(a) Evaluate (int_{1} ^{2} frac{x}{sqrt{5 – x^{2}}} mathrm {d} x)
(b)(i) Evaluate: (begin{vmatrix} 2 & -3 & 1 \ 0 & 1 & -2 \ 1 & 2 & -3 end{vmatrix})
(ii) Using your answer in b(i), solve the simultaneous equations :
(2x – 3y + z = 10)
(y – 2z = -7)
(x + 2y – 3z = -9)
Explanation
(a) (int_{1} ^{2} frac{x}{sqrt{5 – x^{2}}} mathrm {d} x)
Let (u^{2} = 5 – x^{2}).
(2u mathrm {d} u = – 2x mathrm {d} x implies u mathrm {d} u = – x mathrm {d} x)
(int frac{x}{sqrt{5 – x^{2}}} mathrm {d} x = int frac{- u}{sqrt{u^{2}}} mathrm {d} u)
= (- int mathrm {d} u )
= (-[5 – x^{2}]_{1} ^{2})
= ((- (5 – 2^{2}) – (- (5 – 1^{2}))))
= (- 1 + 4 = 3)
(b)(i) (begin{vmatrix} 2 & -3 & 1 \ 0 & 1 & -2 \ 1 & 2 & -3 end{vmatrix})
= (2(-3 + 4) + 3(0 + 2) + 1(0 – 1) = 7)
(ii) (2x – 3y + z = 10)
(0x + y – 2z = -7)
(x + 2y – 3z = -9)
Let A be the matrix of coefficients.
(A = begin{pmatrix} 2 & -3 & 1 \ 0 & 1 & -2 \ 1 & 2 & -3 end{pmatrix})
We write the set of equations in matrix form, i.e Ax = B.
(begin{pmatrix} 2 & -3 & 1 \ 0 & 1 & -2 \ 1 & 2 & -3 end{pmatrix} begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 10 \ -7 \ -9 end{pmatrix}).
We find the inverse of A, A(^{-1}).
Determinant of A = 7.
Co-factors:
(A_{11} = + begin{vmatrix} 1 & -2 \ 2 & -3 end{vmatrix} = +(-3 + 4) = 1)
(A_{12} = – begin{vmatrix} 0 & -2 \ 1 & 2 end{vmatrix} = -(0 + 2) = -2)
(A_{13} = + begin{vmatrix} 0 & 1 \ 1 & 2 end{vmatrix} = +(0 – 1) = -1)
(A_{21} = – begin{vmatrix} -3 & 1 \ 2 & -3 end{vmatrix} = -(9 – 2) = -7)
(A_{22} = + begin{vmatrix} 2 & 1 \ 1 & -3 end{vmatrix} = +(-6 – 1) = -7)
(A_{23} = – begin{vmatrix} 2 & -3 \ 1 & 2 end{vmatrix} = -(4 – (-3)) = -7)
(A_{31} = + begin{vmatrix} -3 & 1 \ 1 & -2 end{vmatrix} = +(6 – 1) = 5)
(A_{32} = – begin{vmatrix} 2 & 1 \ 0 & -2 end{vmatrix} = -(-4 – 0) = 4)
(A_{33} = + begin{vmatrix} 2 & -3 \ 0 & 1 end{vmatrix} = +(2 – 0) = 2)
(therefore C = begin{pmatrix} 1 & -2 & -1 \ -7 & -7 & -7 \ 5 & 4 & 2 end{pmatrix})
(text{adj A} = C^{T} = begin{pmatrix} 1 & -7 & 5 \ -2 & -7 & 4 \ -1 & -7 & 2 end{pmatrix})
(A^{-1} = frac{text{adj A}}{|A|} = frac{1}{7} begin{pmatrix} 1 & -7 & 5 \ -2 & -7 & 4 \ -1 & -7 & 2 end{pmatrix})
(x = A^{-1} b = frac{1}{7} begin{pmatrix} 1 & -7 & 5 \ -2 & -7 & 4 \ -1 & -7 & 2 end{pmatrix} begin{pmatrix} 10 \ -7 \ -9 end{pmatrix})
= (frac{1}{7} begin{pmatrix} 14 \ -7 \ 21 end{pmatrix})
(begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 2 \ -1 \ 3 end{pmatrix})
(x = 2; y = -1 ; z = 3)