Mathematics MCQs for Class 12 on “Application of Determinants – 2”.
1. If f(x) = (begin{vmatrix}1 & a & bc \1 & b & ca \1 & c & ab end {vmatrix}) = (begin{vmatrix}1 & a & a^2 \1 & b & b^2 \1 & c & c^2 end {vmatrix}) then which one among the following is correct?
a) (a – b)(b – c)(c – a)
b) a, b, c are in G.P
c) b, c, a are in G.P
d) a, c, b are in G.P
Answer: b
Clarification: Here, f(x) = (begin{vmatrix}1 & a & bc \1 & b & ca \1 & c & ab end {vmatrix})
Multiplying and diving by abc,
= (1/abc) (begin{vmatrix}a & a^2 & abc \b & b^2 & abc \c & c^2& abc end {vmatrix})
= (begin{vmatrix}1 & a & a^2 \1 & b & b^2 \1 & c & c^2 end {vmatrix})
= (a – b)(b – c)(c – a)
2. What will be the value of f(x) = (begin{vmatrix}p & 2 – i & i + 1 \2 + i & q & 3 + i \1 – i & 3 – i & r end {vmatrix})?
a) Real
b) Imaginary
c) Zero
d) Can’t be predicted
Answer: a
Clarification: Here, f(x) = f’(x)
=> f(x) is purely real
3. If, Si = ai + bi + ci then what is the value of (begin{vmatrix}S0 & S1 & S2 \S1 & S2 & S3 \S2 & S3 & S4 end {vmatrix})?
a) (a + b)2(b – c)2(c – a)2
b) (a – b)2(b – c)2(c + a)2
c) (a – b)2(b – c)2(c – a)2
d) (a – b)2(b + c)2(c – a)2
Answer: c
Clarification: We have, (begin{vmatrix}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2 end {vmatrix})
So, the value of the (begin{vmatrix}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2 end {vmatrix}) = (a – b)(b – c)(c – a)
Now, by circulant determinant,
(begin{vmatrix}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2 end {vmatrix}) X (begin{vmatrix}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2 end {vmatrix}) = (begin{vmatrix}S0 & S1 & S2 \S1 & S2 & S3 \S2 & S3 & S4 end {vmatrix})
Multiplying the determinant in row by row,
We get, (a – b)2(b – c)2(c – a)2
4. Let, α and β be real. Find the set of all values of β for which the system of equation βx + sin α*y + cosα*z = 0, x + cosα * y + sinα * z = 0 , -x + sinα*y – cosα * z = 0 has a non-trivial solution. For β = 1 what are all values of α?
a) 2α = 2nπ ± π/2 + π/2
b) 2α = 2nπ ± π/2 + π/4
c) 2α = 2nπ ± π/4 + π/4
d) 2α = 2nπ ± π/4 + π/2
Answer: c
Clarification: The given system have non-trivial solution if (begin{vmatrix}beta & sin alpha & cos alpha \1 & cos alpha & sin alpha \ -1 & sin alpha & -cos alpha end {vmatrix}) = 0
On opening the determinant we get β = sin 2α + cos 2 α
Therefore, -√2 ≤ β ≤ √2
Now, for β = 1,
sin 2α + cos 2 α = 1
=> (1/√2)sin 2α + (1/√2) cos 2α = (1/√2)
Or, cos(2α – π/4) = 1/√2 = cos(2nπ ± π/4)
=> 2α = 2nπ ± π/4 + π/4
5. Which one among the following is correct if x, y, z are eliminated from, ((frac{bx}{y+z}) = a, (frac{cy}{z+x}) = b, (frac{az}{x+y}) = c)?
a) a2b + b2c + c2a + abc = 0
b) a2b – b2c + c2a + abc = 0
c) a2b + b2c + c2a + 2abc = 0
d) a2b – b2c – c2a – abc = 0
Answer: a
Clarification: (frac{bx}{y+z}) = a bx – ay – az = 0
(frac{cy}{z+x}) = b bx – cy + bz = 0
(frac{az}{x+y}) = c cx + cy – az = 0
(begin{vmatrix}b & -a & -a \b & -c & b \c & c & -a end {vmatrix}) = 0
Or, b(ca – bc) + a(-ab – bc) – a(bc + c2) = 0
or, abc – b2c – a2b – abc – abc – ac2 = 0
or, a2b + b2c + c2a + abc = 0 which is the required eliminate.
6. The co-ordinates of the vertices of a triangle are [m(m + 1), (m + 1)], [(m + 1)(m + 2), (m + 2)] and [(m + 2)(m + 3), (m + 3)]. Then which one among the following is correct?
a) The area of the triangle is dependent on m
b) The area of the triangle is independent on m
c) Answer cannot be predicted
d) Data inadequate
Answer: b
Clarification: The area o the triangle with the given point as vertices is,
1/2 (begin{vmatrix}m(m+1) & (m+1) & 1 \(m+1)(m+2) & (m+2) & 1 \(m+2)(m+3) & (m+3) & 1 end {vmatrix})
= 1/2 (begin{vmatrix}m^2 + m & (m+1) & 1 \m^2 + 3m + 2 & (m+2) & 1 \m^2 + 5m + 6 & (m+3) & 1 end {vmatrix})
Now, by performing the row operation R2 = R2 – R1 and R3 = R3 – R2
= 1/2 (begin{vmatrix}m^2 + m & (m+1) & 1 \2m + 2 & 1 & 0 \2m + 4 & 1 & 0 end {vmatrix})
Now, breaking the determinant we get,
= 1/2 (2m + 2 – 2m – 4)
= -1
Thus, it is independent of m.
7. Which one is correct, the following system of linear equations 2x – 3y + 4z = 7, 3x – 4y + 5z = 8, 4x – 5y + 6z = 9 has?
a) No solutions
b) Infinitely many solutions
c) Unique Solution
d) Can’t be predicted
Answer: b
Clarification: Solving the given system of equation by Cramer’s rule, we get,
x = D1/D, y = D2/D, z = D3/D where,
D = (begin{vmatrix}2 & -3 & 4 \3 & -4 & 5 \4 & -5 & 6 end {vmatrix})
D = –(begin{vmatrix}2 & 3 & 4 \3 & 4 & 5 \4 & 5 & 6 end {vmatrix})
Now, performing, C3 = C3 – C2 and C2 = C2 – C1 we get,
D = –(begin{vmatrix}2 & 1 & 1 \3 & 1 & 1 \4 & 1 & 1 end {vmatrix})
As two columns have identical values, so,
D = 0
Similarly,
D1 = (begin{vmatrix}7 & -3 & 4 \8 & -4 & 5 \9 & -5 & 6 end {vmatrix})
Now, performing, C1 = C1 – C3
D1 = –(begin{vmatrix}3 & -3 & 4 \3 & -4 & 5 \3 & -5 & 6 end {vmatrix})
Now, performing, C3 = C3 – C2
D1 = -3(begin{vmatrix}1 & -3 & 1 \1 & -4 & 1 \1 & -5 & 1 end {vmatrix})
As two columns have identical values, so,
D1 = 0
D2 = (begin{vmatrix}2 & 7 & 4 \3 & 9 & 5 \4 & 8 & 6 end {vmatrix})
Now, performing,
D2 = –(begin{vmatrix}2 & 3 & 2 \3 & 3 & 2 \4 & 3 & 2 end {vmatrix})
Now, performing, C2 = C2 – C3 and C3 = C3 – C1
D2 = 6(begin{vmatrix}2 & 1 & 1 \3 & 1 & 1 \4 & 1 & 1 end {vmatrix})
As two columns have identical values, so,
D2 = 0
D3 = (begin{vmatrix}2 & -3 & 7 \3 & -4 & 9 \4 & -5 & 6 end {vmatrix})
D3 = –(begin{vmatrix}2 & 3 & 4 \3 & 4 & 4 \4 & 5 & 4 end {vmatrix})
Now, performing, C2 = C2 – C2 and C3 = C3 – C2
D3 = -4(begin{vmatrix}2 & 1 & 1 \3 & 1 & 1 \4 & 1 & 1 end {vmatrix})
As two columns have identical values, so,
D3 = 0
Since, D = D1 = D2 = D3 = 0, thus, it has infinitely many solutions.
8. What will be the value of x if (begin{vmatrix}2-x & 2 & 3 \2 & 5-x & 6 \3 & 4 & 10-x end {vmatrix}) = 0?
a) 8 ±√37
b) -8 ± √37
c) 8 ± √35
d) -8 ± √35
Answer: a
Clarification: Here, we have, (begin{vmatrix}2-x & 2 & 3 \2 & 5-x & 6 \3 & 4 & 10-x end {vmatrix}) = 0
Now, replacing C3 = C3 – 3C1, we get,
(begin{vmatrix}2-x & 2 & 3-6 + 3x \2 & 5-x & 6-6 \3 & 4 & 10-x -9 end {vmatrix}) = 0
(begin{vmatrix}2-x & 2 & 3(x-1) \2 & 5-x & 0 \3 & 4 & -(x-1) end {vmatrix}) = 0
Or, (x – 1)(begin{vmatrix} 2-x & 2 & 3 \2 & 5-x & 0 \3 & 4 & -1 end {vmatrix}) = 0
Now, replacing R1 = R1 + 3R3, we get,
Or, (x – 1)(begin{vmatrix}11-x & 14 & 0 \2 & 5-x & 0 \3 & 4 & -1 end {vmatrix}) = 0
Or, (x – 1)[-1 {(11 – x)(5 – x) – 28}] = 0
Or, -(x – 1)(55 – 11x – 5x + x2 – 28)
Or, (x – 1)(x2 – 16x + 27) = 0
Thus, either x – 1 = 0 i.e. x = 1 or x2 – 16x + 27 = 0
Therefore, solving x2 – 16x + 27 = 0 further, we get,
x = 8 ± √37
9. What is the relation between the two determinants f(x) = (begin{vmatrix}–a^2 & ab & ac \ab & -b^2 & bc \ac & bc & -c^2 end {vmatrix}) and g(x) = (begin{vmatrix}0 & c & b \c & 0 & a \b & a & 0 end {vmatrix})?
a) f(x) = g(x)
b) f(x) = (g(x))2
c) g(x) = 2f(x)
d) g(x) = (f(x))2
Answer: b
Clarification: Let, D = (begin{vmatrix}0 & c & b \c & 0 & a \b & a & 0 end {vmatrix})
Expanding D by the 1st row we get,
D = – c(begin{vmatrix}c & a \b & 0 end {vmatrix}) + b(begin{vmatrix}c & 0 \b & a end {vmatrix})
= – c(0 – ab) + b(ac – 0)
= 2abc
Now, we have adjoint of D = D’
= (begin{vmatrix}
begin{vmatrix}0 & a\ a & 0\ end{vmatrix} & – begin{vmatrix}c & a\ b & 0\ end{vmatrix} & begin{vmatrix}c & 0\ b & a\ end{vmatrix}\
– begin{vmatrix}c & b\ a & 0\ end{vmatrix} & begin{vmatrix}0 & b\ b & 0\ end{vmatrix} & – begin{vmatrix}0 & c\ b & 0\ end{vmatrix} \
begin{vmatrix}c & b\ 0 & a\ end{vmatrix} & – begin{vmatrix}0 & b\ c & a\ end{vmatrix} & begin{vmatrix}0 & c\ c & 0\ end{vmatrix} \
end{vmatrix})
Or, D’ = (begin{vmatrix}–a^2 & ab & ac \ab & -b^2 & bc \ac & bc & -c^2 end {vmatrix})
Or, D’ = D2
Or, D’ = D2 = (2abc)2
Mathematics MCQs for Class 12,