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JEE Advanced Maths Determinant Previous Year Questions with Solutions

Determinant JEE Advanced previous year questions with solutions are given on this page. We can say that every square matrix has associated with it a scalar called its determinant. As far as the JEE Advanced exam is concerned, the determinant is an important topic. Students can easily score marks from this topic if they are thorough with the concept. Those who are preparing for the JEE Advanced are recommended to go through the given questions and solutions so that they can be familiar with the question pattern.

Question 1: The total number of distinct x∈R for which

\(\begin{array}{l}\begin{vmatrix} x & x^{2} &1+x^{3} \\ 2x& 4x^{2}& 1+8x^{3}\\ 3x & 9x^{2} &1+27x^{3} \end{vmatrix}=10\end{array} \)
is

(a) 2

(b) 1

(c) 3

(d) none of these

Solution:

Given that

\(\begin{array}{l}\begin{vmatrix} x & x^{2} &1+x^{3} \\ 2x& 4x^{2}& 1+8x^{3}\\ 3x & 9x^{2} &1+27x^{3} \end{vmatrix}=10\end{array} \)

=>

\(\begin{array}{l}x^{3}\begin{vmatrix} 1 & 1 &1+x^{3} \\ 2& 4& 1+8x^{3}\\ 3 & 9 &1+27x^{3} \end{vmatrix}=10\end{array} \)

=> R2 → R2 – 2R1

R3 → R3 – 3R1,

\(\begin{array}{l}x^{3}\begin{vmatrix} 1 & 1 &1+x^{3} \\ 0& 2& 6x^{3}-1\\ 0 & 6 &24x^{3}-2 \end{vmatrix}=10\end{array} \)

=> x3(2(24x3-2) – 6(6x3-1)) = 10

=> x3(48x3 – 4 – 36x3 + 6) = 10

=> x3(12x3 + 2) = 10

=> 12x6 + 2x3 = 10

=> 6x6 + x3 = 5

Put x3 = t

=> 6t2 + t – 5 = 0

=> (6t-5) (t+1) = 0

=> t = 5/6 or t = -1

=> x3 = 5/6 or x3 = -1

=> x = (5/6)1/3 or x = -1

So x has 2 distinct values.

Hence option a is the answer.

Question 2: Given x = -9 is a root of

\(\begin{array}{l}\begin{vmatrix} x & 3 & 7\\ 2 & x &2 \\ 7& 6 & x \end{vmatrix}=0\end{array} \)
, then find its other two roots a and b, and then find ab.

(a) -14

(b) 14

(c) 9

(d) 63

Solution:

Given

\(\begin{array}{l}\begin{vmatrix} x & 3 & 7\\ 2 & x &2 \\ 7& 6 & x \end{vmatrix}=0\end{array} \)

=> x(x2 -12) – 3(2x – 14) + 7(12-7x) = 0

=> x3 – 12x – 6x + 42 + 84 – 49x = 0

=> x3 – 67x + 126 = 0

Given x = -9 is a root.

So (x+9) is a factor of x3 – 67x + 126.

=> x3 – 67x + 126 = 0

=> (x+9) (x2 – 9x + 14) = 0

=> (x+9) (x-2) (x-7) = 0

=> other roots are x = 2, x = 7.

=> a = 2, b = 7

=> ab = 14

Hence option b is the answer.

Question 3: Given x = cy + bz, y = az + cx, z = bx + ay where x, y, z are not all zero. The value of a2 + b2 + c2 + 2abc is

(a) 1

(b) 0

(c) 2

(d) 3

Solution:

Given x = cy + bz, y = az + cx, z = bx + ay

Rewriting, we get

-x + cy + bz = 0

cx – y + az = 0

bx + ay – z = 0

Given x, y, z are not all zero.

Using the condition for concurrency

\(\begin{array}{l}\begin{vmatrix} -1 & c & b\\ c & -1 &a \\ b& a & -1 \end{vmatrix}=0\end{array} \)

=> -1(1-a2) – c(-c – ab) + b(ac + b) = 0

=> a2 – 1 + c2 + abc + abc + b2 = 0

=> a2 + b2 + c2 + 2abc – 1 = 0

=> a2 + b2 + c2 + 2abc = 1

Hence option a is the answer.

Question 4: In a third-order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then, it can be decomposed into n determinants, when n has the value

(a) 1

(b) 9

(c) 24

(d) 16

Solution:

We can divide column one into 2 determinants.

In each of those two determinants, column 2 can be divided into 3 different determinants, thus making a total of 6. Each of these 6 determinants can be divided into 4 determinants, by separating out the 3rd column.

So total determinants formed = 2×3×4 = 24

Hence option c is the answer.

Question 5: System of equation x + 3y + 2z = 6

x + λy + 2z = 7

x + 3y + 2z = μ has

(a) Unique solution if λ = 2, μ ≠ 6

(b) Infinitely many solution if λ = 4, μ = 6

(c) No solution if λ = 5, μ = 7

(d) No solution if λ = 3, μ = 5

Solution:

x + 3y + 2z = 6 ..(i)

x + λy + 2z = 7 ..(ii)

x + 3y + 2z = μ ..(iii)

Here D =

\(\begin{array}{l}\begin{vmatrix} 1 & 3 & 2\\ 1 & \lambda &2 \\ 1& 3 & 2 \end{vmatrix}\end{array} \)

Check the options.

(a) if λ = 2, D = 0, unique solution is not possible.

(b) If λ = 4, μ = 6

x + 4y = 7 – 2z

x + 3y = 6 – 2z

So, y = 1 and x = 3 – 2z.

Substituting in (iii), we get 3 – 2z + 3 + 2z = 6 is satisfied.

Therefore, infinite solutions.

(c) if λ = 5, μ = 7

Consider (ii) and (iii).

x + 5y = 7 – 2z

x + 3y = 7 – 2z

So y = 0, x = 7 – 2z are the solutions.

Substituting in Eq. (i) we have 7 – 2z + 2z = 6 is not satisfied.

Therefore, no solution.

(d) if λ = 3, μ = 5

then (i) and (ii) have no solution.

Therefore, no solution.

Hence options b, c, and d are correct.

Question 6: A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the value of the determinant chosen is positive is

Solution:

The determinant is of the order of 2 with elements 0 or 1 only.

The total number of sample space = 16

The favourable outcomes are

\(\begin{array}{l}\begin{vmatrix} 1 & 0\\ 0& 1 \end{vmatrix}\\\end{array} \)
\(\begin{array}{l}\begin{vmatrix} 1 & 1\\ 0& 1 \end{vmatrix}\\\end{array} \)

And

\(\begin{array}{l}\begin{vmatrix} 1 & 0\\ 1& 1 \end{vmatrix}\\\end{array} \)

Number of favourable outcomes = 3

Hence required probability = 3/16

Question 7: Let M and N be two 3×3 matrices such that MN = NM. Further, if M≠N2 and M2 = N4, then

(a) Determinant of (M2 + MN2) is 0.

(b) There is a 3 × 3 non-zero matrix U such that (M2 + MN2)U is the zero matrix.

(c) Determinant of (M2 + MN2) ≠ 1.

(d) For a 3 × 3 matrix U, if (M2 + MN2)U equals the zero matrix then U is the zero matrix.

Solution:

Given MN = NM, M ≠ N2 and M2 = N4

Then M2 = N4

=> (M+N2)(M-N2) = 0

=> (i) M+N2 = 0 and M-N2≠0

=>(ii) M+N2 = 0 and |M-N2| = 0

In each case, |M+N2|= 0

So (a) is correct and (c) is not correct.

|A| = 0, then there can be many matrices U, such that AU = 0

So (M2 + MN2)U = 0 will be true for many values of U.

So (b) is correct.

Again if AX = 0 and |A| = 0, then X can be non zero.

So (d) is not correct.

Hence option a and b are correct.

Question 8: Let k be an integer such that triangle with vertices (k, -3k), (5, k) and (-k, 2) has area 28 sq.units. Then the orthocentre of this triangle is at the point

(a) (2, ½)

(b) (2, -½)

(c) (1, 3/4)

(d) (1, -3/4)

Solution:

Let A(k, -3k), B(5, k) and C(-k, 2) be the three vertices of the triangle.

Given that area = 28

=>

\(\begin{array}{l}\frac{1}{2}\left | \begin{vmatrix} k & -3k & 1\\ 5& k & 1\\ -k& 2 & 1 \end{vmatrix} \right |=28\end{array} \)

=> 5k2 + 13k – 46 = 0 or 5k2 + 13k + 66 = 0

Consider 5k2 + 13k – 46 = 0

=> k = 2 or k = -23/5

Since k is an integer, k = 2.

Consider 5k2 + 13k + 66 = 0

=> k = (-13±√-1151)/10

So no real solution exists.

So the points are A(2, -6), B(5, 2) and C(-2, 2).

Let orthocentre be H(x, y).

BH is perpendicular to AC.

So ((y-2)/(x-5))(8/-4) = -1

(x-2y) = 1 ..(i)

Also CH perpendicular to AB.

So ((y-2)/(x+2))(8/3) = -1

=> 3x +8y = 10 ..(ii)

Solving (i) and (ii), we get

=> x = 2, y = 1/2

Orthocentre is (2, 1/2)

Hence option a is the answer.

Question 9: If

\(\begin{array}{l}A = \begin{bmatrix} \alpha &2 \\ 2& \alpha \end{bmatrix}\end{array} \)
and if |A3| = 125, then the value of α is

(a) ±1

(b) ±2

(c) ±3

(d) ±5

Solution:

Given

\(\begin{array}{l}A = \begin{bmatrix} \alpha &2 \\ 2& \alpha \end{bmatrix}\end{array} \)

|A| = α2-4

|A3| = 125

=> (α2-4)3 = 125 = 53

=> α2-4 = 5

=> α2 = 9

=> α = ±3

Hence option c is the answer.

Question 10: Which of the following is (are) not the square of a 3×3 matrix with real entries?

(a)

\(\begin{array}{l}\begin{bmatrix} 1 & 0&0 \\ 0& 1& 0\\ 0 & 0 & 1 \end{bmatrix}\end{array} \)

(b)

\(\begin{array}{l}\begin{bmatrix} 1 & 0&0 \\ 0& 1& 0\\ 0 & 0 & -1 \end{bmatrix}\end{array} \)

(c)

\(\begin{array}{l}\begin{bmatrix} 1 & 0&0 \\ 0& -1& 0\\ 0 & 0 & -1 \end{bmatrix}\end{array} \)

(d)

\(\begin{array}{l}\begin{bmatrix} -1 & 0&0 \\ 0& -1& 0\\ 0 & 0 & -1 \end{bmatrix}\end{array} \)

Solution:

Let us check the options.

In options a and c, |A2| = 1

In options b and d, |A2| = -1

|A2| = |A|2

And |A|2 ≠ -1

So given matrices in options b and d cannot be the square of a 3×3 matrix with real entries.

Hence options b and d are the answers.

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