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Byju's Answer

Standard VII

Mathematics

Differences of Squares of Triangular Numbers and Converse

If A and ...

Question

If $A$ and $B$ are two square matrix of order $n$ then prove that : $(A B)^{- 1} = B^{- 1} A^{- 1}$

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Solution

Given A and B are two square matrix of order 'n'
$(A B) (A B)^{- 1} = 1$
Multiplying by $A^{- 1}$ on B.S
$A^{- 1} (A B) (A B)^{- 1} = A^{- 1} .1$
$(A^{- 1} A) B (A B)^{- 1} = A^{- 1}$
$1. B (A B)^{- 1} = A^{- 1}$
Multiplying by $B^{- 1}$ on B.S
$B^{- 1} B (A B)^{- 1} = B^{- 1} A^{- 1}$
$1. (A B)^{- 1} = B^{- 1} A^{- 1}$
$(A B)^{- 1} = B^{- 1} A^{- 1}$
$∴$ Hence proved

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Similar questions

Q. If A, B are two matrix then prove that

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

Q. If

A

and

B

are invertible matrices of the same order, then prove that

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

If A and B are square matrices of the same order such that AB=BA, then prove by induction that A $B^{n}$ = $B^{n}$ A. Further, prove that $(A B)^{n}$ = $A^{n}$ $B^{n}$ for all $n \in N$ .

Q. Consider the following statements:
1. If

A

and

B

are two square matrices of same order, then

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

.
2. If

A

and

B

are two square matrices of same order, then

(A B)^{n} = A^{n} B^{n}

.
3. If

A

and

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are two matrices such that

A B = A

and

B A = B

, then

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and

B

are idempotent.
Which of these is/are not correct?

Q. If A is an invertible matrix of order 3, then which of the following is not true
(a)

|adj A| = {|A|}^{2}

(b)

{(A^{- 1})}^{- 1} = A

B A = C A, than B \neq C

, where B and C are square matrices of order 3
(d)

{(A B)}^{- 1} = B^{- 1} A^{- 1}, where B \neq {[b_{i j}]}_{3 \times 3} and |B| \neq 0

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If A and B are two square matrix of order n then prove that : (AB)−1=B−1A−1

Given A and B are two square matrix of order 'n'(AB)(AB)−1=1Multiplying by A−1 on B.SA−1(AB)(AB)−1=A−1.1(A−1A)B(AB)−1=A−11.B(AB)−1=A−1Multiplying by B−1 on B.SB−1B(AB)−1=B−1A−11.(AB)−1=B−1A−1(AB)−1=B−1A−1∴ Hence proved

If $A$ and $B$ are two square matrix of order $n$ then prove that : $(A B)^{- 1} = B^{- 1} A^{- 1}$