JEE Main 2021 (27 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: 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 :

Q: Wait, why can't we just say $A=B$ or $A^2=B^2$?

That is a classic trap! In matrix algebra, $A^5 = B^5$ does not imply $A=B$. Matrices do not behave like scalars. We must rely on the algebraic properties of the given equations rather than assuming simple equality.

Q: Why does the order of factoring matter so much?

Because matrix multiplication is non-commutative ($AB \neq BA$). If you factor $A^3$ from $A^5 - A^3 B^2$, you must write it as $A^3(A^2 - B^2)$. If you tried to write $(A^2 - B^2)A^3$, you would be fundamentally changing the expression. Always keep the factor on the same side it was originally located.

Q: What if $(A^2 - B^2)$ was singular (not invertible)?

If $(A^2 - B^2)$ were singular, we could not multiply by its inverse to 'cancel' it out. In that case, the product $(A^3 + B^3)(A^2 - B^2) = 0$ would not necessarily imply that $(A^3 + B^3) = 0$. The invertibility condition is the specific key that unlocks the final step of this problem.

Select Answer:

Visualized Solution

Given Conditions

Given matrices $A$ and $B$ are $3 \times 3$ real matrices.
Condition 1: $(A^{2} - B^{2})$ is invertible.
Condition 2: $A^{5} = B^{5}$ .
Condition 3: $A^{3} B^{2} = A^{2} B^{3}$ .
Goal: Find $det (A^{3} + B^{3})$ .

Strategy: Equation Manipulation

We need to find a relationship between $A^{3} + B^{3}$ and the given conditions.
Strategy: Subtract the two power equations to create a factorable expression.
We aim to isolate the term $(A^{2} - B^{2})$ as it is known to be invertible.

Subtracting the Equations

Subtracting Equation 3 from Equation 2:
$A^{5} - A^{3} B^{2} = B^{5} - A^{2} B^{3}$

Factorizing the Left Side

Focus on the left-hand side: $A^{5} - A^{3} B^{2}$ .
Factor out $A^{3}$ from the left:
$A^{3} (A^{2} - B^{2})$

Factorizing the Right Side

Focus on the right-hand side: $B^{5} - A^{2} B^{3}$ .
Factor out $B^{3}$ to get: $B^{3} (B^{2} - A^{2})$ .
Rewrite as: $- B^{3} (A^{2} - B^{2})$ .

Moving to One Side

Rearrange the equation to one side:
$A^{3} (A^{2} - B^{2}) = - B^{3} (A^{2} - B^{2})$
$A^{3} (A^{2} - B^{2}) + B^{3} (A^{2} - B^{2}) = 0$

Final Factorization

Factor out $(A^{2} - B^{2})$ from the right:
$(A^{3} + B^{3}) (A^{2} - B^{2}) = 0$

Invertibility Condition

Given: $(A^{2} - B^{2})$ is invertible.
This implies that $(A^{2} - B^{2})^{- 1}$ exists.
Recall: If $M N = 0$ and $N$ is invertible, then $M = 0$ .

Resulting Matrix Equation

Post-multiply by $(A^{2} - B^{2})^{- 1}$ :
$(A^{3} + B^{3}) (A^{2} - B^{2}) (A^{2} - B^{2})^{- 1} = 0 \cdot (A^{2} - B^{2})^{- 1}$
$A^{3} + B^{3} = 0$

Determinant Calculation

We need to find $det (A^{3} + B^{3})$ .
Since $A^{3} + B^{3} = 0$ (the zero matrix),
$det (A^{3} + B^{3}) = det (0) = 0$ .

Conclusion \& Takeaway

Key Takeaway: The invertibility of $(A^{2} - B^{2})$ allows us to simplify the matrix product to a zero matrix.
Final Answer: The value of the determinant is $0$ .
Challenge: What would happen if $(A^{2} - B^{2})$ was singular (not invertible)?

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The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

We are given two

3 \times 3

real matrices,

A

and

B

. We are provided with three critical conditions: 1.

(A^{2} - B^{2})

is invertible. 2.

A^{5} = B^{5}

. 3.

A^{3} B^{2} = A^{2} B^{3}

Our objective is to determine the value of

det (A^{3} + B^{3})

. While the powers of the matrices appear daunting, we can simplify the expression by identifying hidden algebraic structures.

The Strategy

Creating the Bridge

To connect the given conditions to the target expression, we subtract the third condition from the second:

A^{5} - A^{3} B^{2} = B^{5} - A^{2} B^{3}

This manipulation is strategic, as it allows us to group terms and search for a common factor. Our goal is to isolate the term

(A^{2} - B^{2})

within the equation.

The Art of Factorization

Let us factor the left side of the equation by extracting

A^{3}

A^{3} (A^{2} - B^{2})

Now, consider the right side. By factoring out

B^{3}

, we obtain

B^{3} (B^{2} - A^{2})

. To align this with our left side, we factor out a negative sign:

- B^{3} (A^{2} - B^{2})

Equating the two sides, we have:

A^{3} (A^{2} - B^{2}) = - B^{3} (A^{2} - B^{2})

Moving all terms to one side yields:

A^{3} (A^{2} - B^{2}) + B^{3} (A^{2} - B^{2}) = 0

Factoring out the common term

(A^{2} - B^{2})

from the right, we arrive at the pivotal equation:

(A^{3} + B^{3}) (A^{2} - B^{2}) = 0

The Invertibility Key

We are given that

(A^{2} - B^{2})

is invertible. In matrix algebra, if a product of two matrices

M N = 0

and

N

is invertible, we can multiply by

N^{- 1}

on the right to conclude that

M = 0

Applying this to our equation, we post-multiply by

(A^{2} - B^{2})^{- 1}

(A^{3} + B^{3}) (A^{2} - B^{2}) (A^{2} - B^{2})^{- 1} = 0 \cdot (A^{2} - B^{2})^{- 1}

Since

(A^{2} - B^{2}) (A^{2} - B^{2})^{- 1} = I

, where

I

is the identity matrix, we are left with:

A^{3} + B^{3} = 0

Final Calculation

We have successfully proven that

A^{3} + B^{3}

is the zero matrix. The determinant of the zero matrix is, by definition,

0

Therefore, the final answer is:

0

Common Traps & FAQs

Wait, why can't we just say

A = B

A^{2} = B^{2}

That is a classic trap! In matrix algebra,

A^{5} = B^{5}

does not imply

A = B

. Matrices do not behave like scalars. We must rely on the algebraic properties of the given equations rather than assuming simple equality.

Why does the order of factoring matter so much?

Because matrix multiplication is non-commutative (

A B \neq = B A

). If you factor

A^{3}

from

A^{5} - A^{3} B^{2}

, you must write it as

A^{3} (A^{2} - B^{2})

. If you tried to write

(A^{2} - B^{2}) A^{3}

, you would be fundamentally changing the expression. Always keep the factor on the same side it was originally located.

What if

(A^{2} - B^{2})

was singular (not invertible)?

(A^{2} - B^{2})

were singular, we could not multiply by its inverse to 'cancel' it out. In that case, the product

(A^{3} + B^{3}) (A^{2} - B^{2}) = 0

would not necessarily imply that

(A^{3} + B^{3}) = 0

. The invertibility condition is the specific key that unlocks the final step of this problem.

Animated Solution for Mathematics - Matrices and Determinants: 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 :

Select Answer:

Visualized Solution

Given Conditions

Strategy: Equation Manipulation

Subtracting the Equations

Factorizing the Left Side

Factorizing the Right Side

Moving to One Side

Final Factorization

Invertibility Condition

Resulting Matrix Equation

Determinant Calculation

Conclusion \& Takeaway

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Strategy

The Art of Factorization

The Invertibility Key

Final Calculation

Similar Questions

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:

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:

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 :-

For any $3 \times 3$ matrix $M$ , let $∣ M ∣$ denote the determinant of $M$ . Let $I$ be the $3 \times 3$ identity matrix. Let $E$ and $F$ be two $3 \times 3$ matrices such that $(I - E F)$ is invertible. If $G = (I - E F)^{- 1}$ , then which of the following statements is (are) TRUE ?

Let $θ = \frac{π}{5}$ and $A = (cos θ - sin θ sin θ cos θ)$ . If $B = A + A^{4}$ then $d e t (B)$ :

If $A$ and $B$ are two non-zero $n \times n$ matrices such that $A^{2} + B = A^{2} B$ , then

Let $A, B$ and $C$ be three $2 \times 2$ matrices with real entries such that $B = (I + A)^{- 1}$ and $A + C = I$ . If $B C = [1 - 1 - 5 2]$ and $C B [x_{1} x_{2}] = [12 - 6]$ , then $x_{1} + x_{2}$ is

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

Let $A = [a_{ij}]$ be a real matrix of order $3 \times 3$ , such that $a_{i 1} + a_{i 2} + a_{i 3} = 1$ , for $i = 1, 2, 3$ . Then, the sum of all the entries of the matrix $A^{3}$ is equal to:

Let $A + 2 B = 16 - 5 2 - 3 3 031$ and $2 A - B = 220 - 1 - 1 1 562$ . If $Tr (A)$ denotes the sum of all diagonal elements of the matrix $A$ , then $Tr (A) - Tr (B)$ has value equal to

Select Answer:

Visualized Solution

Given Conditions

Strategy: Equation Manipulation

Subtracting the Equations

Factorizing the Left Side

Factorizing the Right Side

Moving to One Side

Final Factorization

Invertibility Condition

Resulting Matrix Equation

Determinant Calculation

Conclusion \& Takeaway

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Strategy

The Art of Factorization

The Invertibility Key

Final Calculation

Similar Questions

.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:

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.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 A=[aij​] be a real matrix of order 3×3, such that ai1​+ai2​+ai3​=1, for i=1,2,3. Then, the sum of all the entries of the matrix A3 is equal to:

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:

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 $θ = \frac{π}{5}$ and $A = (cos θ - sin θ sin θ cos θ)$ . If $B = A + A^{4}$ then $d e t (B)$ :

If $A$ and $B$ are two non-zero $n \times n$ matrices such that $A^{2} + B = A^{2} B$ , then

Let $A, B$ and $C$ be three $2 \times 2$ matrices with real entries such that $B = (I + A)^{- 1}$ and $A + C = I$ . If $B C = [1 - 1 - 5 2]$ and $C B [x_{1} x_{2}] = [12 - 6]$ , then $x_{1} + x_{2}$ is

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

Let $A = [a_{ij}]$ be a real matrix of order $3 \times 3$ , such that $a_{i 1} + a_{i 2} + a_{i 3} = 1$ , for $i = 1, 2, 3$ . Then, the sum of all the entries of the matrix $A^{3}$ is equal to: