If A is an invertible matrix of order 3 and B is another matrix of the same order as of A, such that |B|=2, Aᵀ|A|B=A|B|Bᵀ. If ||AB⁻¹adj(AᵀB)⁻¹||=K, then the value of 4K is equal to

To solve the problem, we start with the given equation: Aᵀ|A|B = A|B|Bᵀ. Since A is an invertible matrix, we can manipulate this equation using properties of determinants and adjugates.

Step 1: Analyzing the Determinants

We know that the determinant of a product of matrices is the product of their determinants. Thus, we can express the determinants as follows:

• |Aᵀ| = |A| (since the determinant of a transpose is the same as the determinant of the original matrix)
• |B| = 2 (given)

Step 2: Simplifying the Equation

Substituting the determinants into the equation gives us:

|A|²|B| = |A| |B|²

Since |B| = 2, we have:

|A|² * 2 = |A| * 4

This simplifies to:

|A|² = 2|A|, leading to |A|(|A| - 2) = 0.

Since A is invertible, |A| cannot be zero, so |A| must equal 2.

Step 3: Finding K

Next, we need to evaluate ||AB⁻¹adj(AᵀB)⁻¹||. Using properties of determinants:

• adj(A) = |A|A⁻¹, thus adj(AᵀB) = |AᵀB|(AᵀB)⁻¹.
• Since |A| = 2 and |B| = 2, we find |AᵀB| = |A| |B| = 2 * 2 = 4.

Step 4: Calculating ||AB⁻¹adj(AᵀB)⁻¹||

Now we can express ||AB⁻¹adj(AᵀB)⁻¹|| as:

||AB⁻¹ * (1/|AᵀB|) * (AᵀB) = ||AB⁻¹|| * ||(AᵀB)⁻¹||.

Calculating these norms gives us:

• ||AB⁻¹|| = ||A|| * ||B⁻¹|| = 2 * (1/2) = 1.
• ||(AᵀB)⁻¹|| = 1/|AᵀB| = 1/4.

Thus, K = ||AB⁻¹adj(AᵀB)⁻¹|| = 1 * (1/4) = 1/4.

Final Calculation

To find the value of 4K:

4K = 4 * (1/4) = 1.

Therefore, the value of 4K is 1.