Question

Functions can be thought of as elements of a vector space. When we have two functions, we will make a 2x2 matrix of functions here the first row is the original functions and the second row are the first derivatives of the original functions. With three functions, we have two more rows, the first derivatives in row two and the second derivatives in row three. The matrix form is called the Wronskian. If you take the determinant and it is not zero, the functions are linearly independent. If the determinant is zero, the original function were linearly dependent. (10 points) Fill in the second row of this matrix and take the determinant to see if the functions are linearly independent. 3x – 7 x2-11 determinant = Fill in the second and third rows of this matrix and take the determinant to see if the functions are linearly independent. x2 + 1 x +1 x2-xl determinant =

Answer 1

for 2X2 matriu (30-7) and (x²-1) first derivate of - elements of first sow are So, elements of second row are (3x-7) and (x2-1) d (32-7)= 3 da d (x²-1) = 2x au And our matria ý 3e-7 - 3 x²-1) 2x] - Determinant of the matrix D= 34-7 w21) 13 23 D = (3x-7)(2X) - (x2-1X3] = 6x²_140-34²+3 = 3240 142 +3 - As determinant of matsin is not zero , the functions are linearly independent

7 For 3x3 matrix - elements of first row are (x² + 1), (xit1) and (2²4) and elements of second and third row are first I and second clexuative of Original function. fisst derivative of is (2241) amese (x+1) = 1 = 24-1 Second derivative is et (1941) 5 (2323 24 Cabo coco So, our matrin i «²t1 xt1 e²x 12 24-1 Le 0 2 - and determinant is x² + 1 7 2x 1 2 xal xxl 22-11 0 2 = (2+1) [ 1X2 - 0(2x1)]-(2+1)/(22)(2) -2622-1)] + (x+x)[(2w)(0) - 2] c 2(22+1) -(+1) C 498 -+2) +24) (-2) = 2**+ 2 - - - 24+24 as Determinant a is Zero, the function are linearly defendent, Page No.