Matrices & Determinants
Properties of Matrices
Grade 12
Question:
<p>Let \(M\) and \(N\) be two \(3 \times 3\) matrices such that \(MN = NM\). Further, if \(M \neq N^2\) and \(M^2 = N^4\), then</p>
<p>determinant of \((M^2 + MN^2)\) is 0</p>
<p>there is a \(3 \times 3\) non-zero matrix \(U\) such that \((M^2 + MN^2)U\) is the zero matrix</p>
<p>determinant of \((M^2 + MN^2) \geq 1\)</p>
<p>for a \(3 \times 3\) matrix \(U\), if \((M^2 + MN^2)U\) equals the zero matrix then \(U\) is the zero matrix</p>
Step-by-Step Solution
Key Concept: When M and N commute (MN = NM) and M² = N⁴, factor using the commutative property: M² - N⁴ = (M - N²)(M + N²) = 0. Since M ≠ N², we must have M + N² = 0, giving M = -N².
<p><strong>Step 1:</strong> Given MN = NM (matrices commute) and M² = N⁴.</p><p><strong>Step 2:</strong> Rewrite M² = N⁴ as M² - N⁴ = 0.</p><p><strong>Step 3:</strong> Since M and N commute, we can factor: M² - N⁴ = (M - N²)(M + N²) = 0.</p><p><strong>Step 4:</strong> This gives either M - N² = 0 or M + N² = 0.</p><p><strong>Step 5:</strong> We're given M ≠ N², so the first case is rejected.</p><p><strong>Step 6:</strong> Therefore, M + N² = 0, which means <strong>M = -N²</strong>.</p><p>∴ Answer: AB (M + N² = 0)</p>
Correct Answer: AB