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NCERT Exemplar · Q4

Q.Construct a 3×23 \times 2 matrix whose elements are given by aij=eixsin⁡jxa_{ij} = e^{ix}\sin jx.

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The key idea is to treat the formula aij=eixsin⁡jxa_{ij} = e^{ix}\sin jx as a rule that depends only on the row index ii and column index jj. For a 3×23 \times 2 matrix, ii runs from 1 to 3 and jj runs from 1 to 2. The resulting matrix is (exsin⁡xexsin⁡2xe2xsin⁡xe2xsin⁡2xe3xsin⁡xe3xsin⁡2x)\begin{pmatrix} e^{x}\sin x & e^{x}\sin 2x \\ e^{2x}\sin x & e^{2x}\sin 2x \\ e^{3x}\sin x & e^{3x}\sin 2x \end{pmatrix}.

When you see a problem like "construct a matrix whose elements are given by aij=somethinga_{ij} = \text{something}", the first thing to realise is that you are not solving an equation — you are following a recipe. The indices ii and jj are just placeholders for the row number and column number. So the matrix is built by plugging in i=1,2,3i = 1, 2, 3 and j=1,2j = 1, 2 into the expression eixsin⁡jxe^{ix}\sin jx.

The expression eixsin⁡jxe^{ix}\sin jx is a product of two factors: one depends only on ii (the row), the other depends only on jj (the column). This is a special kind of matrix called a separable or rank-1 matrix — but you don't need that term for the exam. What matters is that you can fill each entry independently.

Let's do it step by step.

  1. Identify the dimensions. The matrix is 3×23 \times 2, meaning 3 rows and 2 columns. So ii takes values 1,2,31, 2, 3 and jj takes values 1,21, 2.

  2. Write the general entry. For any row ii and column jj, the entry is aij=eixsin⁡jxa_{ij} = e^{ix}\sin jx. Notice that xx is a variable (presumably a real number or a parameter), so the matrix entries are functions of xx.

  3. Fill the first column (j=1j = 1).

    • Row 1 (i=1i=1): a11=e1⋅xsin⁡(1⋅x)=exsin⁡xa_{11} = e^{1 \cdot x}\sin(1 \cdot x) = e^{x}\sin x
    • Row 2 (i=2i=2): a21=e2xsin⁡xa_{21} = e^{2x}\sin x
    • Row 3 (i=3i=3): a31=e3xsin⁡xa_{31} = e^{3x}\sin x
  4. Fill the second column (j=2j = 2).

    • Row 1 (i=1i=1): a12=exsin⁡(2x)a_{12} = e^{x}\sin(2x)
    • Row 2 (i=2i=2): a22=e2xsin⁡2xa_{22} = e^{2x}\sin 2x
    • Row 3 (i=3i=3): a32=e3xsin⁡2xa_{32} = e^{3x}\sin 2x
  5. Assemble the matrix. Place each entry in its correct position:

(exsin⁡xexsin⁡2xe2xsin⁡xe2xsin⁡2xe3xsin⁡xe3xsin⁡2x)\begin{pmatrix} e^{x}\sin x & e^{x}\sin 2x \\[4pt] e^{2x}\sin x & e^{2x}\sin 2x \\[4pt] e^{3x}\sin x & e^{3x}\sin 2x \end{pmatrix} …

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