Engineering Board Exam Reviewer · Matrix Operations · medium
Given A= bmatrix1&2\3&4 bmatrix, find A2-5A.
- bmatrix-2&0\0&-2 bmatrix
- bmatrix1&0\0&1 bmatrix
- bmatrix2&0\0&2 bmatrix
- bmatrix2&2\2&2 bmatrix
Answer
C. bmatrix2&0\0&2 bmatrix
Solution
A2= bmatrix7&10\15&22 bmatrix and 5A= bmatrix5&10\15&20 bmatrix. Subtracting gives A2-5A= bmatrix2&0\0&2 bmatrix.
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