Question 28 of 58intermediate🔧 ApplyLong Answer4 marks

If A = [31−12]\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, show that A2A^{2} - 5A + 7I = 0. Hence find A−1A^{-1}.

Correct Answer

Given A = [31−12]\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

Step 1: Calculate A²

A² = A.A = [31−12][31−12]=[9−13+2−3−2−1+4]=[85−53]\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}

Step 2: Verify A² - 5A + 7I = O

A² - 5A + 7I = [85−53]−[155−510]+[7007]\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

= [8−15+75−5+0−5+5+03−10+7]=[0000]=O\begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O

Hence proved.

Step 3: Find A⁻¹

From A² - 5A + 7I = O, we get: A² - 5A = -7I

Post-multiplying by A⁻¹: A.A.A⁻¹ - 5A.A⁻¹ = -7I.A⁻¹ A.I - 5I = -7A⁻¹ A - 5I = -7A⁻¹

Therefore, A⁻¹ = (5I - A)/7

= (1/7) [[5005]−[31−12]\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}]

= (1/7) [2−113]\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}

Answer: A⁻¹ = 17[2−113]\frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}

Exercise: EXERCISE 4.4 | Q: 13 | (Page 18)
For More Understanding

Explanation

This question tests the Cayley-Hamilton theorem application. First, we verify the given equation by computing A² through matrix multiplication, then substituting into the polynomial equation. For finding A⁻¹, we rearrange the equation A² - 5A + 7I = O to isolate A⁻¹. The key insight is that from A² - 5A = -7I, post-multiplying by A⁻¹ gives A - 5I = -7A⁻¹, which directly yields the inverse matrix.

Solution Steps

  1. Step 1: Calculate A² by multiplying A with itself using matrix multiplication rules

  2. Step 2: Substitute A², A, and I into the equation A² - 5A + 7I and verify it equals zero matrix

  3. Step 3: Rearrange the equation to A² - 5A = -7I and post-multiply by A⁻¹

  4. Step 4: Simplify to get A - 5I = -7A⁻¹, then solve for A⁻¹ = (5I - A)/7

  5. Step 5: Compute 5I - A and multiply by scalar 1/7 to get the final inverse matrix

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