From the matrix equation $A B = A C$ we can conclude $B = C$ provided

A

is singular

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A

is non-singular

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A

is symmetric

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A

is square

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Solution

The correct option is B $A$ is non-singular
Let $| A | \neq 0$
$∴ A^{- 1}$ exists.
Given $A B = A C$
Now, pre multiplying by $A^{- 1}$ on both sides. we get,
$A^{- 1} A B = A^{- 1} A C$
$\Rightarrow I B = I C$
$\Rightarrow B = C$
So, $A$ is non singular.
Option $B$ is correct.

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From the matrix equation AB=AC we can conclude B=C provided

The correct option is B A is non-singularLet |A|≠0∴A−1 exists.Given AB=ACNow, pre multiplying by A−1 on both sides. we get,A−1AB=A−1AC⇒IB=IC⇒B=CSo, A is non singular.Option B is correct.

From the matrix equation $A B = A C$ we can conclude $B = C$ provided

The correct option is B $A$ is non-singular
Let $| A | \neq 0$
$∴ A^{- 1}$ exists.
Given $A B = A C$
Now, pre multiplying by $A^{- 1}$ on both sides. we get,
$A^{- 1} A B = A^{- 1} A C$
$\Rightarrow I B = I C$
$\Rightarrow B = C$
So, $A$ is non singular.
Option $B$ is correct.