Given,
S={M=[aij],aij∈{0,1,2},{1≤i,j≤2}}
So, let M = \left[\begin{cases}a & b \\ c & d\end{cases}\right] where a,b,c,d∈{0,1,2}
So, total sample space will be 34=81
Now we will find P(A¯) or possibilities for ∣M∣=0
So, for ∣M∣=0 cases will be,
1.\left[{1111\right]or \left[\begin{cases}0 & 0 \\ 0 & 0\end{cases}\right] or \left[{2222\right] \toTotalmatrix= 3<br/>2.Two1 ' sandTwo0 ' s \toTotalmatrix= 4<br/>3.Two2 ' sandTwo0 ' s \toTotalmatrix= 4<br/>4.Two1 ' sandTwo2 ' s \toTotalmatrix= 4<br/>5.One1andthree0 ' s \toTotalmatrix= 4<br/>6.One2andThree0 ' s \toTotalmatrix= 4<br/>7.One1andone2andtwo0 ' s \toTotalmatrix= 8</p><p>So,totalcasesforwhich∣M∣=0 is 31,
So probability will be P(A¯)=8131
Hence, P(A)=1−8131=8150