Question

# Construct a 3 × 4 matrix A = [aij] whose elements aij are given by: (i) aij = i + j (ii) aij = i − j (iii) aij = 2i (iv) aij = j (v) aij = $\frac{1}{2}\left|-3i+j\right|$

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Solution

## $\left(i\right)$ ${a}_{ij}=i+j$ Here, ${a}_{11}=1+1=2,{a}_{12}=1+2=3,{a}_{13}=1+3=4,{a}_{14}=1+4=5\phantom{\rule{0ex}{0ex}}{a}_{21}=2+1=3,{a}_{22}=2+2=4,{a}_{23}=2+3=5,{a}_{24}=2+4=6\phantom{\rule{0ex}{0ex}}{a}_{31}=3+1=4,{a}_{32}=3+2=5,{a}_{33}=3+3=6\mathrm{and}{a}_{34}=3+4=7$ So, the required matrix is $\left[\begin{array}{cccc}2& 3& 4& 5\\ 3& 4& 5& 6\\ 4& 5& 6& 7\end{array}\right]$. $\left(ii\right)$ ${a}_{ij}=i-j$ Here, ${a}_{11}=1-1=0,{a}_{12}=1-2=-1,{a}_{13}=1-3=-2,{a}_{14}=1-4=-3\phantom{\rule{0ex}{0ex}}{a}_{21}=2-1=1,{a}_{22}=2-2=0,{a}_{23}=2-3=-1,{a}_{24}=2-4=-2\phantom{\rule{0ex}{0ex}}{a}_{31}=3-1=2,{a}_{32}=3-2=1,{a}_{33}=3-3=0\mathrm{and}{a}_{34}=3-4=-1$ So, the required matrix is $\left[\begin{array}{cccc}0& -1& -2& -3\\ 1& 0& -1& -2\\ 2& 1& 0& -1\end{array}\right]$. $\left(iii\right)$ ${a}_{ij}=2i$ Here, ${a}_{11}=2\left(1\right)=2,{a}_{12}=2\left(1\right)=2,{a}_{13}=2\left(1\right)=2,{a}_{14}=2\left(1\right)=2\phantom{\rule{0ex}{0ex}}{a}_{21}=2\left(2\right)=4,{a}_{22}=2\left(2\right)=4,{a}_{23}=2\left(2\right)=4,{a}_{24}=2\left(2\right)=4\phantom{\rule{0ex}{0ex}}{a}_{31}=2\left(3\right)=6,{a}_{32}=2\left(3\right)=6,{a}_{33}=2\left(3\right)=6\mathrm{and}{a}_{34}=2\left(3\right)=6$ So, the required matrix is $\left[\begin{array}{cccc}2& 2& 2& 2\\ 4& 4& 4& 4\\ 6& 6& 6& 6\end{array}\right]$. $\left(iv\right)$ ${a}_{ij}=j$ Here, ${a}_{11}=1,{a}_{12}=2,{a}_{13}=3,{a}_{14}=4\phantom{\rule{0ex}{0ex}}{a}_{21}=1,{a}_{22}=2,{a}_{23}=3,{a}_{24}=4\phantom{\rule{0ex}{0ex}}{a}_{31}=1,{a}_{32}=2,{a}_{33}=3\mathrm{and}{a}_{34}=4$ So, the required matrix is $\left[\begin{array}{cccc}1& 2& 3& 4\\ 1& 2& 3& 4\\ 1& 2& 3& 4\end{array}\right]$. $\left(v\right)$ ${a}_{ij}=\frac{1}{2}\left|-3i+j\right|$ Here, ${a}_{11}=\frac{1}{2}\left|-3\left(1\right)+1\right|=\frac{1}{2}\left|-2\right|=1,{a}_{12}=\frac{1}{2}\left|-3\left(1\right)+2\right|=\frac{1}{2}\left|-1\right|=\frac{1}{2},{a}_{13}=\frac{1}{2}\left|-3\left(1\right)+3\right|=\frac{1}{2}\left|0\right|=\frac{0}{2}=0,{a}_{14}=\frac{1}{2}\left|-3\left(1\right)+4\right|=\frac{1}{2}\left|1\right|=\frac{1}{2}\phantom{\rule{0ex}{0ex}}{a}_{21}=\frac{1}{2}\left|-3\left(2\right)+1\right|=\frac{1}{2}\left|-5\right|=\frac{5}{2},{a}_{22}=\frac{1}{2}\left|-3\left(2\right)+2\right|=\frac{1}{2}\left|-4\right|=2,{a}_{23}=\frac{1}{2}\left|-3\left(2\right)+3\right|=\frac{1}{2}\left|-3\right|=\frac{3}{2},{a}_{24}=\frac{1}{2}\left|-3\left(2\right)+4\right|=\frac{1}{2}\left|-2\right|=1\phantom{\rule{0ex}{0ex}}{a}_{31}=\frac{1}{2}\left|-3\left(3\right)+1\right|=\frac{1}{2}\left|-8\right|=4,{a}_{32}=\frac{1}{2}\left|-3\left(3\right)+2\right|=\frac{1}{2}\left|-7\right|=\frac{7}{2},{a}_{33}=\frac{1}{2}\left|-3\left(3\right)+3\right|=\frac{1}{2}\left|-6\right|=3\mathrm{and}{a}_{34}=\frac{1}{2}\left|-3\left(3\right)+4\right|=\frac{1}{2}\left|-5\right|=\frac{5}{2}$ So, the required matrix is $\left[\begin{array}{cccc}1& \frac{1}{2}& 0& \frac{1}{2}\\ \frac{5}{2}& 2& \frac{3}{2}& 1\\ 4& \frac{7}{2}& 3& \frac{5}{2}\end{array}\right]$.

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