Determinants NCERT Solutions

NCERT Miscellaneous Exercise

Question:1 .

Prove that the determinant`|(x, sintheta, costheta), (-sintheta, -x, 1), (costheta, 1, x)|`is independent of `theta`.

Solution:

Let A=`|(x, sintheta, costheta), (-sintheta, -x, 1), (costheta, 1, x)|`

Expanding using first row, we get

`x(-x^2-1)-sintheta(-xsintheta-costheta)` `+costheta(-sintheta+xcostheta)`

`=-x^3-x+xsin^2theta` `+sintheta costheta-sintheta costheta+xcos^2theta`

`=-x^3-x+x(sin^2theta+cos^2theta)`

`=-x^3-x+x=-x^3`

Which is independent of theta.

Question:2 .

Without expanding the determinant, prove that `|(a, a^2, bc), (b, b^2, ca), (c, c^2, ab)|`

`=|(1, a^2, a^3), (1, b^2, b^3), (1, c^2, c^3)|`

Solution:

`|(a, a^2, bc), (b, b^2, ca), (c, c^2, ab)|`

Multiplying `R_1`by a, `R_2` by b and `R_3`by c, we get

`=1/(abc)|(a^2, a^3, abc), (b^2, b^3, abc), (c^2, c^3, abc)|`

Taking abc common from `C_3` we get

`=1/(abc)xxabc|(a^2, a^3, 1), (b^2, b^3, 1), (c^2, c^3, 1)|`

Operating `C_2harrC_3`

`=-|(a^2, 1, a^3), (b^2, 1, b^3), (c^2, 1, c^3)|`

Operating `C_1harrC_2`determinant `=|(1, a^2, a^3),(1, b^2, b^3),(1, c^2, c^3)|`

Thus `|(a, a^2, Bc), (b, b^2, ca), (c, c^2, ab)|`

`=|(1, a^2, a^3), (1, b^2, b^3), (1, c^2, c^3)|`

Question:3 .

Evaluate `|(cos alpha cos beta, cos alpha sin beta, -sin alpha), (-sin beta, cos beta, 0), (sin alpha cos beta, sin alpha sin beta, cos alpha)|`

Solution:

Let A`=|(cos alpha cos beta, cos alpha sin beta, -sin alpha), (-sin beta, cos beta, 0), (sin alpha cos beta, sin alpha sin beta, cos alpha)|`

Expanding using first row, we get

`=cos alpha cos beta (cos alpha cos beta-0)-cos alpha sin beta` ` (-sin beta cos alpha-0)-sin alpha`

`(-sin alpha sin^2 beta -sin alpha cos^2beta)`

`=cos^2 alpha cos^2 beta+cos^2 alpha sin^2 beta` `+Sin^2 alpha sin^2 beta +sin^2alpha cos^2 beta`

`=Cos^2 alpha (cos^2 beta +sin^2 beta)+` ` sin^2alpha (sin^2beta +cos ^2 beta)`

`=cos^2 alpha +sin^2 alpha (:. sin ^2 theta+cos^2 theta=1)`

Question:4 .

If a, b and c are real numbers, and

`Delta= |(b+c, c+a, a+b),(c+a, a+b, b+c), (a+b, b+c, c+a)|=0.`

Solution:

Here `Delta= |(b+c, c+a, a+b),(c+a, a+b, b+c), (a+b, b+c, c+a)|=0`

Operating `R_1->R_1 + R_2+R_3`

`Delta =|(2(a+b+c), 2(a+b+c), 2(a+b+c)),(c+a, a+b, b+c), (a+b, b+c, c+a)|=0`

Taking 2(a+b+c)common from `R_1`

`Delta = 2(a+b+c)|(1, 1, 1), (c+a, a+b, b+ c),(a+b, b+c, c+a)|=0`

Operating `C_2->C_2-C_1` and `C_3->C_3-C_1`

`Delta = 2(a+b+c)|(1, 0, 0), (c+a, b-c, b-a),(a+b, c-a, c-b)|=0`

Expanding using first row,

`Delta=2(a+b+c)` `[1{(b-c)(c-b)-(c-a)(b-a)}]=0`

=`2(a+b+c)` `[bc-b^2-c^2+bc-bc+ac+ab-a^2]=0`

`=-2(a+b+c)` `(a^2+b^2+c^2-ab-bc-ac)=0`

`:. a+b+c=0 ` `or a^2+b^2+c^2-ab-bc-ac=0`

Question:5 .

Solve the equation`|(x+a, x, x), (x, x+a, x), (x, x, x+a)|=0,a!=0`

Solution:

Let A=`|(x+a, x, x), (x, x+a, x), (x, x, x+a)|=0`

Operating `R_1->R_1 + R_2+R_3`

`A=|(3x+a, 3x+a, 3x+a), (x, x+a, x), (x, x, x+a)|=0`

Taking (3x+a)common from `R_1`

`=(3x+a)|(1, 1, 1), (x, x+a, x), (x, x, x+a)|=0`

Operating `C_2->C_2-C_1` and `C_3->C_3-C_1`

=`(3x =a)|(1, 0, 0), (x, a, 0), (x, 0, a)|=0`

Expanding using first row,

=`(3x+a)[1(a^2-0)]=0`

=`(3x+a)(a^2)=0`

`:.3x+a=0`

`=>x=a/3`

Question:6 .

Prove that

`|(a^2, bc, ac+c^2), (a^2+ab, b^2, ac), (ab, b^2+bc, c^2)|=4a^2b^2c^2`

Solution:

Let A= `|(a^2, bc, ac+c^2), (a^2+ab, b^2, ac), (ab, b^2+bc, c^2)|=0`

Taking a common from `C_1b`from`C_2 ` and c from `c_3`

`=abc |(a, c, a+c), (a+b, b, a), (b, b+c, c)|`

Operating `C_1->C_1-C_2-C_3`

`=abc|(-2c, c, a+c), (0, b, a), (-2c, b+c, c)|`

Taking `-2c` common from `C_1`

`=-2abc^2|(1, c, a+c), (0, b, a), (1, b+c, c)|`

Operating `R_3->R_3-R_1`

`=-2abc^2|(1, c, a+c),` `(0, b, a), (0, b, -a)|`

Expanding using first column,

`=-2abc^2[1(-ab-ab)]`

`=-2ab^2cxx-2ab=4a^2b^2c^2`

Question:7 .

If `A^-1= [(3, -1, 1), (-15, 6, -5), (5, -2, 2)]` and B = `[(1, 2, -2), (-1, 3, 0), (0, -2, 1)]`find `(AB)^-1`

Solution:

Here `A^-1[(3, -1, 1), (-15, 6, -5), (5, -2, 2)]=0`

and `B= [(1, 2, -2), (-1, 3, 0), (0, -2, 1)]`

`:.|B|= [(1, 2, -2), (-1, 3, 0), (0, -2, 1)]`

`=1(3-0)-2` `(-1-0)-2(2-0)`

`=3+2-4=1`

`B_11=(3-0)`

`=3, B_12=-(-1-0)`

`=1,B_13=(2-0)=2`

`B_21 -(2-4)`

`=2, B_22=(1-0)`

`=1, B_23=-(-2-0)=2`

`B_31 =(0+6)`

`=6, B_32=-(0-2)`

`=2, B_33=(3+2)=5`

`:.adj\ B= [(3, 1, 2), (2, 1, 2), (6, 2, 5)]`

`= [(3, 2, 6), (1, 1, 2), (2, 2, 5)]`

`:. B^-1= 1/|B| adj\ B`

=`1/1 [(3, 2, 6), (1, 1, 2), (2, 2, 5)]`

`[(3, 2, 6), (1, 1, 2), (2, 2, 5)]`

Now `(AB)^-1=B^-1A^-1`

`[(3, 2, 6), (1, 1, 2), (2, 2, 5)][(3, -1, 1), (-15, 6, -5), (5, -2, 2)]`

`=[(9-30+30, -3+12-12, 3-10+12), (3-15+10, -1+6-4, 1-5+4), (6-30+25, -2+12-10, 2-10+10) ]`

=`[(9, -3, 5), (-2, 1, 0), (1, 0, 2)]`

Question:8 .

Let A=`[(1, -2, 1), (-2, 3, 1), (1, 1, 5)]`, verify that

(i) `[adj\ A]^-1= adj\ (A^-1)`

(ii) `(A^-1)^-1=A`

Solution:

Here A= `[(1, -2, 1), (-2, 3, 1), (1, 1, 5)]`

`:. |A|= |(1, -2, 1), (-2, 3, 1), (1, 1, 5)|`

`=1(15-1)-(-2)(-10-1)+1(-2-3)`

`=14-22-5=-13`

`A_11=(15-1)`

`=14,A_12=-(-10-1)`

`=11,A_13=(-2-3)=-5`

`A_21=-(-10-1)`

`=11,A_22=-(5-1)`

`=4,A_23=-(1+2)=-3`

`A_31=(-2-3)`

`=-5,A_32=-(1+2)`

`=-3,A_33=(3+2)=-1`

`=adj A=[(14, 11, -5), (11, 4, -3), (-5, -3, -1)]`

`=[(14, 11, -5), (11, 4, -3), (-5, -3, -1)]`

`:.A^-1=1/|A| adj A`

`=1/13[(14, 11, -5), (11, 4, -3), (-5, -3, -1)]`

`=1/13[(-14, -11, 5), (-11, -4, 3), (5, 3, 1)]`

(i)Let `adj A = B =[(14, 11, -5), (11, 4, -3), (-5, -3, -1)]`

`:. |B|=[(14, 11, -5), (11, 4, -3), (-5, -3, -1)]`

`=14(-4-9)-11(-11-15)` `-5(-33+20)`

`=-182+286+65=169`

`B_11=(-4-9)`

`=-13, B_12=-(-11-15)`

`=-26,B_13=(-33+20)=20-13`

`=B_21 =-(-11-15)`

`=-26,B_22 =(-14-25)`

`=-39,B_23=-(-42+55)=-13`

`B_31=(-33+20)`

`=-13,B_32=-(42+55)`

`=-13,B_33=56-121=-65`

`:.adj B=[(-13, 26, -13), (-26, -39, -3), (-13, -13, -65)]`

`=[(-13, -26, -13), (26, -39, -13), (-13, -13, -65)]`

`:. B^-1=1/|B| adj B`

`=1/169 [(-13, -26, -13), (26, -39, -13), (-13, -13, -65)]`

`1/13 [(-1, -2, -1), (2, -3, -1), (-1, -1, -5)]`

`:. (adj A)^-1 =1/13[(-1, -2, -1), (2, -3, -1), (-1, -1, -5)]`

Let `A^-1=C=1/13[(-14, -11, 5), (-11, -4, 3), (-5, 3, 1)]`

`:.|C|=|(-14/13, -11/13, 5/13), (-11/13, -4/13, 3/13), (5/13, 3/13, 1/13) |`

`=-14/13[(-4/169-9/169)]` `+11/13(-11/169-15/169)` `+5/13(-33/169+20/169)`

`=-14/13xx13/169` `xx11/13xx-26/169+5/13xx-13/169`

`=14/169-22/269-5/169`

`=-13/169=-1/13`

`C_11=-4/169-9/169`

`=-13/169C_12`

`=-(-11/169-15/169)` `=26/269`

`C_13=-33/169+20/169`

`=-13/169C_21`

`=-(-11/169-15/169)`

`=-26/129`

`C_22=(-14/169-25/169)`

`=-39/169 C_23`

`=-(-42/169+55/169)`

`=-13/169`

`C_31=(-33/169+20/169)`

`=-13/169 C_32`

`=-(-42/169+55/169)`

`=-13/169`

`C_33=(56/169-121/169)`

`=65/169`

`:. adj C=[(-1/13, 2/13, -1/13), (-2/13, -3/13, -1/13), (-1/13, -1/13, -5/13)]`

`=[(-1/13, 2/13, -1/13), (-2/13, -3/13, -1/13), (-1/13, -1/13, -5/13)]`

`1/13 [(-1, -2, -1), (2, -3, -1), (-1, -1, -5)]`

Thus `(adj\ A)^-1 =adj\ (A^-1)`

(ii) Now `|A^-1|=1/13 ` and `adj(A^-1)=1/13 [(-1, -2, -1), (2, -3, -1), (-1, -1, -5)]`

`:.(A^-1)^-1=1/|A^-1| adj(A^-1)`

`=1/1/-13 xx1/13[(-1, -2, -1), (2, -3, -1), (-1, -1, -5)]`

`=[(1, 2, 1), (-2, 3, 1), (1, 1, 5)]=A`

Thus `(A^_1)^-1=A`

12-math-home

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