Question

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

Answer 1

$\left(i\right)$

$a_{ij}=i+j$

Here,
$$\begin{gathered} a_{11}=1+1=2,a_{12}=1+2=3,a_{13}=1+3=4,a_{14}=1+4=5 \\ {a}_{21}=2+1=3,a_{22}=2+2=4,a_{23}=2+3=5,a_{24}=2+4=6 \\ {a}_{31}=3+1=4,a_{32}=3+2=5,a_{33}=3+3=6\mathrm{and}{a}_{34}=3+4=7 \end{gathered}$$

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,
$$\begin{gathered} a_{11}=1-1=0,a_{12}=1-2=-1,a_{13}=1-3=-2,a_{14}=1-4=-3 \\ {a}_{21}=2-1=1,a_{22}=2-2=0,a_{23}=2-3=-1,a_{24}=2-4=-2 \\ {a}_{31}=3-1=2,a_{32}=3-2=1,a_{33}=3-3=0\mathrm{and}{a}_{34}=3-4=-1 \end{gathered}$$

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,
$$\begin{gathered} a_{11}=2\left(1\right)=2,a_{12}=2\left(1\right)=2,a_{13}=2\left(1\right)=2,a_{14}=2\left(1\right)=2 \\ {a}_{21}=2\left(2\right)=4,a_{22}=2\left(2\right)=4,a_{23}=2\left(2\right)=4,a_{24}=2\left(2\right)=4 \\ {a}_{31}=2\left(3\right)=6,a_{32}=2\left(3\right)=6,a_{33}=2\left(3\right)=6\mathrm{and}{a}_{34}=2\left(3\right)=6 \end{gathered}$$

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,
$$\begin{gathered} a_{11}=1,a_{12}=2,a_{13}=3,a_{14}=4 \\ {a}_{21}=1,a_{22}=2,a_{23}=3,a_{24}=4 \\ {a}_{31}=1,a_{32}=2,a_{33}=3\mathrm{and}{a}_{34}=4 \end{gathered}$$

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,

$$\begin{gathered} a_{11}=\frac{1}{2}\left|-3\left(1\right)+1\right|=\frac{1}{2}\left|-2\right|=1,a_{12}=\frac{1}{2}\left|-3\left(1\right)+2\right|=\frac{1}{2}\left|-1\right|=\frac{1}{2},a_{13}=\frac{1}{2}\left|-3\left(1\right)+3\right|=\frac{1}{2}\left|0\right|=\frac{0}{2}=0,a_{14}=\frac{1}{2}\left|-3\left(1\right)+4\right|=\frac{1}{2}\left|1\right|=\frac{1}{2} \\ {a}_{21}=\frac{1}{2}\left|-3\left(2\right)+1\right|=\frac{1}{2}\left|-5\right|=\frac{5}{2},a_{22}=\frac{1}{2}\left|-3\left(2\right)+2\right|=\frac{1}{2}\left|-4\right|=2,a_{23}=\frac{1}{2}\left|-3\left(2\right)+3\right|=\frac{1}{2}\left|-3\right|=\frac{3}{2},a_{24}=\frac{1}{2}\left|-3\left(2\right)+4\right|=\frac{1}{2}\left|-2\right|=1 \\ {a}_{31}=\frac{1}{2}\left|-3\left(3\right)+1\right|=\frac{1}{2}\left|-8\right|=4,a_{32}=\frac{1}{2}\left|-3\left(3\right)+2\right|=\frac{1}{2}\left|-7\right|=\frac{7}{2},a_{33}=\frac{1}{2}\left|-3\left(3\right)+3\right|=\frac{1}{2}\left|-6\right|=3\mathrm{and}{a}_{34}=\frac{1}{2}\left|-3\left(3\right)+4\right|=\frac{1}{2}\left|-5\right|=\frac{5}{2} \end{gathered}$$


So, the required matrix is $\left[\begin{array}{cccc}1& \frac{1}{2}& 0& \frac{1}{2}\\ \frac{5}{2}& 2& \frac{3}{2}& 1\\ 4& \frac{7}{2}& 3& \frac{5}{2}\end{array}\right]$.