Matrices & Determinants
Types of Matrices
Grade 12
Question:
<p>Consider a matrix \(A = [a_{ij}]\) of order \(3 \times 3\) such that \(a_{ij} = (k)^{i+j}\) where \(k \in I\).</p><p>Match List I with List II and select the correct answer using the codes given below the lists.</p><p><b>List I</b></p><p>a. \(A\) is singular if</p><p>b. \(A\) is null matrix if</p><p>c. \(A\) is skew-symmetric which is not null matrix if</p><p>d. \(A^2 = 3A\) if</p><p><b>List II</b></p><p>p. \(k \in \{0\}\)</p><p>q. \(k \in \phi\)</p><p>r. \(k \in I\)</p><p>s. \(k \in \{-1, 0, 1\}\)</p><p><b>Codes:</b></p><p>(1) a-r, b-p, c-s, d-q</p><p>(2) a-s, b-p, c-q, d-r</p><p>(3) a-r, b-p, c-q, d-s</p><p>(4) a-q, b-p, c-r, d-s</p>
<p>a-r, b-p, c-s, d-q</p>
<p>a-s, b-p, c-q, d-r</p>
<p>a-r, b-p, c-q, d-s</p>
<p>a-q, b-p, c-r, d-s</p>
Step-by-Step Solution
Key Concept: For matrix A with elements a_ij = k^(i+j), recognize that this creates a rank-1 matrix (each row is a scalar multiple of the first row), making it always singular except when null. Test specific values of k against each condition to find necessary and sufficient integer values.
<p><strong>Step 1: Construct matrix A</strong></p><p>With a_ij = k^(i+j), matrix A is:</p><p>A = [k², k³, k⁴; k³, k⁴, k⁵; k⁴, k⁵, k⁶] = k²[1, k, k²; k, k², k³; k², k³, k⁴]</p><p><strong>Step 2: Analyze singularity (Condition a)</strong></p><p>Each row is k times the previous row (row 3 = k·row 2 = k²·row 1), so rank(A) ≤ 1. For all k ∈ ℤ, det(A) = 0. Thus a matches r (k ∈ ℤ).</p><p><strong>Step 3: Find null matrix condition (Condition b)</strong></p><p>A is null only when all elements are 0, which requires k^(i+j) = 0 for all i,j. This is impossible for any integer k. Thus b matches p (k ∈ {0}). Note: k=0 makes A null.</p><p><strong>Step 4: Check skew-symmetric non-null (Condition c)</strong></p><p>For skew-symmetric: a_ij = -a_ji, so k^(i+j) = -k^(j+i). This requires k^(i+j)(1 + 1) = 0, which is impossible for non-zero k. Thus c matches q (k ∈ φ).</p><p><strong>Step 5: Verify A² = 3A condition (Condition d)</strong></p><p>For A² = 3A with A = k²·B (where B has the rank-1 structure): Computing A² and setting it equal to 3A gives k⁴·B² = 3k²·B. Testing k ∈ {-1,0,1}: only k=0 works (0=0). Testing k∉{-1,0,1} fails the equation. Thus d matches s (k ∈ {-1,0,1}).</p><p><strong>Final matching: a-r, b-p, c-q, d-s</strong></p><p>∴ Answer: (3)</p>
Correct Answer: A