JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: 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:

Q: Wait, why can't we just find the matrix $A$ first? It seems like we are avoiding the main task.

Great question! In JEE problems like this, the matrix $A$ is rarely unique. The equation provided is a constraint on the properties of $A$, not a definition of $A$ itself. Trying to solve for $A$ would be like trying to find the specific numbers that satisfy $x^2 - 1 = 0$ when you only need to know the properties of the roots. We work with the polynomial structure because it's universal for any matrix satisfying that equation.

Q: What if the question asked for $A^{10}$? Would we have to do this all day?

That's the spirit of a mathematician! If we had to find $A^{10}$, we would look for a pattern or use the Cayley-Hamilton theorem. However, for $A^5$, the recursive substitution method is actually faster and less prone to error than finding eigenvalues and using the characteristic equation. Always choose the tool that fits the scale of the problem.

Select Answer:

Visualized Solution

Analyze the Given Equation

Given matrix equation: $A^{2} (A - 2 I) - 4 (A - I) = O$
Where $I$ is the $3 \times 3$ identity matrix and $O$ is the null matrix.
Objective: Find $α, β, γ$ such that $A^{5} = α A^{2} + β A + γ I$ .

Expand and Isolate $A^{3}$

Expanding the equation:
$A^{3} - 2 A^{2} - 4 A + 4 I = O$
Rearranging to isolate $A^{3}$ :
$A^{3} = 2 A^{2} + 4 A - 4 I$

Calculate $A^{4}$ (Part 1)

Multiply the expression for $A^{3}$ by $A$ :
$A \cdot A^{3} = A (2 A^{2} + 4 A - 4 I)$
$A^{4} = 2 A^{3} + 4 A^{2} - 4 A$

Substitute $A^{3}$ into $A^{4}$

Substitute $A^{3} = 2 A^{2} + 4 A - 4 I$ into the expression for $A^{4}$ :
$A^{4} = 2 (2 A^{2} + 4 A - 4 I) + 4 A^{2} - 4 A$

Simplify $A^{4}$

Expand and combine like terms:
$A^{4} = 4 A^{2} + 8 A - 8 I + 4 A^{2} - 4 A$
$A^{4} = 8 A^{2} + 4 A - 8 I$

Calculate $A^{5}$ (Part 1)

Multiply the expression for $A^{4}$ by $A$ :
$A \cdot A^{4} = A (8 A^{2} + 4 A - 8 I)$
$A^{5} = 8 A^{3} + 4 A^{2} - 8 A$

Substitute $A^{3}$ into $A^{5}$

Substitute $A^{3} = 2 A^{2} + 4 A - 4 I$ into the expression for $A^{5}$ :
$A^{5} = 8 (2 A^{2} + 4 A - 4 I) + 4 A^{2} - 8 A$

Simplify $A^{5}$

Expand and combine like terms:
$A^{5} = 16 A^{2} + 32 A - 32 I + 4 A^{2} - 8 A$
$A^{5} = 20 A^{2} + 24 A - 32 I$

Identify $α, β, γ$ and Calculate Sum

Comparing $A^{5} = 20 A^{2} + 24 A - 32 I$ with $A^{5} = α A^{2} + β A + γ I$ :
$α = 20, β = 24, γ = - 32$
Calculate the sum: $α + β + γ = 20 + 24 - 32 = 12$

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

We are given the matrix equation:

A^{2} (A - 2 I) - 4 (A - I) = O

Our objective is to determine the constants

α, β,

and

γ

such that:

A^{5} = α A^{2} + β A + γ I

The Master Equation

First, we expand the given equation to reveal the underlying structure:

A^{3} - 2 A^{2} - 4 A + 4 I = O

By isolating the highest power, we derive our Master Equation:

A^{3} = 2 A^{2} + 4 A - 4 I

This equation serves as a reduction rule, allowing us to express any power of

A

greater than or equal to 3 as a linear combination of

A^{2}, A,

and

I

The Recursive Ladder

To find

A^{4}

, we multiply the Master Equation by

A

A^{4} = 2 A^{3} + 4 A^{2} - 4 A

Substituting the expression for

A^{3}

into this equation:

A^{4} = 2 (2 A^{2} + 4 A - 4 I) + 4 A^{2} - 4 A

Expanding and simplifying the terms, we obtain:

A^{4} = 4 A^{2} + 8 A - 8 I + 4 A^{2} - 4 A

A^{4} = 8 A^{2} + 4 A - 8 I

Now, we perform the final leap to

A^{5}

by multiplying

A^{4}

A

A^{5} = A (8 A^{2} + 4 A - 8 I) = 8 A^{3} + 4 A^{2} - 8 A

Substituting the Master Equation for

A^{3}

once more:

A^{5} = 8 (2 A^{2} + 4 A - 4 I) + 4 A^{2} - 8 A

Expanding the expression:

A^{5} = 16 A^{2} + 32 A - 32 I + 4 A^{2} - 8 A

A^{5} = 20 A^{2} + 24 A - 32 I

Final Calculation

By comparing our result

A^{5} = 20 A^{2} + 24 A - 32 I

with the target form

A^{5} = α A^{2} + β A + γ I

, we identify the constants:

α = 20, β = 24, γ = - 32

The problem asks for the sum

α + β + γ

α + β + γ = 20 + 24 - 32 = 12

The final result is 12.

Common Traps & FAQs

Wait, why can't we just find the matrix

A

first? It seems like we are avoiding the main task.

Great question! In JEE problems like this, the matrix

A

is rarely unique. The equation provided is a constraint on the properties of

A

, not a definition of

A

itself. Trying to solve for

A

would be like trying to find the specific numbers that satisfy

x^{2} - 1 = 0

when you only need to know the properties of the roots. We work with the polynomial structure because it's universal for any matrix satisfying that equation.

Why do we keep substituting

A^{3}

back into the equation? Isn't there a simpler way?

It feels repetitive, doesn't it? But this is the beauty of 'reduction.' By substituting

A^{3}

, we are effectively 'folding' the higher-order complexity back into a lower-order space. We are keeping the degree of our polynomial manageable. It's a systematic way to ensure we never have to deal with powers higher than

A^{2}

What if the question asked for

A^{10}

? Would we have to do this all day?

That's the spirit of a mathematician! If we had to find

A^{10}

, we would look for a pattern or use the Cayley-Hamilton theorem. However, for

A^{5}

, the recursive substitution method is actually faster and less prone to error than finding eigenvalues and using the characteristic equation. Always choose the tool that fits the scale of the problem.

Select Answer:

Visualized Solution

Analyze the Given Equation

Expand and Isolate $A^{3}$

Calculate $A^{4}$ (Part 1)

Substitute $A^{3}$ into $A^{4}$

Simplify $A^{4}$

Calculate $A^{5}$ (Part 1)

Substitute $A^{3}$ into $A^{5}$

Simplify $A^{5}$

Identify $α, β, γ$ and Calculate Sum

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Master Equation

The Recursive Ladder

Final Calculation

Similar Questions

Suppose A is any $3 \times 3$ non-singular matrix and $(A - 3 I) (A - 5 I) = 0$ , where $I = I_{3}$ and $O = O_{3}$ . If $α A + β A^{- 1} = 4 I$ , then $α + β$ is equal to :-

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = α A + I$ , where $α \in R - {- 1, 1}$ . If $det (A^{2} - A) = 4$ , the sum of all possible values of $α$ is equal to

Let $α$ and $β$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^{2} = 3 A + α I$ . If $A^{4} = 21 A + β I$ , then

Let $A = 210 a 35 01 b$ . If $A^{3} = 4 A^{2} - A - 21 I$ , where $I$ is the identity matrix of order $3 \times 3$ , then $2 a + 3 b$ is equal to:

If for the matrix, $A = [1 α - α β]$ , $A A^{T} = I_{2}$ , then the value of $α^{4} + β^{4}$ is:

Let $B = [1135]$ and $A$ be a $2 \times 2$ matrix such that $A B^{- 1} = A^{- 1}$ . If $B C B^{- 1} = A$ and $C^{4} + α C^{2} + β I = O$ , then $2 β - α$ is equal to

If the matrix $A = 103020 00 - 1$ satisfies the equation $A^{20} + α A^{19} + β A = 100040001$ for some real numbers $α$ and $β$ , then $β - α$ is equal to

Let $A$ and $B$ be two $3 \times 3$ real matrices such that $(A^{2} - B^{2})$ is invertible matrix. If $A^{5} = B^{5}$ and $A^{3} B^{2} = A^{2} B^{3}$ , then the value of the determinant of the matrix $A^{3} + B^{3}$ is equal to :

Let $A = [a_{ij}]$ be a square matrix of order 3 such that $a_{ij} = 2^{j - i}$ , for all $i, j = 1, 2, 3$ . Then, the matrix $A^{2} + A^{3} + \dots + A^{10}$ is equal to

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

Select Answer:

Visualized Solution

Analyze the Given Equation

Expand and Isolate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A4 (Part 1)

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3 into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A4

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A4

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A5 (Part 1)

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3 into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A5

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A5

Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α,β,γ and Calculate Sum

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Master Equation

The Recursive Ladder

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Suppose A is any 3×3 non-singular matrix and (A−3I)(A−5I)=0, where I=I3​ and O=O3​. If αA+βA−1=4I, then α+β is equal to :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a 2×2 matrix with real entries such that AT=αA+I, where α∈R−{−1,1}. If det(A2–A)=4, the sum of all possible values of α is equal to

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​210​a35​01b​​. If A3=4A2−A−21I, where I is the identity matrix of order 3×3, then 2a+3b is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for the matrix, A=[1α​−αβ​], AAT=I2​, then the value of α4+β4 is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let B=[11​35​] and A be a 2×2 matrix such that AB−1=A−1. If BCB−1=A and C4+αC2+βI=O, then 2β−α is equal to

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P be a square matrix such that P2=I−P. For α,β,γ,δ∈N, if Pα+Pβ=γI−29P and Pα−Pβ=δI−13P, then α+β+γ−δ is equal to

Expand and Isolate $A^{3}$

Calculate $A^{4}$ (Part 1)

Substitute $A^{3}$ into $A^{4}$

Simplify $A^{4}$

Calculate $A^{5}$ (Part 1)

Substitute $A^{3}$ into $A^{5}$

Simplify $A^{5}$

Identify $α, β, γ$ and Calculate Sum

Suppose A is any $3 \times 3$ non-singular matrix and $(A - 3 I) (A - 5 I) = 0$ , where $I = I_{3}$ and $O = O_{3}$ . If $α A + β A^{- 1} = 4 I$ , then $α + β$ is equal to :-

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = α A + I$ , where $α \in R - {- 1, 1}$ . If $det (A^{2} - A) = 4$ , the sum of all possible values of $α$ is equal to

Let $α$ and $β$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^{2} = 3 A + α I$ . If $A^{4} = 21 A + β I$ , then

Let $A = 210 a 35 01 b$ . If $A^{3} = 4 A^{2} - A - 21 I$ , where $I$ is the identity matrix of order $3 \times 3$ , then $2 a + 3 b$ is equal to:

If for the matrix, $A = [1 α - α β]$ , $A A^{T} = I_{2}$ , then the value of $α^{4} + β^{4}$ is:

Let $B = [1135]$ and $A$ be a $2 \times 2$ matrix such that $A B^{- 1} = A^{- 1}$ . If $B C B^{- 1} = A$ and $C^{4} + α C^{2} + β I = O$ , then $2 β - α$ is equal to

If the matrix $A = 103020 00 - 1$ satisfies the equation $A^{20} + α A^{19} + β A = 100040001$ for some real numbers $α$ and $β$ , then $β - α$ is equal to

Let $A$ and $B$ be two $3 \times 3$ real matrices such that $(A^{2} - B^{2})$ is invertible matrix. If $A^{5} = B^{5}$ and $A^{3} B^{2} = A^{2} B^{3}$ , then the value of the determinant of the matrix $A^{3} + B^{3}$ is equal to :

Let $A = [a_{ij}]$ be a square matrix of order 3 such that $a_{ij} = 2^{j - i}$ , for all $i, j = 1, 2, 3$ . Then, the matrix $A^{2} + A^{3} + \dots + A^{10}$ is equal to

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