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# Let A,B and C be square matrices such that AB=AC , If A neq 0 , then B=C. Is this True or False?Explain the reasosing behind the answer.

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Matrices
asked 2021-01-31
Let A,B and C be square matrices such that AB=AC , If $$A \neq 0$$ , then B=C.
Is this True or False?Explain the reasosing behind the answer.

## Answers (1)

2021-02-01
Step 1
Let A, B, and C be square matrices such that AB=AC.
If $$A \neq 0$$, then B=C.
Determine true or false.
Step 2
Let A, B, and C be square matrices such that AB=AC.
If $$A \neq 0$$ , then B=C.
This is true only if matrix A is invertible.
If A is invertible then $$A^{-1}$$ exist.
Multiply given equationAB=AC by $$A^{-1}$$ on both sides,
$$A^{-1}AB=A^{-1}AC$$
$$(A^{-1}A)B=(A^{-1}A)C$$
$$I \times B = I \times C$$
B=C
If A is not invertible then this statement is false.
Counter example:
$$A=\begin{bmatrix}2 & -3 \\-4 & 6 \end{bmatrix} , B=\begin{bmatrix}8 & 4 \\5 & 5 \end{bmatrix} , C=\begin{bmatrix}5 & -2 \\3 & 1 \end{bmatrix}$$
Here, $$det|A|=2 \times 6 -(-4) \times (-3)$$
=12-12
=0
That is A is not invertible.
Now, find AB and AC,
$$AB=\begin{bmatrix}2 & -3 \\-4 & 6 \end{bmatrix}\begin{bmatrix}8 & 4 \\5 & 5 \end{bmatrix}$$
$$=\begin{bmatrix}16-15 & 8-15 \\-32+30 & -16+30 \end{bmatrix}$$
$$=\begin{bmatrix}1 & -7 \\-2 & 14 \end{bmatrix}$$
$$AC=\begin{bmatrix}2 & -3 \\-4 & 6 \end{bmatrix}\begin{bmatrix}5 & -2 \\3 & 1 \end{bmatrix}$$
$$=\begin{bmatrix}10-9 & -4-3 \\-20+18 & 8+6 \end{bmatrix}$$
$$=\begin{bmatrix}1 & -7 \\-2 & 14 \end{bmatrix}$$
That is, AB=AC but B is not the same as C.

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