Let A be a 3 × 3 matrix having entries from the set \ - 1, 0, 1\ ....

Given Matrix 3 \times 3 = \left[\begin{cases}- & - & - \\ - & - & - \\ - & - & -\end{cases}\right]

Here we can see we have to find nine elements whose sum is $5$

Case 1: When five 1's are there & four O's are there

Now by division & distribution Method

We get $⇒ \dfrac{9 !}{5 ! 4 !} = 126 . . . \left(i\right)$

Case 2: when six 1's are there & one \left\{- 1\right\}& two O's are there.</p><p>Again by division & distribution Method</p><p>we get⇒ \dfrac{9 !}{6 ! 2 ! 1 !} = 252 . . . \left(ii\right)$</p><p><strong>Case 3:</strong> When seven 1's are there are two \left\{- 1\right\}$ are there,

By division & distribution we get

$⇒ \dfrac{9 !}{7 ! 2 !} = 36 . . . \left(iii\right)$

Now adding equation \left(i\right) + \left(ii\right) + \left(iii\right)$</p><p>We get$126 + 252 + 36 = 414$</p><p>So total$414$ ways will be there.