which has the solution 21(k) = A(-1)*(k – 1)!, where A is an arbitrary function. We now have (E – k)yk = A(-1)*(k – 1)!, | or Yk+1 – kyk = A(-1)* (k – 1)!. The solution to this latter equation is k = (k – 1)! (A(-1)*(i – 1)! i! +B Therefore, the general solution to equation (3.159) is k-1 Yk = | A£ (-1)* +B (k – 1)!. i i=1

Let a, b and c be constant numbers. Show that 1 1 1 a b C = (b− a)(ca)(c - b) a² b² c² You can only use row or column operations to transform the above determinant into a lower or upper triangular form, then use a theorem from the lecture notes to find the determinant. Explain your steps or theorems you are using.

Evaluate the following fourth-order determinate: 2 1 0 2 3 4 -1 D = 5 2 3 -2 1 2 0 3.

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