Let A-B and C be square matrices of order 3- 3- If A is invertible and A-BC-BA-1- then

We have, (A B)C=BA^ 1 ⇒ (A B)C BA^ 1+AA^ 1=I, where I is identity matrix. ⇒ (A B)C+(A B)A^ 1=I ⇒ (A B)(C+A^ 1)=I Therefore, C+A^ 1 is the inverse of A B. So, (C ...

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Byju's Answer

Standard XII

Mathematics

Square Matrix

Let A,B and C...

Question

Let $A, B$ and $C$ be square matrices of order $3 \times 3.$ If $A$ is invertible and $(A - B) C = B A^{- 1},$ then

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

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

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

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All of the above

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Solution

The correct option is A $C (A - B) = A^{- 1} B$
We have, $(A - B) C = B A^{- 1}$
$\Rightarrow (A - B) C - B A^{- 1} + A A^{- 1} = I,$ where $I$ is identity matrix.
$\Rightarrow (A - B) C + (A - B) A^{- 1} = I$
$\Rightarrow (A - B) (C + A^{- 1}) = I$
Therefore, $C + A^{- 1}$ is the inverse of $A - B .$
So, $(C + A^{- 1}) (A - B) = I$
$\Rightarrow C (A - B) + A^{- 1} A - A^{- 1} B = I$
$\Rightarrow C (A - B) = A^{- 1} B$

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Let A,B and C be square matrices of order 3×3. If A is invertible and (A−B)C=BA−1, then

The correct option is A C(A−B)=A−1BWe have, (A−B)C=BA−1 ⇒(A−B)C−BA−1+AA−1=I, where I is identity matrix. ⇒(A−B)C+(A−B)A−1=I ⇒(A−B)(C+A−1)=I Therefore, C+A−1 is the inverse of A−B. So, (C+A−1)(A−B)=I ⇒C(A−B)+A−1A−A−1B=I ⇒C(A−B)=A−1B

Let $A, B$ and $C$ be square matrices of order $3 \times 3.$ If $A$ is invertible and $(A - B) C = B A^{- 1},$ then