JEE Advanced 2016

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $P = 1416014001$ and $I$ be the identity matrix of order 3. If $Q = [q_{ij}]$ is a matrix such that $P^{50} - Q = I$ , then $\frac{q _{31} + q _{32}}{q _{21}}$ equals

Q: How do I know $A^3$ is zero without calculating everything?

Great question! Look at the structure of $A$. It is a strictly lower triangular matrix. When you multiply such matrices, the 'non-zero' region effectively shifts downwards. By the time you reach $A^3$, there is no 'room' left for non-zero elements. It is a geometric property of the matrix, not just a coincidence.

Select Answer:

Progressive Hints

Hint 1

Look closely at the structure of matrix

P

. It is a lower triangular matrix with ones on the diagonal. In JEE problems, whenever you see a matrix like this, immediately think of decomposing it into

P = I + A

, where

I

is the identity matrix and

A

is the strictly lower triangular part. This is your first step to simplifying the problem.

Hint 2

Once you have defined

A

, calculate its powers. Find

A^{2}

and then

A^{3}

. You will notice that

A

is a nilpotent matrix, meaning

A^{k} = O

for some integer

k

. This property is the 'secret weapon' that will collapse the infinite-looking binomial expansion into a simple polynomial.

Hint 3

Now, apply the Binomial Theorem to expand

P^{50} = (I + A)^{50}

. Because

A^{3} = O

, all terms involving

A^{3}, A^{4}, \dots, A^{50}

will vanish. You are left with only

P^{50} = I + 50 A + \frac{50 \times 49}{2} A^{2}

. Use this to find the matrix

Q

and solve for the required ratio.

Visualized Solution

Analyze Matrix $P$

Given $P = 1416014001$
We can decompose $P$ as $P = I + A$
Where $I$ is the identity matrix and $A = 0416004000$

Powers of Matrix $A$

To find $P^{50}$ , we need the powers of $A$ .
Calculate $A^{2} = A \cdot A$
$A^{2} = 0016000000$

Nilpotent Matrix

Calculate $A^{3} = A^{2} \cdot A$
$A^{3} = 000000000 = O$
Since $A^{3} = O$ , matrix $A$ is nilpotent.
Therefore, $A^{n} = O$ for all $n \geq 3$ .

Binomial Expansion of $P^{50}$

Using Binomial Theorem: $P^{50} = (I + A)^{50}$
$P^{50} = I + 50 A + \frac{50 \times 49}{2} A^{2} + \frac{50 \times 49 \times 48}{3 \times 2 \times 1} A^{3} + \dots$
Since $A^{3} = O$ , the series terminates.
$P^{50} = I + 50 A + 1225 A^{2}$

Defining Matrix $Q$

The problem defines $Q = P^{50} - I$
Substitute our expansion for $P^{50}$ :
$Q = (I + 50 A + 1225 A^{2}) - I$
$Q = 50 A + 1225 A^{2}$

Calculate $q_{21}$ and $q_{32}$

We need specific elements of $Q$ : $q_{21}$ and $q_{32}$ .
$q_{21} = 50 (a_{21}) + 1225 (a_{21}^{2})$
From matrices: $a_{21} = 4$ and $a_{21}^{2} = 0$
$q_{21} = 50 (4) + 1225 (0) = 200$
Similarly, $q_{32} = 50 (4) + 1225 (0) = 200$

Calculate $q_{31}$

Now calculate $q_{31} = 50 (a_{31}) + 1225 (a_{31}^{2})$
From matrices: $a_{31} = 16$ and $a_{31}^{2} = 16$
$q_{31} = 50 (16) + 1225 (16)$
$q_{31} = 800 + 19600 = 20400$

Final Ratio Calculation

Required value: $\frac{q _{31} + q _{32}}{q _{21}}$
Substitute the values: $\frac{20400 + 200}{200}$
$\frac{20600}{200} = 103$
Final Answer: $103$

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Matrices

Solution Diagram

Analyzing the Setup

When faced with a high power of a matrix like

P^{50}

, do not attempt direct multiplication. Instead, recognize the structure of the matrix

P = 1416014001

This is a lower triangular matrix with ones on the diagonal. We can decompose it as

P = I + A

, where

I

is the identity matrix and

A

is defined as:

A = 0416004000

The Nilpotent Property

The identity matrix

I

commutes with any matrix, allowing us to use the Binomial Theorem. To proceed, we examine the powers of

A

A^{2} = 0016000000

A^{3} = 000000000

Since

A^{3} = O

, the matrix

A

is nilpotent of index 3. This implies that all higher powers

A^{n}

for

n \geq 3

are zero matrices.

The Master Equation

Using the Binomial Theorem for

P^{50} = (I + A)^{50}

, we expand the expression:

P^{50} = I + 50 A + \frac{50 \times 49}{2} A^{2} + \frac{50 \times 49 \times 48}{6} A^{3} + \dots

Because all terms involving

A^{3}

and higher vanish, the expression simplifies significantly:

P^{50} = I + 50 A + 1225 A^{2}

Given

Q = P^{50} - I

, we substitute the expansion to find:

Q = 50 A + 1225 A^{2}

Final Calculation

We now extract the specific elements

q_{21}

q_{32}

, and

q_{31}

from the matrix

Q

q_{21} = 50 (4) + 1225 (0) = 200

q_{32} = 50 (4) + 1225 (0) = 200

q_{31} = 50 (16) + 1225 (16) = 800 + 19600 = 20400

The final ratio is calculated as follows:

\frac{q _{31} + q _{21}}{q _{32}} = \frac{20400 + 200}{200} = \frac{20600}{200} = 103

The final result is 103.

Common Traps & FAQs

Wait, why can't we just multiply

P

by itself 50 times? It is just matrix multiplication!

I understand the temptation! But in JEE Advanced, time is your most precious currency. If a question asks for the 50th power, it is never asking you to perform brute-force multiplication. It is testing your ability to spot patterns. Always look for the structure—the 'nilpotency'—before you start calculating.

How do I know

A^{3}

is zero without calculating everything?

Great question! Look at the structure of

A

. It is a strictly lower triangular matrix. When you multiply such matrices, the 'non-zero' region effectively shifts downwards. By the time you reach

A^{3}

, there is no 'room' left for non-zero elements. It is a geometric property of the matrix, not just a coincidence.

Why do we reject the higher terms in the binomial expansion?

We do not 'reject' them; they simply become zero! The Binomial Theorem states

(I + A)^{n} = I + n A + \frac{n ( n - 1 )}{2} A^{2} + \dots

. If

A^{3} = O

, then

A^{3}, A^{4}, \dots

are all the zero matrix. Adding the zero matrix changes nothing, so the series naturally terminates. It is not an approximation; it is an exact equality.

Animated Solution for Mathematics - Matrices and Determinants: Let $P = 1416014001$ and $I$ be the identity matrix of order 3. If $Q = [q_{ij}]$ is a matrix such that $P^{50} - Q = I$ , then $\frac{q _{31} + q _{32}}{q _{21}}$ equals

Select Answer:

Visualized Solution

Analyze Matrix $P$

Powers of Matrix $A$

Nilpotent Matrix

Binomial Expansion of $P^{50}$

Defining Matrix $Q$

Calculate $q_{21}$ and $q_{32}$

Calculate $q_{31}$

Final Ratio Calculation

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Nilpotent Property

The Master Equation

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Analyze Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Powers of Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Nilpotent Matrix

Binomial Expansion of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P50

Defining Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q21​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q32​

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q31​

Final Ratio Calculation

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Nilpotent Property

The Master Equation

Final Calculation

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Defining Matrix $Q$

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Let the matrix $A = 110001010$ satisfy $A^{n} = A^{n - 2} + A^{2} - I$ for $n \geq 3$ . Then the sum of all the elements of $A^{50}$ is :-

Let $P$ be a square matrix such that $P^{2} = I - P$ . For $α, β, γ, δ \in N$ , if $P^{α} + P^{β} = γ I - 29 P$ and $P^{α} - P^{β} = δ I - 13 P$ , then $α + β + γ - δ$ is equal to

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