Question 30 of 58advanced🔧 ApplyLong Answer4 marks

For the matrix A = [11112−32−13]\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} Show that A3A^{3} - 6A26A^{2} + 5A + 11I = 0. Hence, find A−1A^{-1}.

Correct Answer

Given matrix A = [11112−32−13]\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}

Step 1: Calculate A² A² = A·A = [11112−32−13][11112−32−13]=[421−38−147−314]\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}

Step 2: Calculate A³ A³ = A²·A = [421−38−147−314][11112−32−13]=[871−2327−6932−1358]\begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}

Step 3: Verify A³ - 6A² + 5A + 11I = 0 A³ - 6A² + 5A + 11I = [871−2327−6932−1358]−[24126−1848−8442−1884]+[555510−1510−515]+[110001100011]=[000000000]\begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix} - \begin{bmatrix} 24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84 \end{bmatrix} + \begin{bmatrix} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15 \end{bmatrix} + \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

Step 4: Find A⁻¹ From A³ - 6A² + 5A + 11I = 0, we get: A(A² - 6A + 5I) = -11I

Post-multiplying by A⁻¹: A² - 6A + 5I = -11A⁻¹

Therefore, A⁻¹ = (6A - A² - 5I)/11

A⁻¹ = 111[−3459−1−45−3−1]\frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}

Exercise: EXERCISE 4.4 | Q: 15 | (Page 18)
For More Understanding

Explanation

This question tests the concept of Cayley-Hamilton Theorem application. The student must first compute A² and A³ through matrix multiplication, then verify the given polynomial equation equals zero. Finally, rearranging the equation to isolate A⁻¹ allows finding the inverse without computing the adjoint and determinant separately. The key insight is that if A satisfies a polynomial equation, we can express A⁻¹ in terms of lower powers of A.

Solution Steps

  1. Step 1: Calculate A² by multiplying A with itself

  2. Step 2: Calculate A³ by multiplying A² with A

  3. Step 3: Substitute A³, A², A, and I into the equation A³ - 6A² + 5A + 11I and verify it equals zero matrix

  4. Step 4: Rearrange the equation to isolate A⁻¹: from A³ - 6A² + 5A + 11I = 0, factor out A to get A(A² - 6A + 5I) = -11I

  5. Step 5: Post-multiply by A⁻¹ to get A² - 6A + 5I = -11A⁻¹, then solve for A⁻¹ = (6A - A² - 5I)/11

PreviousNext