Let $$ \mathbf{A}=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array} ight] $$ and $$ \mathbf{B}=\left[\begin{array}{ll}0 & 1 \ 1 & 0\end{array} ight] $$. Identify $$ \mathbf{A} \wedge \mathbf{B} $$.

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Let $$ \mathbf{A}=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight] $$ and $$ \mathbf{B}=\left[\begin{array}{ll}0 & 1 \ 1 & 0\end{array}ight] $$. Identify $$ \mathbf{A} \wedge \mathbf{B} $$.

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Alright, I need to identify the operation $$ \mathbf{A} \wedge \mathbf{B} $$ where $$ \mathbf{A} $$ and $$ \mathbf{B} $$ are given matrices. First, let's recall what the wedge product $$ \wedge $$ typically represents in linear algebra. The wedge product is often used to denote the exterior product of two vectors or, more generally, the exterior algebra of a vector space. However, in the context of matrices, the wedge product isn't as commonly used as other operations like addition, multiplication, or the determinant.

Given that $$ \mathbf{A} $$ and $$ \mathbf{B} $$ are both $$ 2 \times 2 $$ matrices, I should consider if there's a specific definition or operation that uses the wedge symbol in this context. One possibility is that $$ \mathbf{A} \wedge \mathbf{B} $$ represents the wedge product of the column vectors of $$ \mathbf{A} $$ and $$ \mathbf{B} $$. Let's explore this.

First, let's write down the matrices:

$$
\mathbf{A} = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}
$$

The columns of $$ \mathbf{A} $$ are $$ \mathbf{a}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} $$ and $$ \mathbf{a}_2 = \begin{bmatrix} 1 \ 1 \end{bmatrix} $$. Similarly, the columns of $$ \mathbf{B} $$ are $$ \mathbf{b}_1 = \begin{bmatrix} 0 \ 1 \end{bmatrix} $$ and $$ \mathbf{b}_2 = \begin{bmatrix} 1 \ 0 \end{bmatrix} $$.

The wedge product $$ \mathbf{a}_1 \wedge \mathbf{a}_2 $$ is a bivector in $$ \mathbb{R}^2 $$, which can be represented as a scalar multiple of the area element. Specifically, in two dimensions, the wedge product of two vectors $$ \mathbf{u} = \begin{bmatrix} u_1 \ u_2 \end{bmatrix} $$ and $$ \mathbf{v} = \begin{bmatrix} v_1 \ v_2 \end{bmatrix} $$ is given by:

$$
\mathbf{u} \wedge \mathbf{v} = u_1 v_2 - u_2 v_1
$$

Applying this to $$ \mathbf{a}_1 $$ and $$ \mathbf{a}_2 $$:

$$
\mathbf{a}_1 \wedge \mathbf{a}_2 = (1)(1) - (0)(1) = 1
$$

Similarly, for $$ \mathbf{b}_1 $$ and $$ \mathbf{b}_2 $$:

$$
\mathbf{b}_1 \wedge \mathbf{b}_2 = (0)(0) - (1)(1) = -1
$$

Now, if $$ \mathbf{A} \wedge \mathbf{B} $$ is defined as the wedge product of the column vectors of $$ \mathbf{A} $$ and $$ \mathbf{B} $$, then it would involve combining these bivectors in some way. One possible interpretation is to take the wedge product of corresponding columns:

$$
\mathbf{A} \wedge \mathbf{B} = (\mathbf{a}_1 \wedge \mathbf{b}_1) \wedge (\mathbf{a}_2 \wedge \mathbf{b}_2) = (1) \wedge (-1) = -1
$$

However, this interpretation seems a bit convoluted. Another approach is to consider the wedge product of the entire matrices $$ \mathbf{A} $$ and $$ \mathbf{B} $$ as elements of the exterior algebra. In this case, $$ \mathbf{A} \wedge \mathbf{B} $$ would be a higher-dimensional object, but since we're dealing with $$ 2 \times 2 $$ matrices, the result would likely be a scalar.

Alternatively, if $$ \mathbf{A} \wedge \mathbf{B} $$ is defined as the determinant of the matrix formed by placing $$ \mathbf{A} $$ and $$ \mathbf{B} $$ side by side, then:

$$
\mathbf{A} \wedge \mathbf{B} = \begin{bmatrix} \mathbf{A} & \mathbf{B} \end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 & 1 \ 0 & 1 & 1 & 0 \end{bmatrix}
$$

The determinant of this $$ 2 \times 4 $$ matrix isn't defined, so this interpretation doesn't hold.

Another possibility is that $$ \mathbf{A} \wedge \mathbf{B} $$ represents the tensor product of $$ \mathbf{A} $$ and $$ \mathbf{B} $$, but the tensor product of two $$ 2 \times 2 $$ matrices results in a $$ 4 \times 4 $$ matrix, which doesn't align with the context of the question.

Given the ambiguity in the definition of $$ \mathbf{A} \wedge \mathbf{B} $$, it's challenging to determine the exact operation without additional context. However, based on the most straightforward interpretation of the wedge product of column vectors, $$ \mathbf{A} \wedge \mathbf{B} $$ would be $$-1$$.

$$ \mathbf{A} \wedge \mathbf{B} = -1 $$

Let \( \mathbf{A}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] \) and \( \mathbf{B}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \). Identify \( \mathbf{A} \wedge \mathbf{B} \).

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