Let a∈R and A be a matrix of order 3×3 such that det(A)=−4 and A+I=12aa11102, where I is the 3×3 identity. If det((a+1)adj((a−1)A))=2m3n,m,n∈{0,1,2,…,20}, then m+n is equal to:
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Step-by-Step Solution
Step 1: Calculate the determinant of (A+I)
First, we need to find the value of a. We are given the matrix A+I. We can calculate its determinant using the cofactor expansion method. Expanding along the second row is convenient due to the zero entry.
Step 2: Solve for 'a'
We calculate the determinant of A+I as 3a. We know that det(A+I)=det(A)+sum of principal minors+det(I) is not generally true. Instead, we use the property that if A+I=B, then det(B)=3a. We are given det(A)=−4. However, we need to find A first to relate det(A+I) to det(A). This is a common mistake. Let's re-evaluate. The given matrix is A+I. We need to find A from A+I. Then we can calculate det(A) from the expression for A. But we are already given det(A)=−4. This means we need to find a such that det(A)=−4. Let's find A first.
Step 3: Determine matrix A and its determinant
Since A+I=12aa11102, we can find A by subtracting the identity matrix I=100010001 from A+I. This gives us the matrix A. Now we can calculate det(A).
Step 4: Calculate det(A) and solve for 'a'
We calculate the determinant of A by expanding along the first row. We get det(A)=−2a+2. We are given that det(A)=−4. Equating these two expressions, we can solve for a.
Step 5: Find the value of 'a'
Solving the equation −2a+2=−4 for a, we find that a=3. Now we have the value of a, which is crucial for the next part of the problem.
Step 6: Evaluate the determinant expression
We use the property det(kA)=kndet(A) for an n×n matrix A. Here, k=(a+1) and the matrix is adj((a−1)A). Since A is a 3×3 matrix, adj((a−1)A) is also a 3×3 matrix, so n=3.
Step 7: Apply adjoint determinant property
We use the property det(adj(B))=(det(B))n−1 for an n×n matrix B. Here, B=(a−1)A and n=3. So, det(adj((a−1)A))=(det((a−1)A))2.
Step 8: Simplify further using determinant properties
Again, we use the property det(kA)=kndet(A). Here, k=(a−1) and the matrix is A. Since A is a 3×3 matrix, n=3. So, det((a−1)A)=(a−1)3det(A). Substituting this into the expression, we get the simplified form.
Step 9: Substitute values and calculate the final result
Now we substitute the value a=3 and det(A)=−4 into the simplified expression. We then simplify the powers of 2 to get the final result in the form 2m3n. Since there is no factor of 3, n=0.
Step 10: Determine m and n, and calculate m+n
From the calculation, we have 216. Comparing this with 2m3n, we find m=16 and n=0. Therefore, m+n=16+0=16.