Question 29 of 58intermediate🔧 ApplyLong Answer4 marks

For the matrix A = [3211]\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}, find the numbers a and b such that A2A^{2} + aA + bI = 0.

Correct Answer

a = -4 and b = 1

Exercise: EXERCISE 4.4 | Q: 14 | (Page 18)
For More Understanding

Explanation

This question tests the Cayley-Hamilton theorem concept where a matrix satisfies its own characteristic equation. Students need to compute A², then set up the matrix equation A² + aA + bI = O and solve the resulting system of linear equations. The context provides similar examples (Question 13 and Example 15) showing how to verify matrix equations of this form.

Solution Steps

  1. Step 1: Calculate A² = A × A = [[3, 2], [1, 1]] × [[3, 2], [1, 1]] = [[11, 8], [4, 3]]

  2. Step 2: Substitute into the equation: A² + aA + bI = [[11, 8], [4, 3]] + a[[3, 2], [1, 1]] + b[[1, 0], [0, 1]] = [[11 + 3a + b, 8 + 2a], [4 + a, 3 + a + b]]

  3. Step 3: For this to equal zero matrix O, each element must be zero. From element (2,1): 4 + a = 0, so a = -4

  4. Step 4: From element (1,2): 8 + 2a = 0, substituting a = -4 gives 8 - 8 = 0 ✓

  5. Step 5: From element (1,1): 11 + 3a + b = 0, substituting a = -4 gives 11 - 12 + b = 0, so b = 1

  6. Step 6: Verification: A² - 4A + I = [[11, 8], [4, 3]] - [[12, 8], [4, 4]] + [[1, 0], [0, 1]] = [[0, 0], [0, 0]] ✓

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