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(B) Let the number of $1$s be $x$,the number of $-1$s be $y$,and the number of $0$s be $z$. Since the matrix is $3 	imes 3$,we have $x + y + z = 9$.T

Vedclass · Accepted Answer

$414$

Vedclass · Answer

(B) Let the number of $1$s be $x$,the number of $-1$s be $y$,and the number of $0$s be $z$. Since the matrix is $3 	imes 3$,we have $x + y + z = 9$.The sum of entries is $1(x) + (-1)(y) + 0(z) = 5$,which simplifies to $x - y = 5$,or $x = y + 5$.Substituting $x$ in the first equation: $(y + 5) + y + z = 9 \implies 2y + z = 4$.We analyze the possible non-negative integer solutions for $(x, y, z)$:Case-$I$: If $y = 0$,then $z = 4$ and $x = 5$. Number of matrices $= \frac{9!}{5!0!4!} = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$.Case-$II$: If $y = 1$,then $z = 2$ and $x = 6$. Number of matrices $= \frac{9!}{6!1!2!} = \frac{9 	imes 8 	imes 7}{2 	imes 1} = 252$.Case-$III$: If $y = 2$,then $z = 0$ and $x = 7$. Number of matrices $= \frac{9!}{7!2!0!} = \frac{9 	imes 8}{2 	imes 1} = 36$.Total number of matrices $= 126 + 252 + 36 = 414$.

Let $A$ be a $3 \times 3$ matrix having entries from the set $\{-1, 0, 1\}$. The number of all such matrices $A$ having the sum of all the entries equal to $5$ is:

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