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InterviewSolution
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The value of the determinant `|(1,1,1),(.^(m)C_(1),.^(m +1)C_(1),.^(m+2)C_(1)),(.^(m)C_(2),.^(m +1)C_(2),.^(m+2)C_(2))|` is equal toA. 1B. -1C. 0D. none of these |
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Answer» Correct Answer - A `|{:(1,,1,,1),(.^(m)C_(1),,.^(m+1)C_(1),,.^(m+2)C_(1)),(.^(m)C_(2),,.^(m+1)C_(2),,.^(m+2)C_(2)):}|` `=|{:(1,,1,,1),(.^(m)C_(1),,.^(m+1)C_(1),,.^(m+1)C_(0)+.^(m+1)C_(1)),(.^(m)C_(1),,.^(m+1)C_(2),,.^(m+1)C_(1)+.^(m+1)C_(2)):}| |{:(1,,1,,1),(.^(m)C_(1),,.^(m+1)C_(1),,.^(m+1)C_(0)),(.^(m)C_(2),,,^(m+1)C_(2),,.^(m+1)C_(1)):}|" ""[Applying " C_(3) to C_(3) -C_(2)"]"` `=|{:(1,,1,,1),(.^(m)C_(1),,.^(m)C_(0).^(m)C_(1),,.^(m+1)C_(0)),(.^(m)C_(2),,.^(m)C_(1)+.^(m)C_(2),,.^(m+1)C_(1)):}|` `=|{:(1,,0,,0),(.^(m)C_(1),,.^(m)C_(0),,.^(m+1)C_(0)),(.^(m)C_(2),,.^(m)C_(1),,.^(m+1)C_(1)):}|" ""[Applying " C_(2) to C_(2) -C_(1)" ]"` `=.^(m)C_(0).^(m+1)C_(1)-^(m+1)C_(0).^(m)C_(1)` `=m +1-m` `=1` |
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