<p>Suppose \(a_1, a_2, \ldots\) are real numbers, with \(a_1 \neq 0\). If \(a_1, a_2, a_3, \ldots\) are in A.P., then</p><p>(1) \(A = \begin{bmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_5 & a_6 & a_7 \end{bmatrix}\) is singular (where \(i = \sqrt{-1}\))</p><p>(2) the system of equations \(a_1 x + a_2 y + a_3 z = 0\), \(a_4 x + a_5 y + a_6 z = 0\), \(a_5 x + a_6 y + a_7 z = 0\) has infinite number of solutions</p><p>(3) \(B = \begin{bmatrix} a_1 & ia_2 \\ ia_2 & a_1 \end{bmatrix}\) is nonsingular</p><p>(4) none of these</p>
<p>\(A = \begin{bmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_5 & a_6 & a_7 \end{bmatrix}\) is singular</p>
<p>the system of equations has infinite number of solutions</p>
<p>\(B = \begin{bmatrix} a_1 & ia_2 \\ ia_2 & a_1 \end{bmatrix}\) is nonsingular</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Since the sequence is in A.P., each term can be written as a_n = a_1 + (n-1)d where d is common difference. This linear relationship creates linear dependence in matrix rows/columns. For matrix B, the determinant a_1² + a_2² is always positive (sum of squares of real numbers), making it nonsingular.
<p><strong>Step 1: Express terms using A.P. property</strong></p><p>Let a_n = a_1 + (n-1)d where d is common difference. Then:</p><p>a_1, a_2 = a_1+d, a_3 = a_1+2d, a_4 = a_1+3d, a_5 = a_1+4d, a_6 = a_1+5d, a_7 = a_1+6d</p><p><strong>Step 2: Check singularity of Matrix A</strong></p><p>Matrix A becomes:</p><p>A = [a_1, a_1+d, a_1+2d; a_1+3d, a_1+4d, a_1+5d; a_1+4d, a_1+5d, a_1+6d]</p><p>Perform R₃ → R₃ - R₂: Third row becomes [d, d, d]</p><p>Notice R₂ - R₁ = [3d, 3d, 3d] = 3(d, d, d), so R₂ - R₁ is proportional to the transformed R₃.</p><p>Therefore rows are linearly dependent ⟹ <strong>det(A) = 0, Matrix A is singular. Statement (1) is TRUE</strong></p><p><strong>Step 3: Check the system of equations</strong></p><p>The homogeneous system has coefficient matrix A which is singular (det(A) = 0).</p><p>A singular coefficient matrix in a homogeneous system guarantees infinitely many solutions (beyond the trivial solution).</p><p><strong>Statement (2) is TRUE</strong></p><p><strong>Step 4: Check singularity of Matrix B</strong></p><p>B = [a_1, ia_2; ia_2, a_1]</p><p>det(B) = a_1² - (ia_2)(ia_2) = a_1² - i²a_2² = a_1² - (-1)a_2² = a_1² + a_2²</p><p>Since a_1 ≠ 0 and a_2 are real numbers, a_1² + a_2² > 0</p><p>Therefore det(B) ≠ 0 ⟹ <strong>Matrix B is nonsingular. Statement (3) is TRUE</strong></p><p>∴ Answer: A, B, C (Statements 1, 2, and 3 are correct)</p>
Correct Answer: A,B,C