Let M denote the set of all real matrices of order 3×3 and let S={−3,−2,−1,1,2}. Let S1={A=[aij]∈M:A=AT and aij∈S,∀i,j}, S2={A=[aij]∈M:A=−AT and aij∈S,∀i,j}, S3={A=[aij]∈M:a11+a22+a33=0 and aij∈S,∀i,j}. If n(S1∪S2∪S3)=125α, then α equals
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Step-by-Step Solution
Step 1: Determine the number of elements in S
The set S contains 5 distinct real numbers: {−3,−2,−1,1,2}. This means that each entry aij of a matrix A can take any of these 5 values.
Step 2: Calculate the number of matrices in S1 (Symmetric Matrices)
For a 3×3 symmetric matrix, the elements aij must be equal to aji. This means we only need to choose the values for the diagonal elements (a11,a22,a33) and the upper triangular elements (a12,a13,a23). The remaining elements are determined by these choices. There are 3 diagonal elements and 3 upper triangular elements, making a total of 3+3=6 independent choices. Since each choice can be any of the 5 elements in S, the number of such matrices is 56.
Step 3: Calculate the number of matrices in S2 (Skew-Symmetric Matrices)
For a 3×3 skew-symmetric matrix, aij=−aji. This implies that the diagonal elements must be zero (aii=−aii⟹2aii=0⟹aii=0). However, 0∈/S. Therefore, no 3×3 skew-symmetric matrix can be formed using elements only from S. Thus, n(S2)=0.
Step 4: Calculate the number of matrices in S3 (Matrices with trace zero)
For matrices in S3, the sum of the diagonal elements must be zero. There are 9 elements in a 3×3 matrix. The values of a11 and a22 can be chosen independently in 5×5=25 ways. For each pair of a11 and a22, a33 is uniquely determined as a33=−(a11+a22). We need to check if this a33 is in S. The possible sums a11+a22 range from −3+(−3)=−6 to 2+2=4. The possible values for a33 would then range from −4 to 6. The values in S are {−3,−2,−1,1,2}. We need to count the pairs (a11,a22) such that −(a11+a22)∈S. There are 13 such pairs. The remaining 9−3=6 off-diagonal elements can be chosen independently in 56 ways. So, n(S3)=13×56.
Step 5: Calculate the number of matrices in S1 intersection S3
For matrices in S1∩S3, the matrix must be symmetric and its trace must be zero. We have 6 independent elements for a symmetric matrix (a11,a22,a33,a12,a13,a23). The condition a11+a22+a33=0 means that a11 and a22 can be chosen in 5×5=25 ways, and a33 is determined. As before, there are 13 pairs (a11,a22) such that a33∈S. The remaining 3 independent off-diagonal elements (a12,a13,a23) can be chosen in 53 ways. So, n(S1∩S3)=13×53.
Step 6: Calculate n(S1 union S2 union S3)
We use the Principle of Inclusion-Exclusion. Since n(S2)=0, all terms involving S2 become zero. Therefore, n(S1∪S2∪S3)=n(S1)+n(S3)−n(S1∩S3).
n(S1∪S2∪S3)=56+13×56−13×53=14×56−13×53=14×15625−13×125=218750−1625=217125.
Step 7: Find the value of alpha
Given that n(S1∪S2∪S3)=125α, we can substitute the calculated value and solve for α. Dividing 217125 by 125 gives α=1737.