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:

\(A_{11} = + \begin{vmatrix} 1 & -2 \\ 2 & -3 \end{vmatrix} = +(-3 + 4) = 1\)

\(A_{12} = - \begin{vmatrix} 0 & -2 \\ 1 & 2 \end{vmatrix} = -(0 + 2) = -2\)

\(A_{13} = + \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = +(0 - 1) = -1\)

\(A_{21} = - \begin{vmatrix} -3 & 1 \\ 2 & -3 \end{vmatrix} = -(9 - 2) = -7\)

\(A_{22} = + \begin{vmatrix} 2 & 1 \\ 1 & -3 \end{vmatrix} = +(-6 - 1) = -7\)

\(A_{23} = - \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} = -(4 - (-3)) = -7\)

\(A_{31} = + \begin{vmatrix} -3 & 1 \\ 1 & -2 \end{vmatrix} = +(6 - 1) = 5\)

\(A_{32} = - \begin{vmatrix} 2 & 1 \\ 0 & -2 \end{vmatrix} = -(-4 - 0) = 4\)

\(A_{33} = + \begin{vmatrix} 2  & -3 \\ 0 & 1 \end{vmatrix} = +(2 - 0) = 2\)

\(\therefore C = \begin{pmatrix} 1 & -2 & -1 \\ -7 & -7 & -7 \\ 5 & 4 & 2 \end{pmatrix}\)

\(\text{adj A} = C^{T} = \begin{pmatrix} 1 & -7 & 5 \\ -2 & -7 & 4 \\ -1 & -7 & 2 \end{pmatrix}\)

\(A^{-1} = \frac{\text{adj A}}{|A|} = \frac{1}{7} \begin{pmatrix} 1 & -7 & 5 \\ -2 & -7 & 4 \\ -1 & -7 & 2 \end{pmatrix}\)

\(x = A^{-1} b = \frac{1}{7} \begin{pmatrix} 1 & -7 & 5 \\ -2 & -7 & 4 \\ -1 & -7 & 2 \end{pmatrix} \begin{pmatrix} 10 \\ -7 \\ -9 \end{pmatrix}\)

= \(\frac{1}{7} \begin{pmatrix} 14 \\ -7 \\ 21 \end{pmatrix}\)

\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\)

\(x = 2; y = -1 ; z = 3\)

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