_1+R_2+R_3`,
`=|(2x+2y, 2x+2y, 2x+2y), (y, x+y, x), (x+y, x, y)|`
Taking 2(x+y)Common from `R_1`
`=2(x+y)|(1, 1, 1), (y, x+y, x), (x+y, x, y)|`
Operating`C_2->C_2-C_1` and `C_3->C_3-C_1`
`Delta -2(x+y)|(1, 0, 0), (y, x, x-y), (x+y, -y, -x)|`
Expanding using first row.
`=2(x+y)[(-x^2+xy-y^2)]`
`=-2(x+y)(x^2-y^2-xy)` `-2(x^3+y^3)`
Question:10 .
Evaluate`|(1, x, y), (1,x+y, y), (1, x, x+y)|`
Solution:
Let A= `|(1, x, y), (1, x+y, y), (1, x, x+y)|`
Operating`R_2->R_2-R_1` and `R_3->R_3-R_1`
`Delta= |(1, x, y), (0, y, 0),(0, 0, x)|`
Expanding using first column.
`=(xy - 0)= xy`
Using properties of determinants in Exercises 11 to 15 prove that:
Question:11 .
`|(alpha, alpha^2, beta+gamma), (beta, beta^2, gamma+ alpha), (gamma, gamma^2, alpha+beta)|`
`=(beta-gamma)(gamma- alpha)(alpha-beta)(alpha+beta+gamma)`
Solution:
Let A=` |(alpha, alpha^2, beta+gamma), (beta, beta^2, gamma+ alpha), (gamma, gamma^2, alpha+beta)|`
Operating `C_3->C_3+C_1`
`=|(alpha, alpha^2, alpha+beta+gamma), (beta, beta^2, alpha+beta+gamma), (gamma, gamma^2, alpha+beta+gamma)|`
Taking `(alpha+gamma)` common from `C_3`
`=(alpha+beta+gamma)` `|(alpha, alpha^2, 1), (beta, beta^2, 1), (gamma, gamma^2, 1)|`
Operating`R_2->R_2-R_1 and R_3->R_3-R_1`
`=(alpha+beta+gamma)` `|(alpha, alpha^2, 1), (beta-alpha, beta^2-alpha^2, 0), (gamma-alpha, gamma^2-alpha^2, 0)|`
Expanding using third column
`=(alpha+beta+gamma)` `[(beta-alpha)(gamma^2-alpha^2)-(gamma-alpha)(beta^2-alpha^2)]`
`=(alpha+beta+gamma)` `[(beta-alpha)(gamma+alpha)(gamma-alpha)-(gamma-alpha)(beta+alpha^2)(beta-alpha)]`
`=(alpha+beta+gamma)` `(beta-alpha)(gamma-alpha)[(gamma+alpha)-(beta+alpha)]`
`=(alpha+beta+gamma)` `(beta-alpha)(gamma-alpha)[(gamma+alpha-beta-alpha)]`
`=(alpha+beta+gamma)` `(beta-alpha)(gamma-alpha)-(gamma-beta)`
`=(alpha+beta+gamma)` `(alpha-beta)(beta-gamma)(gamma-alpha)`
Question:12 .
`|(x, x^2, 1+px^3), (y, y^2, 1+py^3), (z, z^2, 1+pz^3)|`
=`(1+pxyz)(x-y)(y-z(z-x)`
Solution:
Let `A=
`|(x, x^2, 1+px^3), (y, y^2, 1+py^3), (z, z^2, 1+pz^3)|`
=`|(x, x^2, 1), (y, y^2, 1), (z, z^2, 1)|+|(x, x^2, 1+px^3), (y, y^2, 1+py^3), (z, z^2, 1+pz^3)|`
In first deteerminant `C_1harrC_3`and in second determinant taking xcommon from `R_1,y` from `R_2 ` and zfrom `R_3`
`|(1, x^2, x), (1, y^2, y), (1, z^2, z)|` `+xyz|(1, x, px^2), (1, y, py^2), (1, z, pz^2)|`
In first deteerminant `C_2harrC_3`and in second determinant taking xcommon from`C_3`
`=-|(1, x, x^2), (1, y, y^2), (1, z, z^2)|` `+pxyz|(1, x, x^2), (1, y, y^2), (1, z, z^2)|`
=`(1+pxyz)|(1, x, x^2), (1, y, y^2), (1, z, z^2)|`
Operating`R_2->R_2-R_1` and `R_3->R_3-R_1`
`=(1+pxyz)|(1, x, x^2), (0, y-x, y^2-x^2), (0, z-z, z^2-x^2)|`
Expanding using first column
`(1+pxyz)[(y-x)(z^2-x^2)` `-(z-x)(y^2-x^2)]`
`(1+pxyz)[(y-x)(z+x)(z-x)` `-(z-x)(y+x)(y-x)]`
`(1+pxyz)(y-x)(z-x)` `[(z+x)-(y+x)]`
`(1+pxyz)(y-x)(z-x)` `[z+x-y-x]`
`(1+pxyz)(y-x)(z-x)` `-(z-y)`
`(1+pxyz)(x-y)(y-z)(z-x)`
Question: 13.
` |(3a, -a+b, -a+c), (-b+a, 3b, -b+c), (-c+a, -c+b, 3c)|`
`=3(a+b+c)(ab+bc+ca)`
Solution:
Let A=` |(3a, -a+b, -a+c), (-b+a, 3b, -b+c), (-c+a, -c+b, 3c)|`
Operating `C_1->C_1+C_2+C_3`
`|(a+b+c, -a+b, -a+c), (a+b+c, 3b, -b+c), (a+b+c, -c+b, 3c)|`
Taking (a+b+c)common from `C_1`
`=(a+b+c)|(1, -a+b, -a+c), (1, 3b, -b+c), (1, -c+b, 3c)|`
Operating`R_2->R_2-R_1` and `R_3->R_3-R_1`
`=(a+b+c)|(1, -a+b, -a+c), (0, a+2b, -b+a), (0, -c+a, 2c+a)|`
Expanding using first column
`=(a+b_c)[(a+2b)(2c+a)-(-c+a)(-b+a)]`
`=(a+b+c)[2ac+a^2+4bc+2ab-(bc-ac-ab+a^2)]`
`=(a+b+c)[2ac+a^2+4bc+2ab-bc+ac+ab-a^2]`
`=(a+b+c)(3ab+3bc+3ac)`
`=3(a+b+c)(ab+bc+ca)`
Question:14 .
Solution:
Let A= `|(1, 1+p, 1+p+q), (2, 3+2p, 4+3P+2q), (3, 6+3p, 10+6p+3q)|=1`
`|(1, 1+p, 1+p+q), (2, 3+2p, 4+3P+2q), (3, 6+3p, 10+6p+3q)|`
Operating`R_2->R_2-2R_1` and `R_3->R_3-3R_1`
=`|(1, 1+p, 1+p+q), (0, 1, 2+p), (0, 3, 7+3p)|`
Expanding using first column
`=[(7+3p)-3(2+p)]`
`(7+3p-6-3p)=1`
Question:15 .
`|(sin alpha, Cos alpha, Cos(alpha+delta)), (sin beta, cos beta, cos(beta+delta)), (sin gamma, cos gamma, cos(gamma+delta))|=0`
Solution:
Let A=
`|(sin alpha, Cos alpha, Cos(alpha+delta)), (sin beta, cos beta, cos(beta+delta)), (sin gamma, cos gamma, cos(gamma+delta))|`
`=|(sin alpha, Cos alpha, Cos alpha cos delta-sin alpha sin delta), (sin beta, cos beta, cos beta cos delta- sin beta sin delta), (sin gamma, cos gamma, cosgamma cos-sin gamma sin delta)|`
Operating `C_3->C_3-cos deltaC_2+sin deltaC_1`
=`|(sin alpha, cos alpha, 0), (sin beta, cos beta, 0), (sin gamma, cos gamma, 0)|`=0
Question:16 .
Solve the system of the following equations:
`2/x+3/y+10/z=4;` `4/x-6/y+5/z=1;` `6/x+9/y-20/z=2`
Solution:
`2/x+3/y+10/z=4;` `4/x-6/y+5/z=1;` `6/x+9/y-20/z=2`
put `1/a=a, 1/y=b and 1/z=c`, we get
`2a+3b+10c=4;4a-6b+5c=` `1;6a+9b-20c=2`
The system of equations can be written inthe form AX=B, where
A=`[(2, 3, 10), (4, -6, 5), (6, 9, -20)],X`
`=[(a), (b), (c)] and B=[(4), (1), (2)]`
Now `|A| =[(2, 3, 10), (4, -6, 5), (6, 9, -20)]`
`2(120-45)-3(-80-30)` `+10(36+36)`
`=150+330+720-1200`
`A_11=(120-45)`
`=75, A_12=-(-80-30)`
`=110,A_13=(36+36)=72`
`A_21=(-60-90)`
`=150, A_22=(-40-60)`
`=-100,A_23=-(18-18)=0`
`A_31=(15+60)`
`=75, A_32=-(-10-40)`
`=30,A_33=(36+36)=-24`
`:.adj A=[(75, 110, 72), (150, -100, 0), (75, 30, -24)]`
=`[(75, 150, 75) (110, -100, 30), (72, 0, -24)]`
Now `A^-1=1/|A|(adj A)`
=`1/1200[(75, 150, 75), (110, -100, 30), (72, 0, -24)]`
`X=A^-1B`
`=[(a), (b), (c)]` `=1/1200[(75, 150, 75), (110, -100, 30), (72, 0, -24)][(4), (1), (2)]`
`=1/1200[(300+150+150), (440-100+60), (288+0-48)]` `1/1200[(600), (400), (240)]`
` =[(1/2), (1/3), (1/5)]`
`:.a=1/2, b=1/3, c=1/5`
But`x=1/a=2,y=1/b`
`=3and z=1/c=5`
Choose the correct answer in Exercises 17 to 19
Question:17 .
If a, b, care in A, p, then the determinant`|(x+2, x+3, x+2a), (x+3, x+4, x+2b), (x+4, x+5, x+2c)|` is
(A)0
(B)1
(c)x
(D)2x
Solution:
Let A= `|(x+2, x+3, x+2a), (x+3, x+4, x+2b), (x+4, x+5, x+2c)|`is
:Operating`R_2->2R_2-R_1-R_3`
`=1/2|(x+2, x+3, x+2a), (0, 0, 2b-a-c), (x+4, x+5, x+2c)|`
But a, b, care in AP,`:. 2b=a+c`
`=1/2|(x+2, x+3, x+2a), (0, 0, 0), (x+4, x+5, x+2c)|=0`
Thus Answer is (A)
Question:18 .
If x, y, z are non zero real numbers, then the inverse of matrix `A=[(x, 0, 0), (0, y, 0), (0, 0, z)]` is
(A)`[(x^-1, 0, 0), (0, y^-1, 0), (0, 0, z^-1)]`
(B)xyz`[(x^-1, 0, 0), (0, y^-1, 0), (0, 0, z^-1)]`
(C)`1/(xyz)[(x, 0, 0), (0, y, 0), (0, 0, z)]`
(D)`1/(xyz)[(1, 0, 0), (0, 1, 0), (0, 0, 1)]`
Solution:
Here `A=[(x, 0, 0), (0, y, 0), (0, 0, z)]`
`|A|[(x, 0, 0), (0, y, 0), (0, 0, z)]`
=`x(yz-0)=xyz`
`A_11=(yz-0)`
`=yz,A_12=-(0-0)`
`=0,A_13=(0-0)=0`
`A_21=-(0-0)`
`=0,A_22=-(xz-0)`
`=xz,A_23=(0-0)=0`
`A_31=(0-0)`
`=0,A_32=-(0-0)`
`=0,A_33=(xy-0)=xy`
`:. adj A= [(yz, 0, 0), (0, xz, 0), (0, 0, xy)]`
=`[(yx, 0, 0), (0, xz, 0), (0, 0, xy)]`
`:.A^-1=1/|A| (adj A)`
`=1/(xyz)[(yz, 0, 0), (0, xz, 0), (0, 0, xy)]`
`[(1/x, 0, 0), (0, 1/y, 0), (0, 0,1/z)]`
`[(x^-1, 0, 0), (0, y^-1, 0), (0, 0, z^-1)]`
Thus answer is (A)
Question:19 .
Let A `[(1, sin theta, 1), (-sin theta, 1, sin theta), (-1, -sin theta, 1)]`, Where `0<=\theta\<=\2pi` then
(A) Det (A) = 0
(B) Det (A) `in (2,oo)`
(C) Det (A) `in (2,6)`
(D) Det (A) `in (2,4)`
Solution:
Here, `A = [(1, sin theta, 1), (-sin theta, 1, sin theta), (-1, sin theta, 1) ]`
`|A| = |(1, sin theta, 1), (-sin theta, 1, sin theta), (-1, sin theta, 1) |`
`=1(1+sin^2 theta)` `-sintheta(-sintheta+sintheta)` `+1(sin^2 theta+1)`
`=1+sin^2 theta + sin^2 theta +1`
`=2+2sin^2theta `
`=2(1+sin^2theta)`
For `theta = 0, pi, 2pi`
`|A| = 2(1+0)=2`
For `theta = pi/2, (3pi)/2`
`|A| = 2(1+1) = 4`
`:. |A| in [2,4]`
Thus, option (D) Det (A) `in (2,4)` is the correct Answer
Reference: