<p>Let M be a 2 × 2 symmetric matrix with integer entries. Then M is invertible, if</p>
<p>(a) the first column of M is the transpose of the second row of M</p>
<p>(b) the second row of M is the transpose of the first column of M</p>
<p>(c) M is a diagonal matrix with non-zero entries in the main diagonal</p>
<p>(d) the product of entries in the main diagonal of M is not the square of an integer</p>
Step-by-Step Solution
Key Concept: A matrix is invertible if and only if its determinant is non-zero. For a 2×2 symmetric matrix with integer entries, we need to find which condition guarantees det(M) ≠ 0.
<p><strong>Step 1:</strong> Let M be a 2×2 symmetric matrix with integer entries: M = [[a, b], [b, c]] where a, b, c are integers.</p><p><strong>Step 2:</strong> The determinant of M is: det(M) = ac - b²</p><p><strong>Step 3:</strong> Analyze Option A and B: If the first column equals the transpose of the second row (or vice versa), then [a, b]ᵀ = [b, c], giving a = b and b = c, so a = b = c. This makes det(M) = a² - a² = 0, so M is NOT invertible.</p><p><strong>Step 4:</strong> Analyze Option D: If ac is not a perfect square, this doesn't guarantee ac - b² ≠ 0. For example, a = 2, c = 3, b = 1 gives ac = 6 (not a perfect square), but det(M) = 6 - 1 = 5 ≠ 0. However, a = 3, c = 3, b = 1 gives ac = 9 (perfect square), and det(M) = 9 - 1 = 8 ≠ 0. This condition is insufficient.</p><p><strong>Step 5:</strong> Analyze Option C: If M is diagonal with non-zero entries, then M = [[a, 0], [0, c]] where a ≠ 0 and c ≠ 0. Then det(M) = ac ≠ 0 since both a and c are non-zero integers. Therefore M is invertible.</p><p><strong>Step 6:</strong> Option C is the ONLY condition that guarantees invertibility for all 2×2 symmetric matrices with integer entries.</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C