Let A = [1 0 0; 3 1 0; 9 3 1] and B = [bij], 1 <= i, j <= 3. If B = A^99 - I, then the value of (b31-b21)/b32 is:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q5 – JEE-MAIN 2026

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Let

A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}]

and

B = [b_{i j}], 1 \leq i, j \leq 3

. If

B = A^{99} - I

, then the value of

\frac{b_{31} - b_{21}}{b_{32}}

is:

Clue 1 of 3: Matrix Decomposition

Recognize that matrix $A$ can be expressed as the sum of the identity matrix $I$ and a nilpotent matrix $N$ .

🧠 Full Solution Path

Step 1: Decompose Matrix A and Find Powers of N✦ Active

Express matrix $A$ as the sum of the identity matrix $I$ and a matrix $N$ . Then, calculate the powers of $N$ to determine its nilpotency index.

A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}] = I + N

where

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

and

N = [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}]

. Calculate powers of

N

N^{2} = [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}] [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{matrix}]

N^{3} = N^{2} \cdot N = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{matrix}] [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]

Since

N^{3} = 0

N

is a nilpotent matrix of index 3.

Step 2: Calculate

A^{99}

and Matrix B○ Expand

Use the binomial theorem for matrices to calculate $A^{99}$ . Since $I$ and $N$ commute, $(I + N)^{99}$ can be expanded. Due to $N^{3} = 0$ , only the first few terms will be non-zero. Then, calculate $B = A^{99} - I$ .

A^{99} = (I + N)^{99} = I + (\binom{99}{1}) N + (\binom{99}{2}) N^{2} + (\binom{99}{3}) N^{3} + \dots

Since

N^{3} = 0

, the expansion simplifies to:

A^{99} = I + 99 N + \frac{99 (99 - 1)}{2} N^{2} = I + 99 N + \frac{99 \times 98}{2} N^{2} = I + 99 N + 4851 N^{2}

Substitute the matrices

I

N

, and

N^{2}

A^{99} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] + 99 [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}] + 4851 [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{matrix}]

A^{99} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] + [\begin{matrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 891 & 297 & 0 \end{matrix}] + [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 43659 & 0 & 0 \end{matrix}]

A^{99} = [\begin{matrix} 1 & 0 & 0 \\ 297 & 1 & 0 \\ 891 + 43659 & 297 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 297 & 1 & 0 \\ 44550 & 297 & 1 \end{matrix}]

Now, calculate

B = A^{99} - I

B = [\begin{matrix} 1 & 0 & 0 \\ 297 & 1 & 0 \\ 44550 & 297 & 1 \end{matrix}] - [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 44550 & 297 & 0 \end{matrix}]

Step 3: Evaluate the Expression○ Expand

Extract the required elements $b_{31}$ , $b_{21}$ , and $b_{32}$ from matrix $B$ and substitute them into the given expression.

From matrix

B

, we have:

b_{31} = 44550

b_{21} = 297

b_{32} = 297

The expression to evaluate is

\frac{b_{31} - b_{21}}{b_{32}}

\frac{44550 - 297}{297} = \frac{44253}{297} = 149

💡 Teacher's Secret Hint

Ensure careful calculation of the division to avoid errors.

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Maths Question 5 – JEE-MAIN 2026

Clue 1 of 3: Matrix Decomposition

🧠 Full Solution Path