For a 3×3 matrix M, let trace(M) denote the sum of all the diagonal elements of M. Let A be a 3×3 matrix such that ∣A∣=21 and trace(A)=3. If B=adj(adj(2A)), then the value of ∣B∣+trace(B) equals:
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Step-by-Step Solution
Step 1: Calculate the determinant of 2A
For an n×n matrix A and a scalar k, the determinant of kA is given by kn∣A∣. In this problem, A is a 3×3 matrix, so n=3. We are given that ∣A∣=21.
Step 2: Calculate the determinant of adj(2A)
The determinant of the adjoint of a matrix M is equal to the determinant of M raised to the power of (n−1), where n is the order of the matrix. Here, M=2A and n=3. We found ∣2A∣=4.
Step 3: Calculate the determinant of B
We need to find the determinant of B=adj(adj(2A)). Using the property ∣adj(M)∣=∣M∣n−1 twice, we get ∣B∣=∣adj(2A)∣n−1. Since ∣adj(2A)∣=∣2A∣n−1, we have ∣B∣=(∣2A∣n−1)n−1=∣2A∣(n−1)2. For n=3, this simplifies to ∣2A∣(3−1)2=∣2A∣22=∣2A∣4. We know ∣2A∣=4.
Step 4: Calculate the trace of B
For an n×n matrix M, the adjoint of the adjoint of M is given by ∣M∣n−2M. Here, M=2A and n=3. So, B=adj(adj(2A))=∣2A∣3−2(2A)=∣2A∣(2A). We know ∣2A∣=4.
Step 5: Calculate ∣B∣+trace(B)
We have calculated ∣B∣=256 and trace(B)=24. Now we sum these two values.