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Q1. a) Solve the following set of linear equations by the matrix method. X1+2X2+X3=4, 3X1−4X2−2X3=2, 5X1+3X2+5X3=-1. b) Find the values of x for which the matrix [[2x, -2],[-4, x]] is singular.

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Q1 a) Solve the following set of linear equations by the matrix method.
The given system of equations is:
X1+2X2+X3=4X_1 + 2X_2 + X_3 = 4
3X1−4X2−2X3=23X_1 - 4X_2 - 2X_3 = 2
5X1+3X2+5X3=−15X_1 + 3X_2 + 5X_3 = -1

  1. Write the system in matrix form AX=BAX = B.

    A=[1213−4−2535],X=[X1X2X3],B=[42−1]A = \begin{bmatrix} 1 & 2 & 1 \\ 3 & -4 & -2 \\ 5 & 3 & 5 \end{bmatrix}, \quad X = \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix}, \quad B = \begin{bmatrix} 4 \\ 2 \\ -1 \end{bmatrix}

  2. Calculate the determinant of matrix AA.

    det⁡(A)=1((−4)(5)−(−2)(3))−2((3)(5)−(−2)(5))+1((3)(3)−(−4)(5))\det(A) = 1((-4)(5) - (-2)(3)) - 2((3)(5) - (-2)(5)) + 1((3)(3) - (-4)(5)) det⁡(A)=1(−20+6)−2(15+10)+1(9+20)\det(A) = 1(-20 + 6) - 2(15 + 10) + 1(9 + 20) det⁡(A)=1(−14)−2(25)+1(29)\det(A) = 1(-14) - 2(25) + 1(29) det⁡(A)=−14−50+29=−35\det(A) = -14 - 50 + 29 = -35

  3. Calculate the adjoint of matrix AA.

    First, find the cofactor matrix CC:

    C=[(−4)(5)−(−2)(3)−((3)(5)−(−2)(5))(3)(3)−(−4)(5)−((2)(5)−(1)(3))(1)(5)−(1)(5)−((1)(3)−(2)(5))((2)(−2)−(1)(−4))−((1)(−2)−(1)(3))((1)(−4)−(2)(3))]C = \begin{bmatrix} (-4)(5)-(-2)(3) & -((3)(5)-(-2)(5)) & (3)(3)-(-4)(5) \\ -((2)(5)-(1)(3)) & (1)(5)-(1)(5) & -((1)(3)-(2)(5)) \\ ((2)(-2)-(1)(-4)) & -((1)(-2)-(1)(3)) & ((1)(-4)-(2)(3)) \end{bmatrix} C=[−14−2529−70705−10]C = \begin{bmatrix} -14 & -25 & 29 \\ -7 & 0 & 7 \\ 0 & 5 & -10 \end{bmatrix}

    The adjoint matrix is the transpose of the cofactor matrix:

    adj(A)=CT=[−14−70−2505297−10]\text{adj}(A) = C^T = \begin{bmatrix} -14 & -7 & 0 \\ -25 & 0 & 5 \\ 29 & 7 & -10 \end{bmatrix}

  4. Calculate the inverse of matrix AA.

    A−1=1det⁡(A)adj(A)=1−35[−14−70−2505297−10]A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{-35} \begin{bmatrix} -14 & -7 & 0 \\ -25 & 0 & 5 \\ 29 & 7 & -10 \end{bmatrix}

  5. Solve for XX using X=A−1BX = A^{-1}B.

    X=1−35[−14−70−2505297−10][42−1]X = \frac{1}{-35} \begin{bmatrix} -14 & -7 & 0 \\ -25 & 0 & 5 \\ 29 & 7 & -10 \end{bmatrix} \begin{bmatrix} 4 \\ 2 \\ -1 \end{bmatrix} X=1−35[(−14)(4)+(−7)(2)+(0)(−1)(−25)(4)+(0)(2)+(5)(−1)(29)(4)+(7)(2)+(−10)(−1)]X = \frac{1}{-35} \begin{bmatrix} (-14)(4) + (-7)(2) + (0)(-1) \\ (-25)(4) + (0)(2) + (5)(-1) \\ (29)(4) + (7)(2) + (-10)(-1) \end{bmatrix} X=1−35[−56−14+0−100+0−5116+14+10]X = \frac{1}{-35} \begin{bmatrix} -56 - 14 + 0 \\ -100 + 0 - 5 \\ 116 + 14 + 10 \end{bmatrix} X=1−35[−70−105140]=[−70−35−105−35140−35]=[23−4]X = \frac{1}{-35} \begin{bmatrix} -70 \\ -105 \\ 140 \end{bmatrix} = \begin{bmatrix} \frac{-70}{-35} \\ \frac{-105}{-35} \\ \frac{140}{-35} \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ -4 \end{bmatrix}

    The solution is:

    X1=2,X2=3,X3=−4\boxed{X_1 = 2, X_2 = 3, X_3 = -4}

    Q1 b) Find the values of xx for which the matrix [2x−2−4x]\begin{bmatrix} 2x & -2 \\ -4 & x \end{bmatrix} is singular.

  6. Understand the condition for a singular matrix.

    A matrix is singular if its determinant is equal to zero.

  7. Calculate the determinant of the given matrix.

    For a 2×22 \times 2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, the determinant is ad−bcad - bc.

    det⁡[2x−2−4x]=(2x)(x)−(−2)(−4)\det \begin{bmatrix} 2x & -2 \\ -4 & x \end{bmatrix} = (2x)(x) - (-2)(-4) det⁡=2x2−8\det = 2x^2 - 8

  8. Set the determinant to zero and solve for xx.

    2x2−8=02x^2 - 8 = 0 2x2=82x^2 = 8 x2=82x^2 = \frac{8}{2} x2=4x^2 = 4 x=±4x = \pm \sqrt{4} x=±2\boxed{x = \pm 2}

    Q2 a) (i) Evaluate the following ∫124cos⁡t dt\int_{1}^{2} 4 \cos t \,dt.

  9. Find the antiderivative of 4cos⁡t4 \cos t.

    The antiderivative of cos⁡t\cos t is sin⁡t\sin t.
    So, the antiderivative of 4cos⁡t4 \cos t is 4sin⁡t4 \sin t.

  10. Evaluate the definite integral using the Fundamental Theorem of Calculus.

    ∫124cos⁡t dt=[4sin⁡t]12\int_{1}^{2} 4 \cos t \,dt = [4 \sin t]_{1}^{2} =4sin⁡(2)−4sin⁡(1)= 4 \sin(2) - 4 \sin(1) =4(sin⁡(2)−sin⁡(1))= 4(\sin(2) - \sin(1)) 4(sin⁡(2)−sin⁡(1))\boxed{4(\sin(2) - \sin(1))}

    (Note: The angles are in radians.)

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