Question

Construct a 5 × 5 real matrix A such that its elements aij are given by the relation

aij = 0 ∀ i = j

aij = 1 ∀ i #j​

Answer 1

Answer:

0 1 1 1 1

1 0 1 1 1

1 1 0 1 1

1 1 1 0 1

1 1 1 1 0

Answer 2

$\large\underline{\sf{Solution-}}$

Consider 5 × 5 matrix as

$\begin{gathered}\sf Let\:A=\left[\begin{array}{ccccc}a_{11}&a_{12}&a_{13}&a_{14}&a_{15}\\a_{21}&a_{22}&a_{23}&a_{24}&a_{25}\\a_{31}&a_{32}&a_{33}&a_{34}&a_{35} \\ a_{41}&a_{42}&a_{43}&a_{44}&a_{45} \\ a_{51}&a_{52}&a_{53}&a_{54}&a_{55}\end{array}\right]\end{gathered}$

According to statement,

$\begin{gathered}\begin{gathered}\bf\: a_{ij} = \begin{cases} &\sf{0, \: if \: i \: = \: j} \\ &\sf{1, \: if \: \: i \ne \: j} \end{cases}\end{gathered}\end{gathered}$

$\bf\implies \:a_{11} = a_{22} = a_{33} = a_{44} = a_{55} = 0$

and

$\bf\implies \:a_{12} = a_{13} = a_{14} = a_{15} = 1$

$\bf\implies \:a_{21} = a_{23} = a_{24} = a_{25} = 1$

$\bf\implies \:a_{31} = a_{32} = a_{34} = a_{35} = 1$

$\bf\implies \:a_{41} = a_{42} = a_{43} = a_{45} = 1$

$\bf\implies \:a_{51} = a_{52} = a_{53} = a_{54} = 1$

Hence,

$\begin{gathered}\sf \:A=\left[\begin{array}{ccccc}0&1&1&1&1\\1&0&1&1&1\\1&1&0&1&1 \\ 1&1&1&0&1 \\ 1&1&1&1&0\end{array}\right]\end{gathered}$

Additional Information :-

Order of matrix is defined as

• number of rows × number of columns.

If order of a matrix is m × n, it provide 3 information :-

• 1. Number of elements in matrix, i.e mn

• 2. Number of elements in each row = n

• 3. Number of elements in each column = m.