Using the properties of determinants in Exercises 1 to 6, evaluate:
Q.1.
Ans.
Let
C_{1} → C_{1} - C_{2}
= (x + 1) (x^{2} – 2x + 2) – 0
= x^{3} – 2x^{2} + 2x + x^{2} – 2x + 2 = x^{3} – x^{2} + 2
Q.2.
Ans.
Let
C_{1} → C_{1} + C_{2} + C_{3}
(Taking a + x + y + z common)
R_{1} → R_{1} - R_{2}, R_{2} → R_{2} - R_{3}
Expanding along C_{1} = (a + x+y +z) = a^{2}(a + x + y + z)
Q.3.
Ans.
Let
Taking x^{2}, y^{2} and z^{2} common from C_{1}, C_{2} and C_{3} respectively
Expanding along R_{1}
= x^{2}y^{2}z^{2} [- x(0 - yz) + x( yz - 0)]
= x^{2}y^{2}z^{2}(xyz + xyz) = x^{2}y^{2}z^{2}(2xyz) = 2x^{3}y^{3}z^{3}
Q.4.
Ans.
Let
C_{1} → C_{1} + C_{2} + C_{3}
Taking (x + y + z) common from C_{1}
R_{1} → R_{1} - R_{2}, R_{2} → R_{2} - R_{3}
Expanding along C_{1}
= (x+y+ z) [(-x - 2y)(-y - 2z) -(2y +z)(-x +y)]
= (x + y + z) (xy + 2zx + 2y^{2} + 4yz + 2xy – 2y^{2} + zx – zy)
= (x + y + z) (3xy + 3zx + 3yz) = 3(x + y + z) (xy + yz + zx)
Q.5.
Ans.
Let
C_{1} → C_{1} + C_{2} + C_{3}
Taking (3x + 4) common from C_{1}
R_{1} → R_{1} - R_{2}, R_{2} → R_{2} - R_{3}
Expanding along C_{1}
Q.6.
Ans.
Let
R_{1} → R_{1} + R_{2} + R_{3}
Taking (a + b + c) common from R_{1}
C_{1} → C_{1} - C_{2}, C_{2} → C_{2} - C_{3}
Taking (b + c + a) from C_{1} and C_{2}
Expanding along R_{1}
Using the properties of determinants in Exercises 7 to 9, prove that:
Q.7.
Ans.
L.H.S. =
R_{1} → xR_{1}, R_{2} → yR_{2}, R_{3} → zR_{3} and dividing the determinant by xyz.
Taking xyz common from C_{1} and C_{2}
C_{3} → C_{3} + C_{1}
Taking (xy + yz + zx) common from C_{3}
[∵ C_{2} and C_{3} are identical]
L.H.S. = R.H.S. Hence proved.
Q.8.
Ans.
L.H.S. =
C_{1} → C_{1} - (C_{2} + C_{3})
Taking –2 common from C_{1}
R_{2} → R_{2} - R_{3}
Expanding along C_{1}
= - 2 [ x - zy- zy ] = – 2(– 2xyz) = 4xyz R.H.S.
L.H.S. = R.H.S.
Hence, proved.
Q.9.
Ans.
L.H.S.=
R_{1} → R_{1} - R_{2}, R_{2} → R_{2} - R_{3}
Taking (a – 1) common from C_{1} and C_{2}
Expanding along C_{3}
= (a – 1)^{2} (a + 1 – 2) = (a – 1)^{2} (a – 1) = (a – 1)^{3} R.H.S.
L.H.S.= R.H.S.
Hence, proved.
Q.10. If A + B + C = 0, then prove that
Ans.
L.H.S. =
Expanding along C_{1}
= 1(1 – cos^{2} A) – cos C(cos C – cos A cos B) + cos B (cos A cos C – cos B)
= sin^{2} A – cos^{2 }C + cos A cos B cos C + cos A cos B cos C – cos^{2} B
= sin^{2} A – cos^{2} B – cos^{2} C + 2 cos A cos B cos C
= – cos (A + B) × cos (A - B) - cos^{2} C + 2 cos A cos B cos C
[∵ sin^{2} A - cos^{2} B= - cos (A +B)× cos (A - B)]
= - cos ( - C) × cos (A - B) + cos C (2 cos A cos B - cos C) [∵ A + B + C = 0]
= – cos C(cos A cos B + sin A sin B) + cos C(2 cos A cos B – cos C)
= – cos C(cos A cos B + sin A sin B – 2 cos A cos B + cos C)
= – cos C(– cos A cos B + sin A sin B + cos C)
= cos C(cos A cos B – sin A sin B – cos C)
= cos C[cos(A + B) – cos C ]
= cos C[cos (– C) – cos C] [∵ A + B = - C]
= cos C[cos C - cos C] = cos C × 0 = 0 R.H.S.
L.H.S. = R.H.S.
Hence, proved.
Q.11. If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), then
Ans.
Area of triangle whose vertices are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})
Let⇒
But area of equilateral triangle whose side is ‘a‘ =
Hence, proved.
Q.12. Find the value of θ satisfying
Ans.
Let
C_{1} → C_{1} - C_{2}
⇒
Taking 7 common from C_{1}
⇒
⇒
Expanding along C_{1}
⇒
⇒ –2 + 7 sin 3θ + 2 (cos 2θ – 3 sin 3θ) = 0
⇒ – 2 + 7 sin 3θ + 2 cos 2θ – 6 sin 3θ = 0
⇒ – 2 + 2 cos 2θ + sin 3θ = 0
⇒ – 2 + 2 (1 – 2 sin^{2} θ) + 3 sin θ – 4 sin^{3} θ = 0
⇒ – 2 + 2 – 4 sin^{2} θ + 3 sin θ – 4 sin^{3} θ = 0
⇒ – 4 sin^{3} θ – 4 sin^{2} θ + 3 sin θ = 0
⇒ – sin θ(4 sin^{2} θ + 4 sin θ – 3) = 0
sin θ = 0 or 4 sin^{2} θ + 4 sin θ – 3 = 0
∴ θ = nπ or 4 sin^{2} θ + 6 sin θ – 2 sin θ – 3 = 0 when n ∈ I
⇒ 2 sin θ(2 sin θ + 3) - 1 (2 sin θ + 3) = 0
⇒ (2 sin θ + 3)(2 sin θ – 1) = 0
⇒ 2 sin θ + 3 = 0 or 2 sin θ - 1 = 0
is not possible as – 1 ≤ x ≤ 1
Hence,
Q.13. If, then find values of x.
Ans.
Let
R_{1} → R_{1} + R_{2} + R_{3}
⇒
Taking (12 + x) common from R_{1},
⇒
C_{1} → C_{1} - C_{2}, C_{2} → C_{2} - C_{3}
Expanding along R_{1}
⇒
(12 + x) (4x^{2} – 0) = 0 ⇒ 12 + x = 0 or 4x^{2} = 0
⇒ x = -12 or x = 0
Q.14. If a_{1}, a_{2}, a_{3}, …, a_{r} are in G.P., then prove that the determinant
is independent of r.
Ans.
If a_{1}, a_{2}, a_{3}, … a_{r }be the terms of G.P., then
a_{n} = AR^{n–1}
(where A is the first term and R is the common ratio of the G.P. )
∴a_{r+1} = AR^{r+1–1} = AR^{r}; a_{r+5 }= AR^{r+5–1} = AR^{r+4}
a_{r+9} = AR^{r +9-1} = AR^{r +8} ; a_{r+7} = AR^{r+7–1} = AR^{r+6}
a_{r+11} = AR^{r+11–1} = AR^{r+10}; a_{r+15} = AR^{r+15–1} = AR^{r+14}
a_{r+17} = AR ^{r +17-1} = AR^{r +16} ; a_{r+21} = AR^{r+21–1} = AR_{r+20 }
∴ The determinant becomes
Taking AR^{r}, AR^{r+6} and AR^{r+10} common from R_{1}, R_{2} and R_{3} respectively.
[∵ R_{1} and R_{2} are identical rows]
= 0
Hence, the given determinant is independent of r.
Q.15. Show that the points (a + 5, a – 4), (a – 2, a + 3) and (a, a) do not lie on a straight line for any value of a.
Ans.
If the given points lie on a straight line, then the area of the triangle formed by joining the points pairwise is zero.
R_{1} → R_{1} - R_{2}, R_{2} → R_{2} - R_{3}
⇒
Expanding along C_{3}
As 7 ≠ 0. Hence,, the three points do not lie on a straight line for any value of a.
Q.16. Show that the ∆ABC is an isosceles triangle if the determinant
Ans.
C_{1 }→ C_{1} - C_{2}, C_{2} → C_{2} - C_{3}
Taking (cos A – cos B) and (cos B – cos C) common from C_{1} and C_{2} respectively.
⇒ (cos A - cos B) (cos B- cos C) [cos B + cos C + 1 - cos A - cos B - 1] = 0
⇒ (cos A – cos B) (cos B – cos C) (cos C – cos A) = 0
cos A – cos B = 0 or cos B – cos C = 0
or cos C – cos A = 0
⇒ cos A = cos B or cos B = cos C or cos C = cos A
⇒ ∠A = ∠C or ∠B = ∠C ⇒ ∠A = ∠B
Hence, ΔABC is an isosceles triangle.
Q.17. Find A^{–1} ifand show that
Ans.
Here,
= 0 – 1 (0 – 1) + 1 (1 – 0)
= 1 + 1 = 2 ≠ 0 (non-singular matrix.)
Now, co-factors,
∴
Now, A^{2} =
Hence, A^{2} =
Now, we have to prove that
Hence, proved.
Long Answer (L.A.)
Q.18. Iffind A^{–1}. Using A^{–1}, solve the system of linear equations x – 2y = 10 , 2x – y – z = 8 , –2y + z = 7.
Ans.
Given that
= 1(–1 – 2) – 2(–2 – 0) + 0
= – 3 + 4 = 1 ≠ 0 (non-singular matrix.)
Now co-factors,
∴
⇒
Now, the system of linear equations is given by x – 2y = 10, 2x – y – z = 8 and – 2y + z = 7, which is in the form of CX = D.
where
∵ (A^{T})^{–1} = (A^{–1})^{T}
Hence, x = 0, y = – 5 and z = – 3
Q.19. Using matrix method, solve the system of equation
3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2.
Ans.
Given that
3x + 2y – 2z = 3
x + 2y + 3z = 6
2x – y + z = 2
Let
= 3(2 + 3) – 2(1 – 6) – 2(– 1 – 4)
= 15 + 10 + 10 = 35 ≠ 0 non-singular matrix
Now, co-factors,
Hence, x = 1, y = 1 and z = 1.
Q.20. Givenfind BA an⇒d use this to solve the system of equations y + 2z = 7, x – y = 3, 2x + 3y + 4z = 17.
Ans.
We have,
BA =
∴
The given equations can be re-write as,
x – y = 3, 2x + 3y + 4z = 17 and y + 2z = 7
∴
⇒
Hence, x = 2, y = – 1 and z = 4
Q.21. If a + b + c ≠ 0 andthen prove that a = b = c.
Ans.
Given that: a + b + c ≠ 0 and
C_{1} → C_{1} + C_{2} + C_{3}
⇒
⇒(Taking a + b + c common from C_{1})
⇒
R_{1} → R_{1} - R_{2} and R_{2} → R_{2 }- R_{3}
⇒
Expanding along C_{1}
⇒
⇒ (b – c) (a – b) – (c – a)2 = 0
⇒ ab – b^{2} – ac + bc – c^{2 }– a^{2} + 2ac = 0
⇒ – a^{2} – b^{2} – c^{2} + ab + bc + ac = 0
⇒ a^{2} + b^{2} + c^{2} – ab – bc – ac = 0
⇒ 2a^{2} + 2b^{2} + 2c^{2} – 2ab – 2bc – 2ac = 0
(Multiplying both sides by 2)
⇒ (a^{2} + b^{2} – 2ab) + (b^{2} + c^{2} – 2bc) + (a^{2} + c^{2} – 2ac) = 0
⇒ (a – b)^{2} + (b – c)^{2} + (a – c)^{2} = 0
It is only possible when (a – b)^{2} = (b – c)^{2} = (a – c)^{2} = 0
∴ a = b = c Hence, proved.
Q.22. Prove thatis divisible by a + b + c and find the quotient.
Ans.
Let
C_{1} → C_{1} + C_{2} + C_{3}
⇒
Taking ab + bc + ac – a^{2} – b^{2} – c^{2} common from C_{1}
R_{1} → R_{1} - R_{2} and R_{2} → R_{2} - R_{3}
Expanding along C_{1}
⇒ (a +b + c)^{2} (ab + bc + ac - a^{2} - b^{2} - c^{2} )
⇒ (a + b + c)^{2}(ab + bc + ac - a^{2} - b^{2} - c^{2})[(c - b) (b - a)-(a- c)^{2}]
⇒ (a + b + c)^{2} (ab + bc + ac - a^{2}- b^{2}- c^{2} )(bc - ca - b2+ ab - a^{2}- c^{2} + 2ac)
⇒ (a + b + c)^{2} (ab + bc + ac - a^{2}- b^{2}- c^{2} )(ab + bc + ca - a^{2}- b^{2}- c^{2} )
⇒ (a + b + c)^{2} (ab + bc + ac - a^{2} - b^{2} - c^{2})^{2}
⇒ (a + b + c) (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ac)^{2}
Hence, the given determinant is divisible by a + b + c and the quotient is
(a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ac)^{2}
⇒ (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ac) (a^{2} + b^{2} + c^{2} - ab - bc - ac)
⇒ (a^{3} + b^{3} + c^{3} - 3abc) (a^{2} + b^{2} + c^{2} - ab - bc - ac)
⇒ (a^{3} + b^{3} + c^{3}- 3abc) (2a^{2} + 2b^{2} + 2c^{2} - 2ab - 2bc - 2ac)
Q.23. If x + y + z = 0, prove that
Ans.
L.H.S.
Let
Expanding along R_{1}
⇒ xa(yza^{2} – x^{2}bc) – yb(y^{2}ac – xzb^{2}) + zc(xyc^{2} – z^{2}ab)
⇒ xyza^{3} – x^{3}abc – y^{3}abc + xyzb^{3} + xyzc^{3} – z^{3}abc
⇒ xyz(a^{3} + b^{3} + c^{3}) – abc(x^{3} + y^{3} + z^{3})
⇒ xyz(a^{3} + b^{3} + c^{3}) – abc(3xyz)
[(∵ x + y + z = 0) (∴ x^{3} + y^{3} + z^{3} = 3xyz)]
⇒ xyz(a^{3} + b^{3} + c^{3} – 3abc)
R_{1} → R_{1} + R_{2} + R_{3}
(Taking a + b + c common from R_{1})
C_{1} → C_{1} - C_{2}, C_{2} → C_{2} - C_{3}
Expanding along R_{1}
⇒ xyz(a + b+ c) [(c - a)^{2} - (b- c) (a - b)]
⇒ xyz(a + b + c)(c ^{2 }+ a^{2}- 2ca - ab + b^{2}+ ac - bc)
⇒ xyz(a + b+ c)( a^{2} + b^{2} + c^{2} - ab - bc - ca)
⇒ xyz(a^{3} + b^{3} + c^{3} - 3abc)
[a^{3} + b^{3} + c^{3} - 3abc = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ca)]
L.H.S. = R.H.S.
Hence, proved.