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If A⁻¹ = [[1, 2], [3, 4]] and B⁻¹ = [[-2, 4], [6, 8]], find (a) (A^T)⁻¹, (b) (AB)⁻¹, (c) (2B)⁻¹. If P = [[3, -2], [0, 2]], find P⁻¹.

Mathematics

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Answers

3 parts
a

(1324)\begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}

b

(10123044)\begin{pmatrix} 10 & 12 \\ 30 & 44 \end{pmatrix}

c

(−1234)\begin{pmatrix} -1 & 2 \\ 3 & 4 \end{pmatrix}

(1313012)\begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ 0 & \frac{1}{2} \end{pmatrix}

Step-by-step solution

9 steps

Here are the solutions to the matrix problems.

1) Given A−1=(1234)A^{-1} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} and B−1=(−2468)B^{-1} = \begin{pmatrix} -2 & 4 \\ 6 & 8 \end{pmatrix}.

Part a· 2 steps

Find (AT)−1(A^T)^{-1}

  1. Use the property (AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T.

    (AT)−1=((1234))T(A^T)^{-1} = \left( \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \right)^T
  2. Transpose the matrix A−1A^{-1} by swapping its rows and columns.

    (AT)−1=(1324)(A^T)^{-1} = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}

    The result is (1324)\boxed{\begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}}.

Part b

(10123044)\begin{pmatrix} 10 & 12 \\ 30 & 44 \end{pmatrix}

Find (AB)−1(AB)^{-1}

  1. Use the property (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}.

    (AB)−1=(−2468)(1234)(AB)^{-1} = \begin{pmatrix} -2 & 4 \\ 6 & 8 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}
  2. Perform the matrix multiplication.

    (AB)−1=((−2)(1)+(4)(3)(−2)(2)+(4)(4)(6)(1)+(8)(3)(6)(2)+(8)(4))(AB)^{-1} = \begin{pmatrix} (-2)(1) + (4)(3) & (-2)(2) + (4)(4) \\ (6)(1) + (8)(3) & (6)(2) + (8)(4) \end{pmatrix} (AB)−1=(−2+12−4+166+2412+32)(AB)^{-1} = \begin{pmatrix} -2 + 12 & -4 + 16 \\ 6 + 24 & 12 + 32 \end{pmatrix} (AB)−1=(10123044)(AB)^{-1} = \begin{pmatrix} 10 & 12 \\ 30 & 44 \end{pmatrix}

    The result is (10123044)\boxed{\begin{pmatrix} 10 & 12 \\ 30 & 44 \end{pmatrix}}.

Part c

(−1234)\begin{pmatrix} -1 & 2 \\ 3 & 4 \end{pmatrix}

Find (2B)−1(2B)^{-1}

  1. Use the property (kB)−1=1kB−1(kB)^{-1} = \frac{1}{k} B^{-1}.

    Here k=2k=2.

    (2B)−1=12B−1(2B)^{-1} = \frac{1}{2} B^{-1} (2B)−1=12(−2468)(2B)^{-1} = \frac{1}{2} \begin{pmatrix} -2 & 4 \\ 6 & 8 \end{pmatrix}
  2. Multiply each element of B−1B^{-1} by 12\frac{1}{2}.

    (2B)−1=(−22426282)(2B)^{-1} = \begin{pmatrix} \frac{-2}{2} & \frac{4}{2} \\ \frac{6}{2} & \frac{8}{2} \end{pmatrix} (2B)−1=(−1234)(2B)^{-1} = \begin{pmatrix} -1 & 2 \\ 3 & 4 \end{pmatrix}

    The result is (−1234)\boxed{\begin{pmatrix} -1 & 2 \\ 3 & 4 \end{pmatrix}}.

    2) If P=(3−202)P = \begin{pmatrix} 3 & -2 \\ 0 & 2 \end{pmatrix}, find P−1P^{-1}.

  3. Calculate the determinant of PP.

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

    det⁡(P)=(3)(2)−(−2)(0)=6−0=6\det(P) = (3)(2) - (-2)(0) = 6 - 0 = 6
  4. Use the formula for the inverse of a 2×22 \times 2 matrix

    P−1=1det⁡(P)(d−b−ca)P^{-1} = \frac{1}{\det(P)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.

    P−1=16(2−(−2)−03)P^{-1} = \frac{1}{6} \begin{pmatrix} 2 & -(-2) \\ -0 & 3 \end{pmatrix} P−1=16(2203)P^{-1} = \frac{1}{6} \begin{pmatrix} 2 & 2 \\ 0 & 3 \end{pmatrix}
  5. Multiply each element by 16\frac{1}{6}.

    P−1=(26260636)P^{-1} = \begin{pmatrix} \frac{2}{6} & \frac{2}{6} \\ \frac{0}{6} & \frac{3}{6} \end{pmatrix} P−1=(1313012)P^{-1} = \begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ 0 & \frac{1}{2} \end{pmatrix}

    The inverse of PP is (1313012)\boxed{\begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ 0 & \frac{1}{2} \end{pmatrix}}.

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