JEE Main 2023 (24 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: If $A$ and $B$ are two non-zero $n \times n$ matrices such that $A^{2} + B = A^{2} B$ , then

Select Answer:

Visualized Solution

Analyze the Given Equation

Given equation for non-zero $n \times n$ matrices $A$ and $B$ :
$A^{2} + B = A^{2} B$

Rearranging the Terms

Rearrange the equation by moving all terms to one side:
$A^{2} B - A^{2} - B = 0$

The Identity Matrix Trick

Add the Identity matrix $I$ to both sides to facilitate factorization:
$A^{2} B - A^{2} - B + I = I$

Grouping for Factorization

Group the first two terms and factor out $A^{2}$ :
$A^{2} (B - I) - B + I = I$

Completing the Factors

Factor out a negative sign from the remaining terms and simplify:
$A^{2} (B - I) - (B - I) = I$
$(A^{2} - I) (B - I) = I$

The Property of Commuting Inverses

Use the property: If $X Y = I$ for square matrices, then $Y X = I$ .
Therefore, $(B - I) (A^{2} - I) = I$

Expanding the New Product

Expand the product on the left side:
$B A^{2} - B - A^{2} + I = I$

Simplifying the Expansion

Cancel $I$ from both sides and rearrange:
$B A^{2} - B - A^{2} = 0$
$B A^{2} = A^{2} + B$

Final Conclusion

From the original equation: $A^{2} + B = A^{2} B$
From our derivation: $A^{2} + B = B A^{2}$
Therefore, $A^{2} B = B A^{2}$

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The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

We are given the matrix relationship:

A^{2} + B = A^{2} B

where

A

and

B

are

n \times n

matrices. Our objective is to explore the properties of this relationship by manipulating the matrix algebra.

The Identity Trick

To simplify the expression, we first move all terms to one side:

A^{2} B - A^{2} - B = 0

This expression resembles the algebraic expansion

(x - 1) (y - 1) = x y - x - y + 1

. To complete the factorization, we add the Identity matrix

I

to both sides:

A^{2} B - A^{2} - B + I = I

Factoring the Expression

We can now group the terms to facilitate factorization. Factoring

A^{2}

from the first two terms and

- 1

from the remaining terms yields:

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

By factoring out the common matrix

(B - I)

, we arrive at the elegant result:

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

Proving Commutativity

The equation

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

implies that the matrices

(A^{2} - I)

and

(B - I)

are inverses of each other. A fundamental property of inverse matrices is that they must commute. Therefore:

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

Expanding this product, we obtain:

B A^{2} - B - A^{2} + I = I

Final Conclusion

By canceling

I

from both sides, we are left with:

B A^{2} - B - A^{2} = 0 ⟹ B A^{2} = A^{2} + B

Comparing this to our original equation

A^{2} + B = A^{2} B

, we conclude that: $A^{2} B = B A^{2}$
This confirms that

A^{2}

and

B

commute, demonstrating the underlying symmetry of the given matrix relationship.

Common Traps & FAQs

Why can't we just divide by \

A^2\

to isolate \

B\

This is a classic trap! In matrix algebra, division is not defined. You cannot simply 'divide' by a matrix. You can only multiply by an inverse, but you do not know if \

A^2\

is invertible. Always treat matrices as distinct algebraic entities that do not follow the same rules as scalar numbers.

Why does \

XY=I\

imply \

YX=I\

This is a fundamental property of square matrices. If the product of two square matrices is the identity matrix, they are inverses of each other. By definition, if \

X\

is the inverse of \

Y\

, then \

XY = YX = I\

. They must commute.

Is \

A^2 B = B A^2\

always true for any matrices?

Absolutely not! Matrix multiplication is generally non-commutative, meaning \

AB \neq BA\

. However, in this specific problem, the algebraic structure we derived forces them to commute. It is a beautiful result of the constraints given in the problem.

Animated Solution for Mathematics - Matrices and Determinants: If $A$ and $B$ are two non-zero $n \times n$ matrices such that $A^{2} + B = A^{2} B$ , then

Select Answer:

Visualized Solution

Analyze the Given Equation

Rearranging the Terms

The Identity Matrix Trick

Grouping for Factorization

Completing the Factors

The Property of Commuting Inverses

Expanding the New Product

Simplifying the Expansion

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Identity Trick

Factoring the Expression

Proving Commutativity

Final Conclusion

Similar Questions

If $A$ and $B$ are square matrices of size $n \times n$ such that $A^{2} - B^{2} = (A - B) (A + B)$ , then which of the following will be always true?

If $A = [a b b a]$ and $A^{2} = [α β β α]$ , then

If $A$ and $B$ are square matrices of equal degree, then which one is correct among the followings?

Let $A = [1324]$ and $B = [a 0 0 b]$ , $a, b \in N$ . Then

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 :

If $A$ is a $3 \times 3$ non-singular matrix such that $A A^{'} = A^{'} A$ and $B = A^{- 1} A^{'}$ , then $B B^{'}$ equals:

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

Let $A = [12 - 1 α]$ and $B = [β 1 10], α, β \in R$ . Let $α_{1}$ be the value of $α$ which satisfies $(A + B)^{2} = A^{2} + [2222]$ and $α_{2}$ be the value of $α$ which satisfies $(A + B)^{2} = B^{2}$ . Then $∣ α_{1} - α_{2} ∣$ 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 $A = [a_{ij}]_{2 \times 2}$ , where $a_{ij} \neq = 0$ for all $i, j$ and $A^{2} = I$ . Let $a$ be the sum of all diagonal elements of $A$ and $b = ∣ A ∣$ . Then $3 a^{2} + 4 b^{2}$ is equal to

Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A and B are two non-zero n×n matrices such that A2+B=A2B, then

Select Answer:

Visualized Solution

Analyze the Given Equation

Rearranging the Terms

The Identity Matrix Trick

Grouping for Factorization

Completing the Factors

The Property of Commuting Inverses

Expanding the New Product

Simplifying the Expansion

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Identity Trick

Factoring the Expression

Proving Commutativity

Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A and B are square matrices of size n×n such that A2−B2=(A−B)(A+B), then which of the following will be always true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A=[ab​ba​] and A2=[αβ​βα​], then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A and B are square matrices of equal degree, then which one is correct among the followings?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=[13​24​] and B=[a0​0b​], a,b∈N. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A and B be two 3×3 real matrices such that (A2−B2) is invertible matrix. If A5=B5 and A3B2=A2B3, then the value of the determinant of the matrix A3+B3 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A is a 3×3 non-singular matrix such that AA′=A′A and B=A−1A′, then BB′ equals:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A,B and C be three 2×2 matrices with real entries such that B=(I+A)−1 and A+C=I. If BC=[1−1​−52​] and CB[x1​x2​​]=[12−6​], then x1​+x2​ is

.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=[aij​]2×2​, where aij​=0 for all i,j and A2=I. Let a be the sum of all diagonal elements of A and b=∣A∣. Then 3a2+4b2 is equal to

Animated Solution for Mathematics - Matrices and Determinants: If $A$ and $B$ are two non-zero $n \times n$ matrices such that $A^{2} + B = A^{2} B$ , then

If $A$ and $B$ are square matrices of size $n \times n$ such that $A^{2} - B^{2} = (A - B) (A + B)$ , then which of the following will be always true?

If $A = [a b b a]$ and $A^{2} = [α β β α]$ , then

If $A$ and $B$ are square matrices of equal degree, then which one is correct among the followings?

Let $A = [1324]$ and $B = [a 0 0 b]$ , $a, b \in N$ . Then

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 :

If $A$ is a $3 \times 3$ non-singular matrix such that $A A^{'} = A^{'} A$ and $B = A^{- 1} A^{'}$ , then $B B^{'}$ equals:

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

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 $A = [a_{ij}]_{2 \times 2}$ , where $a_{ij} \neq = 0$ for all $i, j$ and $A^{2} = I$ . Let $a$ be the sum of all diagonal elements of $A$ and $b = ∣ A ∣$ . Then $3 a^{2} + 4 b^{2}$ is equal to