# Mathematics Questions and Answers – Application of Determinants – 2

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This set of Mathematics MCQs for Class 12 focuses 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

Explanation: 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

Explanation: 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

Explanation: 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
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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

Explanation: 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

Explanation: $$\frac{bx}{y+z}$$ = 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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