Mathematics Questions and Answers – Application of Determinants – 1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses 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
View Answer

Answer: c
Explanation: 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
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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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: b
Explanation: 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
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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
View Answer

Answer: b
Explanation: 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 \(\begin{vmatrix}y + z & x & x \\y & z + x & y \\z & z & x + y\end {vmatrix}\) ?
a) 4
b) -4
c) 1
d) 0
View Answer

Answer: a
Explanation: 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.
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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
View Answer

Answer: b
Explanation: 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
View Answer

Answer: c
Explanation: 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
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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 0π/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)
View Answer

Answer: c
Explanation: (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
View Answer

Answer: d
Explanation: 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
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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)
View Answer

Answer: d
Explanation: 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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter