Question 6.1.10: Find det A for A = [1 2 3 4 5 0 2 3 4 5 0 0 3 4 5 0 0 0 4 5 ......

Linear Algebra with Applications [1016048]

Find det A for

$A = \begin{bmatrix}1&2&3&4&5\\0&2&3&4&5\\0&0&3&4&5\\0&0&0&4&5\\0&0&0&0&5\end{bmatrix}$ .

Step-by-Step

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Note that A is an upper triangular matrix. To have a nonzero product, a pattern must contain the first component of the first column, then the second component of the second column, and so on. Thus, only the diagonal pattern P makes a nonzero contribution. We conclude that

$det A = (\text{sgn } P)(\text{prod } P) = (−1)^0 1 · 2 · 3 · 4 · 5 = 5! = 120$ .

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