Matrix A = [[1, 4, 3], [5, 2, 1], [6, 4, 3]], and B = A - A^T, then B is?

Steps to Solve

1. Calculate the Transpose of Matrix A

The transpose of matrix A, denoted as $A^T$, is obtained by flipping the matrix over its diagonal. For matrix

$$ A = \begin{bmatrix} 1 & 4 & 3 \ 5 & 2 & 1 \ 6 & 4 & 3 \end{bmatrix} $$

The transpose $A^T$ is

$$ A^T = \begin{bmatrix} 1 & 5 & 6 \ 4 & 2 & 4 \ 3 & 1 & 3 \end{bmatrix} $$.

2. Subtract the Transpose from Matrix A

Now, we need to calculate matrix B by subtracting $A^T$ from A:

$$ B = A - A^T = \begin{bmatrix} 1 & 4 & 3 \ 5 & 2 & 1 \ 6 & 4 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 5 & 6 \ 4 & 2 & 4 \ 3 & 1 & 3 \end{bmatrix} $$.

Performing the subtraction element-wise gives:

$$ B = \begin{bmatrix} 1 - 1 & 4 - 5 & 3 - 6 \ 5 - 4 & 2 - 2 & 1 - 4 \ 6 - 3 & 4 - 1 & 3 - 3 \end{bmatrix} = \begin{bmatrix} 0 & -1 & -3 \ 1 & 0 & -3 \ 3 & 3 & 0 \end{bmatrix} $$.

3. Determine the Nature of Matrix B

To determine whether matrix B is symmetric, skew-symmetric, diagonal, or scalar, we examine the properties:

• A matrix is symmetric if $B = B^T$,
• A matrix is skew-symmetric if $B = -B^T$.

We calculate the transpose of B:

$$ B^T = \begin{bmatrix} 0 & 1 & 3 \ -1 & 0 & 3 \ -3 & -3 & 0 \end{bmatrix} $$.

Checking if $B = -B^T$, we see:

$$ -B^T = \begin{bmatrix} 0 & -1 & -3 \ 1 & 0 & -3 \ 3 & 3 & 0 \end{bmatrix} = B $$.

Thus, $B$ is skew-symmetric.