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Question

# Construct a 2 × 3 matrix whose elements aij are given by : (i) aij = i . j (ii) aij = 2i − j (iii) aij = i + j (iv) aij = $\frac{\left(i+j{\right)}^{2}}{2}$

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Solution

## $\left(i\right)$ Here, ${a}_{ij}=i.j$, $1\le i\le 2$ and $1\le j\le 3$ ${a}_{11}=1×1=1,{a}_{12}=1×2=2,{a}_{13}=1×3=3\phantom{\rule{0ex}{0ex}}{a}_{21}=2×1=2,{a}_{22}=2×2=4\mathrm{and}{a}_{23}=2×3=6$ Required matrix = A = $\left[\begin{array}{ccc}1& 2& 3\\ 2& 4& 6\end{array}\right]$ $\left(ii\right)$ Here, ${a}_{ij}=2i-j$ ${a}_{11}=2\left(1\right)-1=2-1=1,{a}_{12}=2\left(1\right)-2=2-2=0,{a}_{13}=2\left(1\right)-3=2-3=-1\phantom{\rule{0ex}{0ex}}{a}_{21}=2\left(2\right)-1=4-1=3,{a}_{22}=2\left(2\right)-2=4-2=2\mathrm{and}{a}_{23}=2\left(2\right)-3=4-3=1\phantom{\rule{0ex}{0ex}}$ Required matrix = A = $\left[\begin{array}{ccc}1& 0& -1\\ 3& 2& 1\end{array}\right]$ $\left(iii\right)$ Here, ${a}_{ij}=i+j$ ${a}_{11}=1+1=2,{a}_{12}=1+2=3,{a}_{13}=1+3=4\phantom{\rule{0ex}{0ex}}{a}_{21}=2+1=3,{a}_{22}=2+2=4\mathrm{and}{a}_{23}=2+3=5$ Required matrix = A = $\left[\begin{array}{ccc}2& 3& 4\\ 3& 4& 5\end{array}\right]$ $\left(iv\right)$ Here, ${a}_{ij}=\frac{{\left(i+j\right)}^{2}}{2}$ ${a}_{11}=\frac{{\left(1+1\right)}^{2}}{2}=\frac{{\left(2\right)}^{2}}{2}=\frac{4}{2}=2,{a}_{12}=\frac{{\left(1+2\right)}^{2}}{2}=\frac{{\left(3\right)}^{2}}{2}=\frac{9}{2},{a}_{13}=\frac{{\left(1+3\right)}^{2}}{2}=\frac{{\left(4\right)}^{2}}{2}=\frac{16}{2}=8\phantom{\rule{0ex}{0ex}}{a}_{21}=\frac{{\left(2+1\right)}^{2}}{2}=\frac{{\left(3\right)}^{2}}{2}=\frac{9}{2},{a}_{22}=\frac{{\left(2+2\right)}^{2}}{2}=\frac{{\left(4\right)}^{2}}{2}=\frac{16}{2}=8\mathrm{and}{a}_{23}=\frac{{\left(2+3\right)}^{2}}{2}=\frac{{\left(5\right)}^{2}}{2}=\frac{25}{2}$ Required matrix = A = $\left[\begin{array}{ccc}2& \frac{9}{2}& 8\\ \frac{9}{2}& 8& \frac{25}{2}\end{array}\right]$

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