Let A=⎡⎢⎣−123⎤⎥⎦ and B=[−2−1−4], then matrix (AB)A equals
A
12A
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B
−12A
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C
4A
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D
3A
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Solution
The correct option is C−12A A=⎡⎢⎣−123⎤⎥⎦,B=[−2−1−4] AB=⎡⎢⎣214−4−2−8−6−3−12⎤⎥⎦ (AB)A=⎡⎢⎣214−4−2−8−6−3−12⎤⎥⎦⎡⎢⎣−123⎤⎥⎦ ⇒(AB)A=⎡⎢⎣12−24−36⎤⎥⎦=−12⎡⎢⎣−123⎤⎥⎦=−12A