Question 12 of 58intermediate🔧 ApplyLong Answer4 marks

Let A = [3725]\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} and B = [6879]\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}. Verify that (AB)^{-1} = B−1B^{-1} A−1A^{-1}.

Correct Answer

First, we determine the product AB. AB=[3725][6879]=[18+4924+6312+3516+45]=[67874761]AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 18+49 & 24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}

Next, we find (AB)−1(AB)^{-1}. The determinant is ∣AB∣=67(61)−87(47)=4087−4089=−2|AB| = 67(61) - 87(47) = 4087 - 4089 = -2. Since ∣AB∣≠0|AB| \neq 0, the inverse exists. The adjoint is adj(AB)=[61−87−4767]\text{adj}(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}. (AB)−1=1−2[61−87−4767]=12[−618747−67](AB)^{-1} = \frac{1}{-2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -61 & 87 \\ 47 & -67 \end{bmatrix}

Now, we calculate the inverses of the individual matrices. For A, ∣A∣=3(5)−7(2)=1|A| = 3(5) - 7(2) = 1. Thus, A−1=11[5−7−23]=[5−7−23]A^{-1} = \frac{1}{1} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}. For B, ∣B∣=6(9)−8(7)=−2|B| = 6(9) - 8(7) = -2. Thus, B−1=1−2[9−8−76]=[−9/247/2−3]B^{-1} = \frac{1}{-2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -9/2 & 4 \\ 7/2 & -3 \end{bmatrix}.

Finally, we compute B−1A−1B^{-1}A^{-1}. B−1A−1=[−9/247/2−3][5−7−23]=[−45/2−863/2+1235/2+6−49/2−9]=[−61/287/247/2−67/2]=12[−618747−67]B^{-1}A^{-1} = \begin{bmatrix} -9/2 & 4 \\ 7/2 & -3 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} -45/2 - 8 & 63/2 + 12 \\ 35/2 + 6 & -49/2 - 9 \end{bmatrix} = \begin{bmatrix} -61/2 & 87/2 \\ 47/2 & -67/2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -61 & 87 \\ 47 & -67 \end{bmatrix}

Comparing the results, we see that (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}. Hence verified.

Exercise: EXERCISE 4.4 | Q: 12 | (Chapter: 18)
For More Understanding

Explanation

This solution follows the methodology shown in Example 14 of the provided context. It calculates the matrix product AB first, then finds its inverse. Separately, it finds the inverses of A and B and computes their product in the reverse order (B inverse times A inverse) to demonstrate the property (AB)^{-1} = B−1A−1B^{-1} A^{-1}.

Solution Steps

  1. Step 1: Calculate the matrix product AB.

  2. Step 2: Find the determinant of AB and then (AB)^{-1} using the formula A−1A^{-1} = (1/|A|)adj(A).

  3. Step 3: Find the determinants of A and B individually to confirm their inverses exist.

  4. Step 4: Calculate A−1A^{-1} and B−1B^{-1} using the inverse formula.

  5. Step 5: Calculate the matrix product B−1A−1B^{-1} A^{-1}.

  6. Step 6: Compare the result of (AB)^{-1} with B−1A−1B^{-1} A^{-1} to verify they are equal.

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