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Find the value of det(22A) for the following matrix:\({\rm{A}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 2&3&1\\ { - 1}&0&2\\ { - 3}&1&2 \end{array}} \right]\)1. 13402. -12823. -10884. 1542 |
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Answer» Correct Answer - Option 3 : -1088 Concept: Property of determinant of a matrix: Let A be a matrix of order n × n then det(kA) = kn det(A)
Given: \({\rm{A}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 2&3&1\\ { - 1}&0&2\\ { - 3}&1&2 \end{array}} \right]\) Here the order of the matrix is 3. Now, det(A) = 2(0 - 2) - 3(-2 + 6) + 1(-1 + 0) = 2(-2) - 3(4) + 1(-1) = - 4 - 12 - 1 = -17 Now using the property the value of det(22A) is: det(22A) = det(4A) = 43 det(A) = 64 × -17 = -1088 |
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