Question

Construct a 3×4 matrix, whose elements are given by
I. $a_{ij}=\frac{1}{2}\mid -3i+j\mid$
II. $a_{ij}=2i-j$

Answer 1

In general, a 3 × 4 matrix is given by A=$\begin{bmatrix} a_{11} & a_{12} & a_{13} &a_{14} \\\[0.3em] a_{21} & a_{22} & a_{23}&a_{24} \\\[0.3em] a_{31} & a_{32} & a_{33} &a_{34} \end{bmatrix}$

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(i) $a_{ij}=\mid-3i+j\mid,i=1,2,3\,and\,j=1,2,3,4$
Therefore $a_{11}=\frac{1}{2}|-3\times1+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1$
$a_{21}=\frac{1}{2}|-3 \times 2+1|=\frac{1}{2}|-6+1|=\frac{1}{2}|-5|=\frac{5}{2}$
$a_{31}=\frac{1}{2}|-3\times3+1|=\frac{1}{2}|-9+1|=\frac{1}{2}|-8|=\frac{8}{2}=4$

$a_{12}=\frac{1}{2}|-3\times1+2|=\frac{1}{2}|-3+2|=\frac{1}{2}|-1|=\frac{1}{2}$
$a_{22}=\frac{1}{2}|-3\times2+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{4}{2}=2$
$a_{32}=\frac{1}{2}|-3\times3+2|=\frac{1}{2}|-9+2|=\frac{1}{2}|-7|=\frac{7}{2}$

$a_{13}=\frac{1}{2}|-3\times1+3|=\frac{1}{2}|-3+3|=0$
$a_{23}=\frac{1}{2}|-3\times2+3|=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{3}{2}$
$a_{33}=\frac{1}{2}|-3\times3+3|=\frac{1}{2}|-9+3|=\frac{1}{2}|-6|=\frac{6}{2}=3$

$a_{14}=\frac{1}{2}|-3\times1+4|=\frac{1}{2}|-3+4|=\frac{1}{2}|1|=\frac{1}{2}$
$a_{24}=\frac{1}{2}|-3\times2+4|=\frac{1}{2}|-6+4|=\frac{1}{2}|-2|=\frac{2}{2}=1$
$a_{34}=\frac{1}{2}|-3\times3+4|=\frac{1}{2}|-9+4|=\frac{1}{2}|-5|=\frac{5}{2}$

Therefore, the required matrix is $A=\begin{bmatrix}1& \frac{1}{2}& 0& \frac{1}{2}\\\ \frac{5}{2}& 2& \frac{3}{2} &1 \\\ 4& \frac{7}{2}& 3 &\frac{5}{2}\end{bmatrix}$

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(ii)$a_{ij}$=2i-j i=1,2,3 and j=1,2,3,4
therefore
$a_{11}=2\times1-1=2-1=1$
$a_{21}=2\times2-1=4-1=3$
$a_{31}=2\times3-1=6-1=5$

$a_{12}=2\times1-2=2-2=0$
$a_{22}=2\times2-2=4-2=2$
$a_{32}=2\times3-2=6-2=4$

$a_{13}=2\times1-3=2-3=-1$
$a_{23}=2\times2-3=4-3=1$
$a_{33}=2\times3-3=6-3=3$

$a_{14}=2\times1-4=2-4=-2$
$a_{24}=2\times2-4=4-4=0$
$a_{34}=2\times3-4=6-4=2$
Therefore, the required matrix is $A=\begin{bmatrix}1& 0& -1& -2\\\ 3& 2& 1& 0\\\ 5 &4& 3 &2\end{bmatrix}$