From the matrix equation AB $=$ AC, where A, B, C are the square matrices of same order, we can conclude B $=$ C provided.

A is non-singular

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A is singular

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A is symmetric

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A is skew symmetric

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Solution

The correct option is A A is non-singular
Given $A B = A C$
In order to prove that
$B = C$
We need to prove that $\frac{I}{A}$ or $A^{- 1}$ exists because $A B = A C$
$A B \times \frac{I}{A} = C$
$\Rightarrow B = C$
$∴ A^{- 1}$ should exist
$\Rightarrow d e t A \neq 0$
$\Rightarrow A$ should be non singular
option A is correct

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Similar questions

Q. If A , B are symmetric matrices of same order, then AB − BA is a A. Skew symmetric matrix B. Symmetric matrix C. Zero matrix D. Identity matrix

Q. If A and B are square matrices of same order and B is a skew-symmetric matrix, show that

A^{'} B A

is a skew-symmetric matrix.

Choose the correct answer in the following question:

If A,B are symmetric matrices of same order, then AB-BA is a
(a)skew-symmetric matrix
(b)symmetric matrix
(c)zero matrix
(d)identity matrix

Q. Let A and B are two non-singular square matrices,

A^{T}

and

B^{T}

are the transpose matrices of A and B respectively, then which of the following is correct

Q. If

A

and

B

are skew-symmetric matrices of the same order, prove that

A B + B A

is symmetric matrix.

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From the matrix equation AB=AC, where A, B, C are the square matrices of same order, we can conclude B=C provided.

The correct option is A A is non-singularGiven AB=ACIn order to prove thatB=CWe need to prove that IA or A−1 exists because AB=ACAB×IA=C⇒B=C∴A−1should exist⇒detA≠0⇒Ashould be non singularoption A is correct

From the matrix equation AB $=$ AC, where A, B, C are the square matrices of same order, we can conclude B $=$ C provided.