To solve the problem, we need to find the value of \( x \) such that the equation \( B^T A^T = C \) holds true, where matrices \( A, B, \) and \( C \) are given as:
| Matrix | Expression |
|---|---|
| \( A \) | \(\begin{bmatrix} 2 & x \\ 3 & 1 \end{bmatrix}\) |
| \( B \) | \(\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}\) |
| \( C \) | \(\begin{bmatrix} 3x+2 & 10 \\ 14-x & 10 \end{bmatrix}\) |
Step 1: Transpose the matrices \( A \) and \( B \).
The transpose of matrix \( A \), denoted \( A^T \), is:
\(A^T = \begin{bmatrix} 2 & 3 \\ x & 1 \end{bmatrix}\)
The transpose of matrix \( B \), denoted \( B^T \), is:
\(B^T = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}\)
Step 2: Calculate the product \( B^T A^T \).
We multiply the matrices \( B^T \) and \( A^T \) as follows:
\(B^T A^T = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ x & 1 \end{bmatrix}\)
The product is calculated by taking the dot product of rows of \( B^T \) with columns of \( A^T \):
This gives us the product \( B^T A^T \):
\(B^T A^T = \begin{bmatrix} 6 + x & 10 \\ 4 + 4x & 10 \end{bmatrix}\)
Step 3: Equate \( B^T A^T \) and \( C \) to find \( x \).
We know that \( B^T A^T = C \). Therefore, we set the corresponding elements of the two matrices equal:
From the first equation, solve for \( x \):
\(6 + x = 3x + 2 \Rightarrow 6 - 2 = 3x - x \Rightarrow 4 = 2x \Rightarrow x = 2\)
Verify this value with the second equation:
\(4 + 4(2) = 14 - 2 \Rightarrow 4 + 8 = 12, \Rightarrow 12 = 12\)
This confirms the calculations are correct.
The value of \( x \) is \(2\).
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: