Let the matrix A = [1 0 0; 1 0 1; 0 1 0] satisfy A^n = A^(n-2) + A^2 - I for n >= 3. Then the sum of all the elements of A^50 is:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – JEE-MAIN 2025

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Home Maths MatricesQuestion #300

Let the matrix

A = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]

satisfy

A^{n} = A^{n - 2} + A^{2} - I

for

n \geq 3

. Then the sum of all the elements of

A^{50}

is:

Clue 1 of 3: Matrix Recurrence Relation

Analyze the given recurrence relation $A^{n} = A^{n - 2} + A^{2} - I$ to find a pattern for $A^{n}$ .

🧠 Full Solution Path

Step 1: Calculate

A^{2}

and simplify the recurrence relation✦ Active

First, calculate the square of matrix $A$ :

A = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]

A^{2} = A \cdot A = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}]

The given recurrence relation is $A^{n} = A^{n - 2} + A^{2} - I$ . Let $C = A^{2} - I$ . Then the relation becomes $A^{n} = A^{n - 2} + C$ for $n \geq 3$ . Calculate $C$ :

C = A^{2} - I = [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] - [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{matrix}]

Step 2: Derive the expression for

A^{50}

○ Expand

Using the simplified recurrence $A^{n} = A^{n - 2} + C$ repeatedly for $n = 50$ (an even number):

A^{50} = A^{48} + C

A^{48} = A^{46} + C

Continuing this pattern down to $A^{4}$ :

A^{4} = A^{2} + C

Summing these relations, we find that $A^{50} = A^{2} + k C$ . The number of times $C$ is added is $k = \frac{50 - 2}{2} = 24$ . Thus:

A^{50} = A^{2} + 24 C

Substitute $C = A^{2} - I$ back into the equation:

A^{50} = A^{2} + 24 (A^{2} - I) = A^{2} + 24 A^{2} - 24 I = 25 A^{2} - 24 I

💡 Teacher's Secret Hint

Ensure the correct number of terms for $C$ is calculated when summing the recurrence.

Step 3: Calculate the sum of elements of

A^{50}

○ Expand

Let $S (M)$ denote the sum of all elements of matrix $M$ . We need to find $S (A^{50})$ . Using the property $S (k M + l N) = k S (M) + l S (N)$ :

S (A^{50}) = S (25 A^{2} - 24 I) = 25 S (A^{2}) - 24 S (I)

Calculate the sum of elements for $A^{2}$ and $I$ :

S (A^{2}) = S ([\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}]) = 1 + 0 + 0 + 1 + 1 + 0 + 1 + 0 + 1 = 5

S (I) = S ([\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]) = 1 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 1 = 3

Substitute these values into the expression for $S (A^{50})$ :

S (A^{50}) = 25 (5) - 24 (3) = 125 - 72 = 53

💡 Teacher's Secret Hint

Remember that the sum of elements of a scalar multiple of a matrix is the scalar multiple of the sum of elements.

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Maths Question 4 – JEE-MAIN 2025

Clue 1 of 3: Matrix Recurrence Relation

🧠 Full Solution Path