Home  >>  CBSE XII  >>  Math  >>  Matrices

# Construct a 3 x 4 matrix,whose elements are given by:$(i)\;a_{ij}=\frac{1}{2}| -3i+j\;| \qquad$

Note: This is a 2 part question, split as 2 separate questions here.

Toolbox:
• In general, if $A_{3\times 4}$ is a 3x4 matrix, $A=\begin{bmatrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{21} & a_{22} & a_{23} & a_{24}\\a_{31} & a_{32} & a_{33} & a_{34}\end{bmatrix}$, where i and j can be (1,2,3,4)
Given $a_{ij}=\frac{1}{2}\mid-3i+j\mid \Rightarrow$
• $a_{11} = \frac{1}{2}\mid-3i+j\mid=\frac{1}{2}\mid-3(1)+1\mid=\frac{1}{2}\mid(-3+1)\mid=1+1=1.$
• $a_{12} = \frac{1}{2}\mid-3i+j\mid=\frac{1}{2}\mid-3(1)+2\mid=\frac{1}{2}\mid(-3+2)\mid=\frac{1}{2}\mid-1\mid=\frac{1}{2}.$
• $a_{13} = \frac{1}{2}\mid-3i+j\mid=\frac{1}{2}\mid-3(1)+3\mid=\frac{1}{2}\mid(-3+3)\mid=o.$
• $a_{14} = \frac{1}{2}\mid-3i+j\mid=\frac{1}{2}\mid-3(1)+4\mid=\frac{1}{2}\mid(+1)\mid=\frac{1}{2}.$

• $a_{21} = \frac{1}{2}\mid-3i+j\mid=\frac{1}{2}\mid-3(2)+1\mid=\frac{1}{2}\mid(-6+1)\mid=\frac{1}{2}\mid -5\mid=\frac{5}{2}$
• $a_{22} = \frac{1}{2}\mid-3i+j\mid=\frac{1}{2}\mid-3(2)+2\mid=\frac{1}{2}\mid(-6+2)\mid=\frac{1}{2}\mid -4\mid=\frac{4}{2}=2.$
• $a_{23} = \frac{1}{2}\mid-3i+j\mid=\frac{1}{2}\mid-3(2)+4\mid=\frac{1}{2}\mid(-6+3)\mid=\frac{1}{2}\mid -3\mid=\frac{3}{2}.$
• $a_{24}= \frac{1}{2}\mid-3i+j\mid=\frac{1}{2}\mid-3(2)+4\mid=\frac{1}{2}\mid(-6+4)\mid=\frac{1}{2}\mid -2\mid=1.$

• $a_{31} = \frac{1}{2}\mid-3i+j\mid=\frac{1}{2}\mid-3(3)+2\mid=\frac{1}{2}\mid(-9+1)\mid=\frac{1}{2}\mid -8\mid=4.$
• $a_{32} = \frac{1}{2}\mid-3i+j\mid=\frac{1}{2}\mid-3(3)+2\mid=\frac{1}{2}\mid -9+2\mid=\frac{7}{2}.$
• $a_{33} = \frac{1}{2}\mid-3i+j\mid = \frac{1}{2}\mid-3(3)+3\mid = \frac{1}{2}\mid-9+3\mid = \frac{6}{2} = 3$
• $a_{34} = \frac{1}{2}\mid-3i+j\mid = \frac{1}{2}\mid-3(3)+4\mid = \frac{1}{2}\mid-9+4\mid = \frac{5}{2}.$

Hence the required matrix is given by $A=\begin{bmatrix}1 & \frac{1}{2} & 0 & \frac{1}{2}\\\frac{5}{2} & 2 & \frac{3}{2} & 1\\4 & \frac{7}{2} & 3 & \frac{5}{2}\end{bmatrix}.$