JEE Main 2021 (18 March Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $I$ be an identity matrix of order $2 \times 2$ and $P = [25 - 1 - 3]$ . Then the value of $n \in N$ for which $P^{n} = 5 I - 8 P$ is equal to ___

Enter Numerical Value:

Visualized Solution

Problem Setup

Given matrix $P = [25 - 1 - 3]$
Identity matrix $I = [1001]$
Goal: Find $n \in N$ such that $P^{n} = 5 I - 8 P$

Characteristic Equation

To simplify matrix powers, we use the characteristic equation.
$det (P - λ I) = 0$
$det [2 - λ 5 - 1 - 3 - λ] = 0$

Expanding the Determinant

$(2 - λ) (- 3 - λ) - (- 1) (5) = 0$
$- 6 - 2 λ + 3 λ + λ^{2} + 5 = 0$
$λ^{2} + λ - 1 = 0$

Cayley-Hamilton Theorem

By Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation.
Replacing $λ$ with $P$ : $P^{2} + P - I = 0$
Rearranging gives: $P^{2} = I - P$

Calculating $P^{3}$

Multiply both sides by $P$ : $P^{3} = P \cdot P^{2}$
Substitute $P^{2} = I - P$ : $P^{3} = P (I - P) = P - P^{2}$
Substitute $P^{2}$ again: $P^{3} = P - (I - P) = 2 P - I$

Calculating $P^{4}$

Multiply $P^{3}$ by $P$ : $P^{4} = P \cdot P^{3}$
Substitute $P^{3} = 2 P - I$ : $P^{4} = P (2 P - I) = 2 P^{2} - P$
Substitute $P^{2} = I - P$ : $P^{4} = 2 (I - P) - P = 2 I - 3 P$

Calculating $P^{5}$

Multiply $P^{4}$ by $P$ : $P^{5} = P \cdot P^{4}$
Substitute $P^{4} = 2 I - 3 P$ : $P^{5} = P (2 I - 3 P) = 2 P - 3 P^{2}$
Substitute $P^{2} = I - P$ : $P^{5} = 2 P - 3 (I - P) = 5 P - 3 I$

Calculating $P^{6}$

Multiply $P^{5}$ by $P$ : $P^{6} = P \cdot P^{5}$
Substitute $P^{5} = 5 P - 3 I$ : $P^{6} = P (5 P - 3 I) = 5 P^{2} - 3 P$
Substitute $P^{2} = I - P$ : $P^{6} = 5 (I - P) - 3 P = 5 I - 8 P$

Conclusion

Target expression: $P^{n} = 5 I - 8 P$
Calculated expression: $P^{6} = 5 I - 8 P$
Comparing the two equations, we get $n = 6$ .
Final Answer: $n = 6$

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

We are given the matrix

P = [25 - 1 - 3]

and tasked with finding

n \in N

such that

P^{n} = 5 I - 8 P

Attempting to solve this by direct matrix multiplication is inefficient and prone to error. Instead, we utilize the properties of the characteristic equation.

The Characteristic Equation

We define the characteristic equation by calculating the determinant of

P - λ I

det [2 - λ 5 - 1 - 3 - λ] = 0

Expanding this determinant, we obtain:

(2 - λ) (- 3 - λ) - (- 1) (5) = 0

λ^{2} + λ - 6 + 5 = 0

λ^{2} + λ - 1 = 0

Applying Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation. By replacing

λ

with

P

, we derive the fundamental reduction formula:

P^{2} + P - I = 0 \Rightarrow P^{2} = I - P

We can now use this recursive relationship to find higher powers of

P

without performing full matrix multiplication.

Iterative Reduction

We calculate the powers of

P

sequentially:

For

P^{3}

P^{3} = P \cdot P^{2} = P (I - P) = P - P^{2} = P - (I - P) = 2 P - I

For

P^{4}

P^{4} = P \cdot P^{3} = P (2 P - I) = 2 P^{2} - P = 2 (I - P) - P = 2 I - 3 P

For

P^{5}

P^{5} = P \cdot P^{4} = P (2 I - 3 P) = 2 P - 3 P^{2} = 2 P - 3 (I - P) = 5 P - 3 I

For

P^{6}

P^{6} = P \cdot P^{5} = P (5 P - 3 I) = 5 P^{2} - 3 P = 5 (I - P) - 3 P = 5 I - 8 P

Final Conclusion

By comparing our result

P^{6} = 5 I - 8 P

with the target equation

P^{n} = 5 I - 8 P

, we conclude that:

$n = 6$

Common Traps & FAQs

Why can't I just multiply the matrix

P

by itself six times?

You certainly could, but in a JEE Advanced exam, time is your most precious currency. Brute force multiplication is prone to sign errors and is computationally expensive. The Cayley-Hamilton approach is elegant, faster, and significantly reduces the probability of a calculation error.

What exactly is the Cayley-Hamilton Theorem?

It is a profound result in linear algebra stating that every square matrix is a root of its own characteristic polynomial. It bridges the gap between the abstract world of eigenvalues and the concrete world of matrix algebra, allowing us to reduce powers of matrices to lower-degree polynomials.

How do I know when to stop the reduction process?

You stop when the expression on the right-hand side matches the form

a I + b P

given in the problem statement. In this case, we kept reducing until we reached the form

5 I - 8 P

Animated Solution for Mathematics - Matrices and Determinants: Let $I$ be an identity matrix of order $2 \times 2$ and $P = [25 - 1 - 3]$ . Then the value of $n \in N$ for which $P^{n} = 5 I - 8 P$ is equal to ___

Enter Numerical Value:

Visualized Solution

Problem Setup

Characteristic Equation

Expanding the Determinant

Cayley-Hamilton Theorem

Calculating $P^{3}$

Calculating $P^{4}$

Calculating $P^{5}$

Calculating $P^{6}$

Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Characteristic Equation

Applying Cayley-Hamilton Theorem

Iterative Reduction

Final Conclusion

Similar Questions

Let $A = [21 - 1 1]$ . If the sum of the diagonal elements of $A^{13}$ is $3^{n}$ , then $n$ is equal to_________

Let $A = 0 a 1 10 c 230$ , where $a, c \in R$ . If $A^{3} = A$ and the positive value of $a$ belongs to the interval $(n - 1, n]$ , where $n \in N$ , then $n$ is equal to ____.

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

Let $ω$ be a complex cube root of unity with $ω \neq = 1$ and $P = [p_{ij}]$ be an $n \times n$ matrix with $p_{ij} = ω^{i + j}$ . Then $P^{2} \neq = 0$ , when $n =$

Let $A$ be a $3 \times 3$ real matrix such that $A^{2} (A - 2 I) - 4 (A - I) = O$ , where $I$ and $O$ are the identity and null matrices, respectively. If $A^{5} = α A^{2} + β A + γ I$ , where $α, β$ and $γ$ are real constants, then $α + β + γ$ is equal to:

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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

Let $P = 139013001$ and $Q = [q_{ij}]$ be two $3 \times 3$ matrices such that $Q - P^{5} = I_{3}$ . Then $\frac{q _{21} + q _{31}}{q _{32}}$ is equal to:

Let $A = 100 - 1 10 0 - 1 1$ and $B = 7 A^{20} - 20 A^{7} + 2 I$ , where $I$ is an identity matrix of order $3 \times 3$ . If $B = [b_{ij}]$ , then $b_{13}$ is equal to

Let $A = cos θ 0 sin θ 010 - sin θ 0 cos θ$ . If for some $θ \in (0, π)$ , $A^{2} = A^{T}$ , then the sum of the diagonal elements of the matrix $(A + I)^{3} + (A - I)^{3} - 6 A$ is equal to _____ .

Enter Numerical Value:

Visualized Solution

Problem Setup

Characteristic Equation

Expanding the Determinant

Cayley-Hamilton Theorem

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P3

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P4

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P5

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P6

Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Characteristic Equation

Applying Cayley-Hamilton Theorem

Iterative Reduction

Final Conclusion

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=​1416​014​001​​ and I be the identity matrix of order 3. If Q=[qij​] is a matrix such that P50−Q=I, then q21​q31​+q32​​ equals

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Calculating $P^{3}$

Calculating $P^{4}$

Calculating $P^{5}$

Calculating $P^{6}$

Let $A = [21 - 1 1]$ . If the sum of the diagonal elements of $A^{13}$ is $3^{n}$ , then $n$ is equal to_________

Let $A = 0 a 1 10 c 230$ , where $a, c \in R$ . If $A^{3} = A$ and the positive value of $a$ belongs to the interval $(n - 1, n]$ , where $n \in N$ , then $n$ is equal to ____.

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

Let $ω$ be a complex cube root of unity with $ω \neq = 1$ and $P = [p_{ij}]$ be an $n \times n$ matrix with $p_{ij} = ω^{i + j}$ . Then $P^{2} \neq = 0$ , when $n =$

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

Let $P = 139013001$ and $Q = [q_{ij}]$ be two $3 \times 3$ matrices such that $Q - P^{5} = I_{3}$ . Then $\frac{q _{21} + q _{31}}{q _{32}}$ is equal to:

Let $A = 100 - 1 10 0 - 1 1$ and $B = 7 A^{20} - 20 A^{7} + 2 I$ , where $I$ is an identity matrix of order $3 \times 3$ . If $B = [b_{ij}]$ , then $b_{13}$ is equal to