Matrices & Determinants
Adjoint and Inverse of a Matrix
Grade None
Question:
<p>If \(A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}\) and \(A\,\text{adj}\,A = AA^T\), then \(5a + b\) is equal to:</p>
<p>13</p>
<p>\(-1\)</p>
<p>5</p>
<p>4</p>
Step-by-Step Solution
Key Concept: For a 2×2 matrix, adj(A) relates to det(A) through the property A·adj(A) = det(A)·I, while the right side AA^T is a symmetric product. Equating these requires matching both the scalar multiplier (determinant) and the matrix structure.
<p><strong>Step 1:</strong> Recall that for any square matrix A: A·adj(A) = det(A)·I</p><p><strong>Step 2:</strong> Calculate det(A) = 5a(2) - (-b)(3) = 10a + 3b</p><p><strong>Step 3:</strong> Therefore A·adj(A) = (10a + 3b)·I = <begin>bmatrix</begin>10a+3b & 0\\ 0 & 10a+3b<end>bmatrix></p><p><strong>Step 4:</strong> Compute AA^T = <begin>bmatrix</begin>5a & -b\\ 3 & 2<end>bmatrix><begin>bmatrix</begin>5a & 3\\ -b & 2<end>bmatrix> = <begin>bmatrix</begin>25a²+b² & 15a-2b\\ 15a-2b & 9+4<end>bmatrix> = <begin>bmatrix</begin>25a²+b² & 15a-2b\\ 15a-2b & 13<end>bmatrix></p><p><strong>Step 5:</strong> Since A·adj(A) = AA^T, match entries:</p><p>Off-diagonal: 15a - 2b = 0 ⟹ b = 7.5a</p><p>Diagonal elements: 10a + 3b = 25a² + b² and 10a + 3b = 13</p><p><strong>Step 6:</strong> From 10a + 3b = 13 and b = 7.5a:</p><p>10a + 3(7.5a) = 13 ⟹ 10a + 22.5a = 13 ⟹ 32.5a = 13 ⟹ a = 0.4, b = 3</p><p><strong>Step 7:</strong> Verify: 25(0.16) + 9 = 13 ✓</p><p>∴ 5a + b = 5(0.4) + 3 = 2 + 3 = <strong>5</strong></p>
Correct Answer: A