Question 52 of 58intermediate🔧 ApplyLong Answer4 marks

If A−1A^{-1} = [3−11−156−55−22]\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} and B = [12−2−1300−21]\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}, find (AB)^{-1}.

Correct Answer

(AB)−1=[9−35−210102](AB)^{-1} = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}

Exercise: Miscellaneous Exercises on Chapter 4 | Q: 3 | (Page 24)
For More Understanding

Explanation

This question tests the property (AB)^{-1} = B−1A−1B^{-1} A^{-1}. The student must first find B−1B^{-1} by computing the determinant and adjoint of B, then multiply B−1B^{-1} with the given A−1A^{-1}. The context shows this property being verified in Example 14.

Solution Steps

  1. Step 1: State the property: (AB)^{-1} = B−1A−1B^{-1} A^{-1}

  2. Step 2: Calculate |B| = 1(3-0) - 2(-1-0) + (-2)(2-0) = 3 + 2 - 4 = 1

  3. Step 3: Find adj(B) by calculating cofactors and taking transpose: Cofactor matrix = [312212625]\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \end{bmatrix} adj(B) = [326112225]\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}

  4. Step 4: Since |B| = 1, B−1B^{-1} = adj(B) = [326112225]\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}

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

    [326112225][3−11−156−55−22]\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}

  6. Step 6: Row 1: (3)(3) + 2(-15) + 6(5) = 9, (3)(-1) + 2(6) + 6(-2) = -3, (3)(1) + 2(-5) + 6(2) = 5

  7. Step 7: Row 2: (1)(3) + 1(-15) + 2(5) = -2, (1)(-1) + 1(6) + 2(-2) = 1, (1)(1) + 1(-5) + 2(2) = 0

  8. Step 8: Row 3: (2)(3) + 2(-15) + 5(5) = 1, (2)(-1) + 2(6) + 5(-2) = 0, (2)(1) + 2(-5) + 5(2) = 2

  9. Step 9: (AB)^{-1} = [9−35−210102]\begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}

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