Mathematics Multiple Choice Questions on “Application of Determinants – 1”.
1. What is the value of (begin{vmatrix}-bc & ca + ab & ca + ab \ab + bc & -ca & ab + bc \bc + ca & bc + ca & -ab end {vmatrix}) ?
a) Σab
b) (Σab)2
c) (Σab)3
d) (Σab)4
Answer: c
Clarification: Applying R1 –> R1 + R2 + R3
(begin{vmatrix}Sigma ab & Sigma ab & Sigma ab \ab + bc & -ca & ab + bc \bc + ca & bc + ca & -ab end {vmatrix})
This is equal to,
= Σab (begin{vmatrix}1 & 1 & 1 \ab + bc & -ca & ab + bc \bc + ca & bc + ca & -ab end {vmatrix})
Applying C1 –> C1 – C2 and C2 –> C2 – C3
= Σab (begin{vmatrix}0 & 0 & 1 \ Sigma ab & -Sigma ab & ab + bc \0 & Sigma ca & -ab end {vmatrix})
= (Σab)3
2. If the system of equation 2x + 5y + 8z = 0, x + 4y + 7z = 0, 6x + 9y – αz = 0 has a non trivial solution then what is the value of α?
a) -12
b) 0
c) 12
d) 2
Answer: c
Clarification: Here, in L.H.S we have,
(begin{vmatrix}2 & 5 & 8 \1 & 4 & 7 \6 & 4 & alpha end {vmatrix})
So, for trivial roots the above value is = 0
=>(begin{vmatrix}2 & 5 & 8 \1 & 4 & 7 \6 & 4 & alpha end {vmatrix}) = 0
Solving it further we get α = 12
3. What is the value of x if, (begin{vmatrix}x & 3 & 6 \3 & 6 & x \6 & x & 3 end {vmatrix}) = (begin{vmatrix}2 & x & 7 \x & 7 & 2 \7 & 2 & x end {vmatrix}) = (begin{vmatrix}4 & 5 & x \5 & x & 4 \x & 4 & 6 end {vmatrix})?
a) 9
b) -9
c) 0
d) Can’t be predicted
Answer: b
Clarification: Given that,
(begin{vmatrix}x & 3 & 6 \3 & 6 & x \6 & x & 3 end {vmatrix}) = (begin{vmatrix}2 & x & 7 \x & 7 & 2\7 & 2 & x end {vmatrix}) = (begin{vmatrix}4 & 5 & x \ 5 & x & 4 \x & 4 & 6 end {vmatrix})
So, by circular determinant property,
Sum of the elements of a row = 0
So, x + 3 + 6 = 2 + x + 7 = 4 + 5 + x = 0
=> x = -9
4. Which one of the following is correct if a, b and c are the sides of a triangle ABC and (begin{vmatrix}a^2 & b^2 & c^2 \(a + 1)^2 & (b + 1)^2 & (c + 1)^2 \ (a – 1)^2 & (b – 1)^2 & (c – 1)^2 end {vmatrix}) ?
a) ABC is an equilateral triangle
b) ABC is an isosceles triangle
c) ABC is a right angled triangle
d) ABC is a scalene triangle
Answer: b
Clarification: When a = b or b = c or c = a the determinant reduces to 0
It is not necessary that a = b = c for determinant to be 0
Therefore, the triangle is isosceles.
5. What is the value of k if (begin{vmatrix}y + z & x & x \y & z + x & y \z & z & x + yend {vmatrix}) ?
a) 4
b) -4
c) 1
d) 0
Answer: a
Clarification: Put the value of x, y, z = 1
Thus, putting the value of x = 1, y = 1 and z = 1 on both sides, we get
(begin{vmatrix}2 & 1 & 1 \1 & 2 & 1 \1 & 1 & 2 end {vmatrix}) = k
So, solving the determinant we get k = 4.
6. What will be the value of the given determinant (begin{vmatrix}109 & 102 & 95 \6 & 13 & 20 \1 & -6 & 13 end {vmatrix})?
a) constant other than 0
b) 0
c) 100
d) -1997
Answer: b
Clarification: In the given determinant form the elements are in A.P
Also the common difference of this A.P is 7
Thus the value of the given determinant = 0
7. What is the value of r = 1Σn f(x) if f(r) = (begin{vmatrix}2r & x & n(n + 1) \(6r^2 – 1) & y & n^2 (2n + 3) \(4r^3 – 2nr) & z & n^3(n + 1) end {vmatrix}) where n € N?
a) 1
b) -1
c) 0
d) 2
Answer: c
Clarification: The given determinant is f(r) = (begin{vmatrix}2r & x & n(n + 1) \(6r^2 – 1) & y & n^2(2n + 3) \(4r^3 – 2nr) & z & n^3(n + 1) end {vmatrix})
Now, r = 1Σn (2r) = 2[(n(n + 1))/2] ……….(1)
= n2 + n
r = 1Σn(6r2 – 1) = 6[((n + 1)(2n + 1))/6] – n ……….(2)
= n(2n2 + 2n + n + 1) – n
= 2n3 + 2n2 + n2 + n – n
= 2n3 + 3n2
= r = 1Σn(4r3 – 2nr) = n3 (n + 1) ……….(3)
From (1), (2) and (3) we get
r = 1Σn f(x) = 0
8. If f(x) = (begin{vmatrix}sec x & cos x & sec^2 x + cot x, cosec x \cos^2 x & cos^2 x & cosec^2 x \1 & cos^2 x & cos^2 x end {vmatrix}) then what is the value of 0∫π/2 f(x) dx = (π/4 + 8/15)?
a) (π/4 + 8/15)
b) (π/4 – 8/15)
c) (π/4 + 8/15)
d) (-π/4 + 8/15)
Answer: c
Clarification: (dy/dx) = (dx/dy)-1
So, d2y/dx2 = -(dx/dy)-2 d/dx(dx/dy)
= -(dy/dx)2(d2x/dy2)(dy/dx)
= d2y/dx2 + (dy/dx)3 d2y/dx2 = 0
9. What will be the value of (begin{vmatrix}0 & i – 100 & i – 500 \100 – i & 0 & 1000 – i \500 – i & i – 1000 & 0 end {vmatrix})?
a) 100
b) 500
c) 1000
d) 0
Answer: d
Clarification: The above matrix is a skew symmetric matrix and its order is odd
And we know that for any skew symmetric matrix with odd order has determinant = 0
Therefore, the value of the given determinant = 0
10. If f(x) = (begin{vmatrix}x^n & x^{n+2} & x^{2n} \1 & x^p & p \x^{n+5} & x^{p+6} & x^{2n+5} end {vmatrix}) = 0,then what will be the value of p?
a) xn
b) (n + 1)
c) Either xn or (n + 1)
d) Both xn and (n + 1)
Answer: d
Clarification: Here, C1 and C3 becomes equal when we put p = xn
And R1 and R3 becomes equal when we put p = n + 1
As both of the conditions are satisfied so d is the correct one.