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Question
If M is a 3×3 matrix satisfying
M⎡⎢⎣010⎤⎥⎦=⎡⎢⎣−123⎤⎥⎦,M⎡⎢⎣1−10⎤⎥⎦=⎡⎢⎣11−1⎤⎥⎦;M⎡⎢⎣111⎤⎥⎦=⎡⎢⎣0012⎤⎥⎦
Then, the sum of the diagonal entries of M is
If M is a 3×3 matrix satisfying
M⎡⎢⎣010⎤⎥⎦=⎡⎢⎣−123⎤⎥⎦,M⎡⎢⎣1−10⎤⎥⎦=⎡⎢⎣11−1⎤⎥⎦;M⎡⎢⎣111⎤⎥⎦=⎡⎢⎣0012⎤⎥⎦
Then, the sum of the diagonal entries of M is
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Solution
The correct option is A 9
Let M=⎡⎢⎣a1a2a3b1b2b3c1c2c3⎤⎥⎦ be the matrixAccording to question⎡⎢⎣a1a2a3b1b2b3c1c2c3⎤⎥⎦⎡⎢⎣010⎤⎥⎦=⎡⎢⎣−123⎤⎥⎦⇒⎡⎢⎣a2b2c2⎤⎥⎦=⎡⎢⎣−123⎤⎥⎦⇒a2=−1;b2=2;c2=3Again, ⎡⎢⎣a1a2a3b1b2b3c1c2c3⎤⎥⎦⎡⎢⎣1−10⎤⎥⎦=⎡⎢⎣11−1⎤⎥⎦⇒⎡⎢⎣a1−a2b1−b2c1−c2⎤⎥⎦=⎡⎢⎣11−1⎤⎥⎦⇒a1−a2=1⇒a1=0;b1−b2=1⇒b1=3;c1−c2=−1⇒c1=2Again ⎡⎢⎣a1a2a3b1b2b3c1c2c3⎤⎥⎦⎡⎢⎣111⎤⎥⎦=⎡⎢⎣0012⎤⎥⎦⇒⎡⎢⎣a1+a2+a3b1+b2+b3c1+c2+c3⎤⎥⎦=⎡⎢⎣0012⎤⎥⎦⇒a1+a2+a3=0⇒a3=1b1+b2+b3=0⇒b3=−5c1+c2+c3=12⇒c3=7Sum of diagonal=a1+b2+c3=0+2+7=9
Let M=⎡⎢⎣a1a2a3b1b2b3c1c2c3⎤⎥⎦ be the matrix
According to question
⎡⎢⎣a1a2a3b1b2b3c1c2c3⎤⎥⎦⎡⎢⎣010⎤⎥⎦=⎡⎢⎣−123⎤⎥⎦
⇒⎡⎢⎣a2b2c2⎤⎥⎦=⎡⎢⎣−123⎤⎥⎦⇒a2=−1;b2=2;c2=3
Again, ⎡⎢⎣a1a2a3b1b2b3c1c2c3⎤⎥⎦⎡⎢⎣1−10⎤⎥⎦=⎡⎢⎣11−1⎤⎥⎦
⇒⎡⎢⎣a1−a2b1−b2c1−c2⎤⎥⎦=⎡⎢⎣11−1⎤⎥⎦⇒a1−a2=1⇒a1=0;b1−b2=1⇒b1=3;c1−c2=−1⇒c1=2
Again ⎡⎢⎣a1a2a3b1b2b3c1c2c3⎤⎥⎦⎡⎢⎣111⎤⎥⎦=⎡⎢⎣0012⎤⎥⎦
⇒⎡⎢⎣a1+a2+a3b1+b2+b3c1+c2+c3⎤⎥⎦=⎡⎢⎣0012⎤⎥⎦
⇒a1+a2+a3=0⇒a3=1
b1+b2+b3=0⇒b3=−5
c1+c2+c3=12⇒c3=7
Sum of diagonal=a1+b2+c3
=0+2+7=9
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