Question:

Let $A =\left[ a _{i j}\right], a _{i j} \in Z \cap[0,4], 1 \leq i, j \leq 2$ The number of matrices $A$ such that the sum of all entries is a prime number $p \in(2,13)$ is _____

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For problems involving sums of matrix entries, use generating functions to simplify the calculations and find the possible sums efficiently.

Updated On: Mar 20, 2025

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Correct Answer: 204

Approach Solution - 1

As given

a + b + c + d = 3

or 5 or 7 or 11
if

sum = 3

(1 + x + x^{2} + \dots .. + x^{4})^{4} \to x^{3}

(1 - x^{5})^{4} (1 - x)^{- 4} \to x^{3}

∴^{4 + 3 - 1} C_{3} =^{6} C_{3} = 20

sum = 5

(1 - 4 x^{5}) (1 - x)^{- 4} \to x^{5}

\Rightarrow^{4 + 5 - 1} C_{5} - 4 x^{4.4 + 0 - 1} C_{0} =^{8} C_{5} - 4 = 52

If sum

= 7

(1 - 4 x^{5}) (1 - x)^{- 4} \to x^{7}

\Rightarrow^{4 + 5 - 1} C_{4} -^{4.4 + 0 - 1} C_{0} =^{8} C_{5} - 4 = 52

If sum = 11

(1 - 4 x^{5} + 6 x^{10}) (1 - x)^{- 4} \to x^{11}

\Rightarrow^{4 + 11 - 1} C_{11} - 4 \cdot^{4 + 6 - 4} C_{6} + 6 \cdot^{4 + 1 - 1} C_{1}

=^{14} C_{11} - 4 \cdot^{9} C_{6} + 6.4 = 364 - 336 + 24 = 52

∴ Total matrices = 20 + 52 + 80 + 52 = 204

So, the correct answer is 204.

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Approach Solution -2

We are given that the sum of all entries of the matrix $ A $ is a prime number $ p \in \{2, 13\} $. Each element of the matrix $ a_{ij} $ is chosen from the set $ \{0, 1, 2, 3, 4\} $.
Step 1: We begin by considering the possible sums of the matrix entries. Let the sum of all matrix entries be $ a + b + c + d $. We are given that this sum is either 3, 5, 7, or 11.
Step 2: For $ \text{sum} = 3 $, the generating function is: \[ (1 + x + x^2 + \dots + x^4)^4 \rightarrow x^3. \] Simplifying: \[ (1 - x^5)(1 - x) \rightarrow x^3, \] leading to: \[ 4 \times 3 = 20 \quad (\text{for sum 3}). \] Step 3: For $ \text{sum} = 5 $, the generating function becomes: \[ (1 - 4x^5)(1 - x) \rightarrow x^5, \] giving: \[ \text{Total for sum 5} = 52. \] Step 4: For $ \text{sum} = 7 $, we calculate: \[ (1 - 4x^5)(1 - x) \rightarrow x^7, \] leading to: \[ \text{Total for sum 7} = 52. \] Step 5: For $ \text{sum} = 11 $, we compute: \[ (1 - 4x^5 + 6x^{10})(1 - x) \rightarrow x^{11}, \] giving: \[ \text{Total for sum 11} = 52. \] Step 6: Summing all the possible cases, we get the total number of matrices as: \[ 20 + 52 + 80 + 52 = 204. \] Thus, the total number of matrices is $ 204 $.

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Concepts Used:

Determinants

Definition of Determinant

A determinant can be defined in many ways for a square matrix.

The first and most simple way is to formulate the determinant by taking into account the top-row elements and the corresponding minors. Take the first element of the top row and multiply it by its minor, then subtract the product of the second element and its minor. Continue to alternately add and subtract the product of each element of the top row with its respective min or until all the elements of the top row have been considered.

For example let us consider a 1×1 matrix A.

A=[a₁…….a_n]

Read More: Properties of Determinants

Second Method to find the determinant:

The second way to define a determinant is to express in terms of the columns of the matrix by expressing an n x n matrix in terms of the column vectors.

Consider the column vectors of matrix A as A = [ a₁, a₂, a₃, …a_n] where any element a_jis a vector of size x.

Then the determinant of matrix A is defined such that

Det [ a₁ + a₂ …. ba_j+cv … a_x ] = b det (A) + c det [ a₁+ a₂ + … v … a_x ]

Det [ a₁ + a₂ …. a_ja_j+1… a_x ] = – det [ a₁+ a₂ + … a_j+1 a_j … a_x ]

Det (I) = 1

Where the scalars are denoted by b and c, a vector of size x is denoted by v, and the identity matrix of size x is denoted by I.

Read More: Minors and Cofactors

We can infer from these equations that the determinant is a linear function of the columns. Further, we observe that the sign of the determinant can be interchanged by interchanging the position of adjacent columns. The identity matrix of the respective unit scalar is mapped by the alternating multi-linear function of the columns. This function is the determinant of the matrix.