In a 4 × 4 matrix the sum of each row, column and both the main diagonals is $α$. Then the sum of the four corner elements

is also $α$

Let $A =[a_{ij}]$ be a 4 × 4 matrix. It is given that

$a_{i1} + a_{i2} + a_{i3} + a_{i4} = α$ for $i = 1, 2, 3, 4$

$a_{1j} + a_{2j} +a_{3j} +a_{4j} = α$ for $j = 1, 2, 3, 4$

$a_{11}+ a_{22} + a_{33} +a_{44} = α$

and $a_{14} +a_{23} + a_{32}+ a_{41} = α$

We have to find $a_{11} +a_{14} +a_{41} + a_{44}$.

Clearly,

$(a_{11}+a_{12} +a_{13} +a_{14})+(a_{41} +a_{42} + a_{43} +a_{44})$

$+(a_{11}+ a_{22} + a_{33} +a_{44})+(a_{14} +a_{23} + a_{32}+ a_{41})$

$-(a_{12}+ a_{22} + a_{32} +a_{42})+(a_{13} +a_{23} + a_{33}+ a_{43})$

$=2(a_{11}+a_{14}+a_{41}+a_{44})$

$⇒4α-2α=2(a_{11}+a_{14}+a_{41}+a_{44})$

$⇒a_{11}+a_{14}+a_{41}+a_{44} = α$