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    The $2 \times 2$ determinant is alternating and 2-linear over a commutative ring

    E-HK-52-7

    Area
    Algebra
    Topics
    Determinants, Linear Algebra
    Status
    draft

    Let \(K\) be a commutative ring with identity. Show that the determinant function on \(2 \times 2\) matrices \(A\) over \(K\) is alternating and 2-linear as a function of the columns of \(A\).

    Solution

    1. Definition of the \(2 \times 2\) determinant:

    1.1. For columns \(C_1 = \begin{pmatrix} a_{11} \\ a_{21} \end{pmatrix}, \, C_2 = \begin{pmatrix} a_{12} \\ a_{22} \end{pmatrix} \in K^2\), the determinant is:\[ D(C_1, C_2) = a_{11} a_{22} - a_{12} a_{21}. \]

    standard formula for \(2 \times 2\) determinant.

    2. Proof of 2-linearity:

    2.1. Linearity in the first column:Let \(u = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}, \, v = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} \in K^2\), \(\alpha, \beta \in K\), and \(C_2 = \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} \in K^2\).Then \(\alpha u + \beta v = \begin{pmatrix} \alpha u_1 + \beta v_1 \\ \alpha u_2 + \beta v_2 \end{pmatrix}\).Evaluating \(D\):\[ \begin{aligned} D(\alpha u + \beta v, C_2) &= (\alpha u_1 + \beta v_1) w_2 - w_1 (\alpha u_2 + \beta v_2) \\ &= \alpha(u_1 w_2 - w_1 u_2) + \beta(v_1 w_2 - w_1 v_2) \\ &= \alpha D(u, C_2) + \beta D(v, C_2). \end{aligned} \]

    distributivity and commutativity of multiplication in \(K\).

    2.2. Linearity in the second column:Let \(C_1 = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} \in K^2\), and \(C_2 = \alpha v + \beta w = \begin{pmatrix} \alpha v_1 + \beta w_1 \\ \alpha v_2 + \beta w_2 \end{pmatrix}\).Evaluating \(D\):\[ \begin{aligned} D(C_1, \alpha v + \beta w) &= u_1 (\alpha v_2 + \beta w_2) - (\alpha v_1 + \beta w_1) u_2 \\ &= \alpha(u_1 v_2 - v_1 u_2) + \beta(u_1 w_2 - w_1 u_2) \\ &= \alpha D(C_1, v) + \beta D(C_1, w). \end{aligned} \]Thus \(D\) is linear in each column (2-linear).

    distributivity and commutativity in \(K\).

    3. Proof of alternating property:

    3.1. Let both columns be equal: \(C_1 = C_2 = C = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \in K^2\).Then:\[ D(C, C) = x_1 x_2 - x_1 x_2 = 0. \]Because \(K\) is commutative, the two terms are identical and cancel.Thus \(D\) is alternating.

    subtraction of identical elements in \(K\).

    4. Conclusion:The determinant function on \(2 \times 2\) matrices over a commutative ring \(K\) is 2-linear and alternating as a function of the columns. Q.E.D.

    2 and 3.