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NECO 2022 · Paper 2 · Q4

Find the inverse of the matrix A=(3452)A = \begin{pmatrix} 3 & 4 \\ 5 & 2 \end{pmatrix} and use it to solve the equations 3x+4y=13x + 4y = 1 and 5x+2y=35x + 2y = 3 simultaneously.

  1. (a)

    Find A−1A^{-1}. (Enter its four entries.)

    Separate values with commas, e.g. 3, −2

  2. (b)

    Use A−1A^{-1} to find xx and yy.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1. The determinant is ∣A∣=3×2−4×5|A| = 3 \times 2 - 4 \times 5
    =6−20= 6 - 20
    =−14= -14.
  2. Swap the leading-diagonal entries and change the signs of the other two: the adjoint is (2−4−53)\begin{pmatrix} 2 & -4 \\ -5 & 3 \end{pmatrix}.
  3. Divide by the determinant: A−1=−114(2−4−53)A^{-1} = -\frac{1}{14}\begin{pmatrix} 2 & -4 \\ -5 & 3 \end{pmatrix}
    =(−1727514−314)= \begin{pmatrix} -\frac17 & \frac27 \\ \frac{5}{14} & -\frac{3}{14} \end{pmatrix}.

(b)

  1. Write the equations as A(xy)=(13)A\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}, so (xy)=A−1(13)\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\begin{pmatrix} 1 \\ 3 \end{pmatrix}.
  2. Multiply the adjoint by the column: (2(1)−4(3)−5(1)+3(3))=(−104)\begin{pmatrix} 2(1) - 4(3) \\ -5(1) + 3(3) \end{pmatrix} = \begin{pmatrix} -10 \\ 4 \end{pmatrix}.
  3. Multiply by −114-\frac{1}{14}: x=1014=57x = \frac{10}{14} = \frac57 and y=−414=−27y = -\frac{4}{14} = -\frac27.
  4. Check in the first equation: 3(57)+4(−27)3\left(\frac57\right) + 4\left(-\frac27\right) is 157−87\frac{15}{7} - \frac{8}{7}, which equals 1.
  5. So x=57x = \frac57 and y=−27y = -\frac27.

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