Determinants NCERT Solutions
Solution NCERT Exercise 4.2 Q:9-16
Question: 9 .
`|(x, x^2, yz),(y, y^2, zx), (z, z^2, xy)|`
`=(x-y)(y-z)(z-x)(xy+yz+zx).`
Solution:
Here L.H.S=`|(x, x^2, yz),(y, y^2, zx), (z, z^2, xy)|`
Operating `R_1-> xR_,R_2 ->yR_2 and R_3->zR_3,` we get.
`=1/(xyz)|(x^2, x^3, xyz),(y^2, Y^3, xyz),(z^2, z^3, xyz)|`
Taking xyz common from` C_3'`we get
`=1/(xyz)xx xyz|(x^2, x^3, 1),(y^2, y^3, 1),(Z^2, z^3, 1)|`
Operating`R_2-> R_2- R_1and R_3->R_3-R_1`
`= |(x^2, x^3, 1),(y^2-x^2, y^3-x^3, 0),(z^2-x^2, z^3-x^3, 0),(z^2-x^2, z^3-x^3, 0)|`
Expanding using thirdcolumn, we get
`=1|(y^2-x^2, y^3-x^3),(z^2-x^2, z^3-x^3)|`
`=[(y^2-x^2)(z^3-x^3)` `-(z^2-x^2)(y^2-x^3)]`
`=(y+x)(y-x)(z-x)(z^2+x^2+xz)-(Z+x)(z-x)(y-x(y^2+x^2+xy)`
`=(y-x)(z-x)[(y+x)` `(Z^2+x^2 +xy)-(z+x)(y^2+x^2+xy)]`
`=(y-x)(z-x)[yz^2+x^2y+xyz+xz^2+x^3+x^2z-` `y^2z-x^2z-xyz+-^2y-x^3-x^2y)]`
`=(y-x)(z-x)[yz^2+xz^2-y^2z-xy^2]`
`=(y-x)(z-x)[(yx^2-y^2z)+(xz^2-xy^2)]`
`=(y-x)(z-x)[yz(z-y)+x(z^2-y^2)]`
`=(y-x)(z-x)[yz(z-y)+x(z+y)(z-y)]`
`=(y-x)(z-x)(z-y)[yz+xz+xy]`
`=(x-y)(y-z)(z-x)(xy+yz+xz)`R.H.S.
Question:10 .
(i)`|(x+4, 2x, 2x), (2x, x+4, 2x), (2x, 2x, x+4)|` `-(5x+4)(4-x)^2-a^2`
Solution:(i)Here L.H.S=`|(x+4, 2x, 2x), (2x, x+4, 2x), (2x, 2x, x+4)|`
Operating `R_1->R_1+R_2+R_3,`we get
`|(5x+4, 5x+4, 5x+4), (2x, x+4, 2x), (2x, 2x, x+4)|`
Taking(5x+4)common from `R_1`, we get
`=(5x+4)|(1, 1, 1), (2x, x+4, 2x), (2x, 2x, x+4)|`
Operating `C_2->C_2and C_3->C_3-C_1`
`=(5x+4)|(1, 0, 0), (2x, 4-x,0) ,(2x, 0, 4-x)|`
Expanding using first row, we get
`= (5x+4)[1.|(4-x, 0 ), (0, 4-x)|]`
`=(5x+4)[(4-x)^2-0]`
`=(5x+4)(4-x)^2=R.H.S.`
(ii)`|(y+k, y, y), (y, y+k, y), (y, y, y+k)|` `=k^2(3y+k)`
Solution:Here L.H.S.`|(y+k, y, y), (y, y+k, y), (y, y, y+k)|`
Operating `R_1->R_1+R_2+R_3`,we get
`=|(3y+k, 3y+k, 3y+K),(y, y+k, y),(y, y, y+k)|`
Taking (3y+k)common from`R_1` we get
`= (3y+k)|(1, 1, 1), (y, y+k, y), (y, y, y+k)|`
Operating `C_2->C_2-C_1and C_3-> C_3-C_1`, we get
`(3y+k)|(1, 0, 0),(y, k, 0),(y, 0, k)|`
Expanding using first row, we get
`=(3y+k)[1|(k, 0), (0, k)|]`
`=(3y +k)[k^2-0]`
`=k^2(3y+K)`= R.H.S.
Question:11 .(i)
`|(a-b-c, 2a, 2a), (2b, b-c-a, 2b), (2c, 2c, c-a-b)|` `=(a+b+c)^3`
Solution:(i)
Here L.H.S.`|(a-b-c, 2a, 2a), (2b, b-c-a, 2b), (2c, 2c, c-a-b)|`
Operating `R_1->R_1+R_2+R_3`, we get
`=|(a+b+c, a+b+c, a+b+c), (2b, b-c-a, 2b), (2c, 2c, c-a-b)|`
Taking (a+b+c)common from first row, we get
`=(a+b+c)|(1, 1, 1), (2b, b-c-a, 2b), (2c, 2c, c-a-b)|`
Operating `C_2-> C_2-C_1and C_3-> C_3-C_1`, we get
`=(a+b+c)|(1, 0, 0), (2b, -c-a-b, 0), (2c, 0, -a-b-c)|`
Expanding using first row, we get
`=(a+b+c)[1|(-c-a-b, 0), (0, -a -b-c)|]`
`=(a+b+c)[|(-c-a-b)(-a-b-c)-0|]`
`=(a+b+c)(a+b+c)^2=(a+b+c)^3`=R.H.S.
Solution: (ii)
`|(x+y+2z, x, y), (z, y+z+2x, y), (z, x, z+x+2y)|` `=2(x+y+z)^3`
Solution:
Here L.H.S. =`|(x+y+2z, x, y), (z, y+z+2x, y), (z, x, z+x+2y)|`
Operating `C_1->C_1+C_2+C_3`, we get
`=|(2x+2y+2z, x, y),(2x+2y+2z, y+z+2x, y), (2x+2y+2z, x, z+x+2y)|`
Taking(2x+2y+2z)common from first column, we get
`= 2(x+y+z)` `|(1, x, y), (1, y+z+2x, y), (1, x, z+x+2y)|`
Operating`R_2->R_2-R_1and R_3->R_3-R_1`, we get
Expanding using first column, we get
`=2(x+y+z)` `[1|(x+y+z, 0), (0, x+y+z)|]`
`=2(x+y+z)[(x+y+z)^2-0]`
`=2(x+y+z)^3`=R.H.S.
Question:12 .
`|(1, x, x^2), (x^2, 1, x), (x, x^2, 1)|` `=(1-x^3)^2`
Solution:
Here L.H.S.`|(1, x, x^2), (x^2, 1, x), (x, x^2, 1)|`
operating `R_1->R_1+R_2+R_3`, we get
`=|(1+x+x^2, 1+x+x^2, 1+x+x^2), (x^2, 1, x), (x, x^2, 1)|`
Taking `(1+x+x^2)`common from first row, we get
`=(1+x+x^2)` `|(1, 1, 1), (x^2, 1, x), (x, x^2, 1)|`
Operating`C_2->C_2-C_1and C_3->C_3-C_1`, we get
`=(1+x+x^2)` `|(1, 0, 0), (x^2, 1-x^2, x-x^2), (x, x^2-x, 1-x)|`
Expanding using first row, we get
`(1+x+x^2)[1|(1-x^2, x-x^2), (x^2-x, 1-x)|`]
`(1+x+ x^2)[(1-x^2(1-x)(x-x^2)(x^2-x)]`
`(1+x+ x^2)[(1-x^2(1-x)-x(1-x)(x^2-x)]`
`(1+x+ x^2)(1-x)[1-x^2-x^3+x^2]`
`(1-x^3)(1-x^3)=(1-x^3)^2`=R.H.S.
Question: 13 .
`|(1+a^2-b^2, 2ab, -2b), (2ab, 1-a^2+b^2, 2a), (2b, -2a, 1-a^2-b^2)|` `=(1+a^2+b^2)^3`
Solution:
Here L.H.S.=`|(1+a^2-b^2, 2ab, -2b), (2ab, 1-a^2+b^2, 2a), (2b, -2a, 1-a^2-b^2)|`
Operating `R_1-> R_1+bR_3 and R_2-> R_2-aR_3`, we get
`=|(1+a^2-b^2, 0, -b(1+a^2+b^2)), (0, 1-a^2+b^2, a(1+a^2+b^2)), (2b, -2a, 1-a^2-b^2)|`
Taking `(1+a^2+b^2)`common from `R^1and R_2`, we get
`= (1+a^2+b^2)^2` `|(1, 0, -b), (0, 1, a), (2b, -2a, 1-a^2, -b^2)|`
Expanding using frst row, we get
`=(1+a^2+b^2)^2 ` `[1|(1, a), (-2a, 1-a^2-b^2)|-b|(0, 1), (2b, -2a)|]`
`=(1+a^2+b^2)^2[(1-a^2-b^2+2a^2)-b(0-2b)]`
`=(1+a^2+b^2)^2[1+a^2-b^2+2b^2]`
`=(1+a^2+b^2)^2(1+a^2+b^2)`
`=(1+a^2+b^2)^3=` R.H.S.
Question:14 .
`|(a^2+1, ab, ac), (ab, b^2+1, bc) , (ca, cb, c^2+1)|` `=1+a^2+b^2+c^2`
Here L.H.S
`=|(a^2+1, ab, ac), (ab, b^2+1, bc) , (ca, cb, c^2+1)|`
Taking a common from `R_1,b` from` R_2` and c from `R_3`, we get.
`=abc|((a^2+1)/a, b, c), (a, (b^2+1)/b, c), (a, b, (c^2+1)/1)|`
Multiplying `C_1` by a, `C_2` by b and `C_3 ` by c,we get
`=abcxx 1/(abc)|(a^2+1, b^2, c^2), (a^2, b^2+1, c^2), (a^2, b^2, C^2+1)|`
Operating`C_1->C_1+ C_2+C_3`, we get
`|(1+a^2+b^2+c^2, b^2, c^2), (1+a^2+b^2+c^2, b^2+1, c^2), (1+a^2+b^2+c^2, b^2, c^2+1)|`
Taking `(1+a^2+b^2+c^2)`common from `C_1` we get
`=(1+a^2+b^2+c^2)` `|(1, b^2, c^2), (1, b^2+1, c^2), (1, b^2, c^2+1)|`
Operating `R_2->R_2-R_1 and R_3->R_3-R_1`, we get
`=(1+a^2+b^2+c^2)|(1, b^2, c^2), (0, 1, 0),(0, 0,1)|`
Expanding using first column, we get
`=(1+a^2+b^2+c^2)[1|(1, 0), (0,1)]`
`=(1+a^2+b^2+c^2)(1-0)=1+a^2+b^2+c^2`=R.H.S.
Choose the correct answer in Exercises 15 and 6.
Question:15 Let A be a square matrix of order 3X3, then |kA| is equal to
(A)k|A|
(B)`k^2|A|`
(C)`k^3|A|`
(D) `3k|A|`
Solution:
Let, `A=[(a_1, a_2, a_3), (b_1, b_2, b_3),(c_1, c_2, c_3)]` be a square matrix of order 3.
Now, `kA=k[(a_1, a_2, a_3), (b_1, b_2, b_3),(c_1, c_2, c_3)]` `[(ka_1, ka_2, ka_3), (kb_1, kb_2, kb_3),(kc_1, kc_2, kc_3)]`
`:. |kA|=[(ka_1, ka_2, ka_3), (kb_1, kb_2, kb_3),(kc_1, kc_2, kc_3)]`
Taking k as common from R1, R2 and R3, we get
`|kA|=k xx k xx k [(a_1, a_2, a_3), (b_1, b_2, b_3),(c_1, c_2, c_3)]`
`=k^3|A|`
Thus, answer is (C)`k^3|A|`
Question: 16. Which of the following is correct:
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix
(C) Determinant is a number associated to a square matrix
(D) None of these
Solution:
Since a determinant is a number associated to a square matrix, thus answer is option (C)
Reference: