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3 parts
a

x=3,y=0x=3, y=0

b

x=−1,y=83x=-1, y=\frac{8}{3}

c

x=3,y=0x=3, y=0

Step-by-step solution

12 steps

Here are the solutions to the simultaneous equations using the matrix method:

Part a· 4 steps

Given the equations

2x+y=62x + y = 6 x−y=3x - y = 3

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

    [211−1][xy]=[63]\begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \end{bmatrix}

  2. Calculate the determinant of matrix AA.

    det⁡(A)=(2)(−1)−(1)(1)=−2−1=−3\det(A) = (2)(-1) - (1)(1) = -2 - 1 = -3

  3. Find the inverse of matrix AA, A−1A^{-1}.

    A−1=1det⁡(A)[−1−1−12]=1−3[−1−1−12]=[131313−23]A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} -1 & -1 \\ -1 & 2 \end{bmatrix} = \frac{1}{-3} \begin{bmatrix} -1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{bmatrix}

  4. Multiply A−1A^{-1} by BB to find XX.

    [xy]=[131313−23][63]=[13(6)+13(3)13(6)+(−23)(3)]=[2+12−2]=[30]\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{bmatrix} \begin{bmatrix} 6 \\ 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{3}(6) + \frac{1}{3}(3) \\ \frac{1}{3}(6) + (-\frac{2}{3})(3) \end{bmatrix} = \begin{bmatrix} 2 + 1 \\ 2 - 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \end{bmatrix}

    Thus, x=3x=3 and y=0y=0.
    The solution is x=3,y=0\boxed{x=3, y=0}.

Part b

x=−1,y=83x=-1, y=\frac{8}{3}

Given the equations

5x+3y=35x + 3y = 3 3y−11=3x3y - 11 = 3x

  1. Rearrange the second equation to the standard form ax+by=cax + by = c.

    5x+3y=35x + 3y = 3 −3x+3y=11-3x + 3y = 11

    Write the equations in matrix form AX=BAX = B.

    [53−33][xy]=[311]\begin{bmatrix} 5 & 3 \\ -3 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 11 \end{bmatrix}

  2. Calculate the determinant of matrix AA.

    det⁡(A)=(5)(3)−(3)(−3)=15−(−9)=15+9=24\det(A) = (5)(3) - (3)(-3) = 15 - (-9) = 15 + 9 = 24

  3. Find the inverse of matrix AA, A−1A^{-1}.

    A−1=1det⁡(A)[3−335]=124[3−335]=[324−324324524]=[18−1818524]A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} 3 & -3 \\ 3 & 5 \end{bmatrix} = \frac{1}{24} \begin{bmatrix} 3 & -3 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} \frac{3}{24} & -\frac{3}{24} \\ \frac{3}{24} & \frac{5}{24} \end{bmatrix} = \begin{bmatrix} \frac{1}{8} & -\frac{1}{8} \\ \frac{1}{8} & \frac{5}{24} \end{bmatrix}

  4. Multiply A−1A^{-1} by BB to find XX.

    [xy]=[18−1818524][311]=[18(3)+(−18)(11)18(3)+524(11)]=[38−11838+5524]=[−88924+5524]=[−16424]=[−183]\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{8} & -\frac{1}{8} \\ \frac{1}{8} & \frac{5}{24} \end{bmatrix} \begin{bmatrix} 3 \\ 11 \end{bmatrix} = \begin{bmatrix} \frac{1}{8}(3) + (-\frac{1}{8})(11) \\ \frac{1}{8}(3) + \frac{5}{24}(11) \end{bmatrix} = \begin{bmatrix} \frac{3}{8} - \frac{11}{8} \\ \frac{3}{8} + \frac{55}{24} \end{bmatrix} = \begin{bmatrix} -\frac{8}{8} \\ \frac{9}{24} + \frac{55}{24} \end{bmatrix} = \begin{bmatrix} -1 \\ \frac{64}{24} \end{bmatrix} = \begin{bmatrix} -1 \\ \frac{8}{3} \end{bmatrix}

    Thus, x=−1x=-1 and y=83y=\frac{8}{3}.
    The solution is x=−1,y=83\boxed{x=-1, y=\frac{8}{3}}.

Part c

x=3,y=0x=3, y=0

Given the equations

x+y=3x + y = 3 x−y=3x - y = 3

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

    [111−1][xy]=[33]\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix}

  2. Calculate the determinant of matrix AA.

    det⁡(A)=(1)(−1)−(1)(1)=−1−1=−2\det(A) = (1)(-1) - (1)(1) = -1 - 1 = -2

  3. Find the inverse of matrix AA, A−1A^{-1}.

    A−1=1det⁡(A)[−1−1−11]=1−2[−1−1−11]=[121212−12]A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} = \frac{1}{-2} \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}

  4. Multiply A−1A^{-1} by BB to find XX.

    [xy]=[121212−12][33]=[12(3)+12(3)12(3)+(−12)(3)]=[32+3232−32]=[620]=[30]\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} 3 \\ 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}(3) + \frac{1}{2}(3) \\ \frac{1}{2}(3) + (-\frac{1}{2})(3) \end{bmatrix} = \begin{bmatrix} \frac{3}{2} + \frac{3}{2} \\ \frac{3}{2} - \frac{3}{2} \end{bmatrix} = \begin{bmatrix} \frac{6}{2} \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \end{bmatrix}

    Thus, x=3x=3 and y=0y=0.
    The solution is x=3,y=0\boxed{x=3, y=0}.

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