Construct a 3 × 4 matrix, whose elements are given by
Question:

Construct a $3 \times 4$ matrix, whose elements are given by

(i) $a_{i j}=\frac{1}{2}|-3 i+j|$

(ii) $a_{i j}=2 i-j$

Solution:

In general, a $3 \times 4$ matrix is given by $A=\left[\begin{array}{llll}a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34}\end{array}\right]$

(i) $a_{i j}=\frac{1}{2}|-3 i+j|, i=1,2,3$ and $j=1,2,3,4$

$\therefore a_{11}=\frac{1}{2}|-3 \times 1+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1$

$a_{21}=\frac{1}{2}|-3 \times 2+1|=\frac{1}{2}|-6+1|=\frac{1}{2}|-5|=\frac{5}{2}$

$a_{31}=\frac{1}{2}|-3 \times 3+1|=\frac{1}{2}|-9+1|=\frac{1}{2}|-8|=\frac{8}{2}=4$

 

$a_{12}=\frac{1}{2}|-3 \times 1+2|=\frac{1}{2}|-3+2|=\frac{1}{2}|-1|=\frac{1}{2}$

$a_{22}=\frac{1}{2}|-3 \times 2+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{4}{2}=2$

$a_{32}=\frac{1}{2}|-3 \times 3+2|=\frac{1}{2}|-9+2|=\frac{1}{2}|-7|=\frac{7}{2}$

 

$a_{13}=\frac{1}{2}|-3 \times 1+3|=\frac{1}{2}|-3+3|=0$

$a_{23}=\frac{1}{2}|-3 \times 2+3|=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{3}{2}$

$a_{33}=\frac{1}{2}|-3 \times 3+3|=\frac{1}{2}|-9+3|=\frac{1}{2}|-6|=\frac{6}{2}=3$

 

$a_{14}=\frac{1}{2}|-3 \times 1+4|=\frac{1}{2}|-3+4|=\frac{1}{2}|1|=\frac{1}{2}$

$a_{24}=\frac{1}{2}|-3 \times 2+4|=\frac{1}{2}|-6+4|=\frac{1}{2}|-2|=\frac{2}{2}=1$

$a_{34}=\frac{1}{2}|-3 \times 3+4|=\frac{1}{2}|-9+4|=\frac{1}{2}|-5|=\frac{5}{2}$

 

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

(ii) $a_{i j}=2 i-j, i=1,2,3$ and $j=1,2,3,4$

$\therefore a_{11}=2 \times 1-1=2-1=1$

$a_{21}=2 \times 2-1=4-1=3$

$a_{31}=2 \times 3-1=6-1=5$

 

$a_{12}=2 \times 1-2=2-2=0$

$a_{22}=2 \times 2-2=4-2=2$

$a_{32}=2 \times 3-2=6-2=4$

 

$a_{13}=2 \times 1-3=2-3=-1$

$a_{23}=2 \times 2-3=4-3=1$

$a_{33}=2 \times 3-3=6-3=3$

 

$a_{14}=2 \times 1-4=2-4=-2$

$a_{24}=2 \times 2-4=4-4=0$

$a_{34}=2 \times 3-4=6-4=2$

Therefore, the required matrix is $A=\left[\begin{array}{llll}1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2\end{array}\right]$

 

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