Matrices and Determinants JEE 2010: Worked Solution

What is the correct answer to the matrices and determinants JEE 2010 question?

Option C is correct: the diagonal entry and upper-right entry are free nonzero choices, but the lower-left entry is forced. The matrices and determinants JEE 2010 question therefore has the count

(p - 1)^{2} .

The source is IIT-JEE 2010, Paper 1, Mathematics, listed in the Advanced archive. It is a single-correct question.

Take the matrix family

p is an odd prime, A = (\begin{matrix} a & b \\ c & a \end{matrix}), a, b, c \in {0, 1, \dots, p - 1} .

Count the matrices satisfying both conditions:

p ∤ tr (A), p ∣ det (A) .

The first condition says the prime does not divide the trace. The second says it does divide the determinant. The trace is the sum of the diagonal entries.

The four options are:

Why does the trace condition exclude the zero diagonal entry?

The trace restriction allows every nonzero diagonal entry and excludes zero. Both diagonal positions contain the same variable, so they are not two independent choices.

Calculate the trace first: $tr (A) = a + a = 2 a .$

Since the prime is odd, it cannot divide two. Primality then gives

p ∤ 2, p ∣ 2 a ⟺ p ∣ a .

Within the specified representatives, only zero is divisible by the prime. Hence the required trace restriction becomes

a \in {1, 2, \dots, p - 1} .

The first step of the official solution therefore gives exactly

p - 1 choices for a .

Why must both off-diagonal entries be nonzero?

Both off-diagonal entries must be nonzero because their product must match the square of the nonzero diagonal entry modulo the prime. Establish this restriction before multiplying any counts.

The determinant is

det (A) = a \cdot a - b \cdot c = a^{2} - b c .

Translate the divisibility condition:

p ∣ det (A) ⟺ a^{2} - b c \equiv 0 (\mod p) ⟺ b c \equiv a^{2} (\mod p) .

A prime cannot divide the square without dividing the original integer. Therefore,

a ≢ 0 (\mod p) ⟹ a^{2} ≢ 0 (\mod p) .

If either off-diagonal entry were zero, their product would be zero modulo the prime. That contradicts the required nonzero residue.

Divisible by the prime does not mean equal to zero as an integer. The determinant may be a positive or negative multiple of the prime.

Why is there exactly one possible lower-left entry?

Every allowed diagonal entry and nonzero upper-right entry determines exactly one lower-left entry. Nonzero does not mean independently selectable: the determinant condition ties the off-diagonal entries together.

Choose in the official order:

a : p - 1 choices, b : p - 1 choices .

Every nonzero residue modulo a prime has a multiplicative inverse. Thus, $b b^{- 1} \equiv 1 (\mod p) .$

Multiplying the product congruence by that inverse gives

b c \equiv a^{2} (\mod p) ⟹ c \equiv a^{2} b^{- 1} (\mod p) .

This residue has exactly one representative in the permitted range. It is nonzero because otherwise the required product would be zero.

c \in {0, 1, \dots, p - 1} .

For explicit uniqueness, suppose two candidates work. Multiplying by the inverse gives

b c_{1} \equiv b c_{2} (\mod p) ⟹ c_{1} \equiv c_{2} (\mod p) .

Two congruent representatives in the permitted range are equal. Distinct ordered pairs of diagonal and upper-right entries produce distinct matrices, so there is no double counting.

N = \underset{choose a}{\underset{⏟}{(p - 1)}} \underset{choose b}{\underset{⏟}{(p - 1)}} \underset{forced c}{\underset{⏟}{1}} = (p - 1)^{2} .

Option C follows. As a numerical check, take

p = 5, a = 1, b = 2 .

The inverse and forced entry are

2^{- 1} \equiv 3 (\mod 5), c \equiv 1^{2} \cdot 3 \equiv 3 (\mod 5) .

Therefore,

A = (\begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix}), tr (A) = 2, det (A) = 1 - 6 = - 5 .

The trace is not divisible by five, while the determinant is. The determinant need not equal zero.

How does reversing the determinant condition produce option A?

Option A counts matrices whose trace and determinant are both not divisible by the prime. It answers a different question by counting the invalid off-diagonal pairs while keeping the diagonal entry nonzero.

For each fixed allowed diagonal entry, the unrestricted ordered pairs number

# {(b, c)} = p^{2} .

The worked solution establishes that exactly the following number make the determinant divisible: $p - 1 .$

Subtracting those valid pairs leaves the pairs whose determinant is not divisible:

p^{2} - (p - 1) = p^{2} - p + 1 .

Multiplying by the allowed diagonal choices gives

(p - 1) (p^{2} - p + 1),

which is exactly option A. The subtraction is correct, but it selects the wrong side of the determinant condition.

Write the two requirements separately before counting or taking a complement: $Trace: p ∤ 2 a .$

Determinant: p ∣ (a^{2} - b c) .

How do you solve three related matrix-counting questions?

Use the same matrix family and separate free choices from forced entries. These are original practice variations on the supplied PYQ, not additional verified past-year questions. Keep the prime odd and the entry range unchanged:

A = (\begin{matrix} a & b \\ c & a \end{matrix}), a, b, c \in {0, 1, \dots, p - 1} .

Which matrices qualify when the prime is three?

There are four qualifying matrices under the original trace and determinant conditions. The two allowed diagonal entries have the same square modulo three:

p = 3, a \in {1, 2}, a^{2} \equiv 1 (\mod 3), b c \equiv 1 (\mod 3) .

The complete list of triples is

(a, b, c) \in {(1, 1, 1), (1, 2, 2), (2, 1, 1), (2, 2, 2)} .

Thus the matrices are

(\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}), (\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}), (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}), (\begin{matrix} 2 & 2 \\ 2 & 2 \end{matrix}) .

How many matrices have both trace and determinant divisible by the odd prime?

Set the diagonal entry to zero, then count pairs with at least one zero entry. The trace restriction forces $p ∣ 2 a ⟹ a = 0 .$

Because the modulus is prime, the determinant condition becomes

b c \equiv 0 (\mod p) ⟺ b = 0 or c = 0 .

Each case allows all possible values of the other entry. Subtract the all-zero pair once because both cases include it:

N = p + p - 1 = 2 p - 1, overlap: (b, c) = (0, 0) .

How many matrices have divisible determinant with no trace restriction?

Add the counts for nonzero and zero diagonal entries. These cases are disjoint and include every allowed diagonal entry.

The original PYQ supplies the first count; the preceding variation supplies the second:

N = \underset{a \neq 0}{\underset{⏟}{(p - 1)^{2}}} + \underset{a = 0}{\underset{⏟}{(2 p - 1)}} = p^{2} .

Cover the solutions and redo the three variations. Before multiplying, mark each entry free, forced, or split into cases.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Frequently asked questions

What is the answer to the matrices and determinants JEE 2010 question?

Option C, (p-1)^2, is correct. For A = [[a,b],[c,a]], the trace condition excludes a = 0, while the determinant condition requires bc ≡ a^2 (mod p). There are p-1 choices each for a and b, and every such pair determines exactly one allowed c.

Why is c not an independent choice in the JEE 2010 matrix question?

The determinant condition requires bc ≡ a^2 (mod p), with a and b nonzero. Since p is prime, b has a multiplicative inverse, giving c ≡ a^2b^(-1) (mod p). This fixes exactly one nonzero value of c in the permitted range from 0 to p-1.

Does a determinant divisible by p have to equal zero?

No. Divisibility means the determinant is an integer multiple of p, which can be positive, negative or zero. For p = 5, the matrix [[1,2],[3,1]] has determinant -5 and trace 2, so it satisfies both conditions in the question.

Why is option A wrong in the JEE 2010 matrix-counting question?

Option A counts matrices whose trace and determinant are both not divisible by p. For each nonzero a, subtracting the p-1 valid off-diagonal pairs from all p^2 pairs gives p^2-p+1 pairs whose determinant is not divisible by p. Multiplying by p-1 gives option A, but the question requires the determinant to be divisible by p.