Find the determinant of the matrix with 5 's on the diagonal, 1 's above the diagonal, and 0 's below the
diagonal. (Note that your answer will be a function of .)
For any upper triangular matrix, the determinant is the product of the diagonal elements, regardless of the entries above the diagonal.
Given that all diagonal entries of are , the determinant of is:
Final Answer:
After simplifying, the determinant is five to the nth power. Thus det A = 5ⁿ
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The Deep Dive
To find the determinant of the matrix , we can denote it as an upper triangular matrix with 5’s on the diagonal and 1’s above the diagonal. The matrix takes the form:
Using properties of determinants, specifically for triangular matrices, the determinant is the product of the diagonal entries. However, our configuration has additional non-zero entries above the diagonal. We can utilize a recursive approach or apply a pattern recognition with determinants of smaller sizes.
Let’s denote for an matrix . We can establish a relationship based on the size of the matrix. We consider expanding the determinant:
For , .
For , the matrix becomes .
Thus, .
For larger matrices, we could use the Laplace expansion or other properties, but we note that the determinant relates recursively to . Generally, the pattern continues as:
With initial conditions and . Solving this recurrence relationship gives a general expression.
Ultimately, the solution to the recurrence yields: