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Worked Examples · Example 3

Q.Construct a 3×23 \times 2 matrix whose elements are given by aij=12 ∣ i−3j ∣a_{ij} = \frac{1}{2}\,|\,i - 3j\,|.

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We build a 3×23 \times 2 matrix by plugging each row index ii (1 to 3) and column index jj (1 to 2) into the formula aij=12∣i−3j∣a_{ij} = \frac{1}{2}|i - 3j|, then simplify each absolute value. The final matrix is (152122032)\begin{pmatrix} 1 & \frac{5}{2} \\ \frac{1}{2} & 2 \\ 0 & \frac{3}{2} \end{pmatrix}.

The core idea here is straightforward: a matrix is just a rectangular array of numbers, and each entry has a specific address — row ii, column jj. The formula aij=12∣i−3j∣a_{ij} = \frac{1}{2}|i - 3j| tells us exactly how to compute the number at that address. The absolute value ensures every entry is non-negative, and the factor 12\frac{1}{2} scales the result.

Let’s walk through it systematically.

  1. Identify the range of indices.

    The matrix is 3×23 \times 2, so ii runs from 1 to 3 (rows), and jj runs from 1 to 2 (columns). That gives us 3×2=63 \times 2 = 6 entries to compute.

  2. Compute each entry using the formula.

    For each pair (i,j)(i, j), we evaluate aij=12∣i−3j∣a_{ij} = \frac{1}{2}|i - 3j|.

    Let’s do it row by row.

    Row 1 (i=1i = 1):

    • j=1j = 1: a11=12∣1−3(1)∣=12∣1−3∣=12∣−2∣=12×2=1a_{11} = \frac{1}{2}|1 - 3(1)| = \frac{1}{2}|1 - 3| = \frac{1}{2}|-2| = \frac{1}{2} \times 2 = 1
    • j=2j = 2: a12=12∣1−3(2)∣=12∣1−6∣=12∣−5∣=12×5=52a_{12} = \frac{1}{2}|1 - 3(2)| = \frac{1}{2}|1 - 6| = \frac{1}{2}|-5| = \frac{1}{2} \times 5 = \frac{5}{2}

    Row 2 (i=2i = 2):

    • j=1j = 1: a21=12∣2−3(1)∣=12∣2−3∣=12∣−1∣=12×1=12a_{21} = \frac{1}{2}|2 - 3(1)| = \frac{1}{2}|2 - 3| = \frac{1}{2}|-1| = \frac{1}{2} \times 1 = \frac{1}{2}
    • j=2j = 2: a22=12∣2−3(2)∣=12∣2−6∣=12∣−4∣=12×4=2a_{22} = \frac{1}{2}|2 - 3(2)| = \frac{1}{2}|2 - 6| = \frac{1}{2}|-4| = \frac{1}{2} \times 4 = 2

    Row 3 (i=3i = 3):

    • j=1j = 1: a31=12∣3−3(1)∣=12∣3−3∣=12×0=0a_{31} = \frac{1}{2}|3 - 3(1)| = \frac{1}{2}|3 - 3| = \frac{1}{2} \times 0 = 0
    • j=2j = 2: a32=12∣3−3(2)∣=12∣3−6∣=12∣−3∣=12×3=32a_{32} = \frac{1}{2}|3 - 3(2)| = \frac{1}{2}|3 - 6| = \frac{1}{2}|-3| = \frac{1}{2} \times 3 = \frac{3}{2}
  3. Assemble the matrix. …

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