Let \(V=\mathbb{R}^2.\) The map \(f:V^2\to\mathbb{R}\) defined by \[f(\vec x,\vec y) = \vec x\cdot \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\vec y\] is bilinear and alternating.

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Let \(V=\mathbb{R}^2.\) The map \(f:V^2\to\mathbb{R}\) defined by \[f(\vec x,\vec y) = \vec x\cdot \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\vec y\] satfies \[f\left(\begin{pmatrix}a \\ c \end{pmatrix},\begin{pmatrix}b \\ d \end{pmatrix}\right) = \det\left(\begin{pmatrix}a & b \\ c & d\end{pmatrix}\right).\]

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• False

If \(f:\mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}\) is bilinear and such that \(f(\vec x,\vec y)=-f(\vec y,\vec x)\) for all \(\vec x,\vec y\in \mathbb{R}^n,\) then \(f\) is alternating.

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If \(f:\mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}\) is bilinear and alternating, then \(f(\vec x,\vec y)=-f(\vec y,\vec x)\) for all \(\vec x,\vec y\in \mathbb{R}^n.\)

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The bilinear map map \(f:\mathbb{F}_2^2\to\mathbb{F}_2, (x,y)\mapsto x y\) satisfies \(f(x,y)=-f(y,x)\) for all \((x,y)\in\mathbb{F}_2^2\) but it is not alternating.

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• False

\(\det(\mathbf{1}_{4})=\det(-\mathbf{1}_{4})\)

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Let \(\mathbf{A},\mathbf{B}\in M_{n,n}(\mathbb{R}).\) If \(\det(\mathbf{A})=\det(\mathbf{B}),\) then \(\mathbf{A}= \pm\mathbf{B}.\)

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If \(\mathbf{A}\in M_{n,n}(\mathbb{K}),\) then \(\det(\mathbf{A})=1\) implies \(\mathbf{A}= \mathbf{1}_{n}.\)

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\(\det(\mathbf{1}_{5})=\det(-\mathbf{1}_{5})\)

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• False

It holds that \[\det\begin{pmatrix}0 & 0 & 1 \\ 1 & 2 & 5 \\ 2 & 1 & 3\end{pmatrix}=-3.\]

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If \(\mathbf{A}\in M_{n,n}(\mathbb{R})\) has only rational entries, then \(\det(\mathbf{A})\in\mathbb{Q}.\)

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The set \(\{\mathbf{A}\in M_{n,n}(\mathbb{R})|\det(\mathbf{A})=0\}\) is a subspace of \(M_{n,n}(\mathbb{R}).\)

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• False

If \(\mathbf{B}\in M_{n,n}(\mathbb{K})\) is the reduced row echelon form of \(\mathbf{A}\in M_{n,n}(\mathbb{K}),\) then \(\det(\mathbf{A})=\det(\mathbf{B}).\)

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Let \(\mathbf{A}\in M_{n,n}(\mathbb{K}).\) If all minors \(\mathbf{A}^{(k,l)}\in M_{n-1,n-1}(\mathbb{K}), 1\leqslant k,l\leqslant n\) have zero determinant, then \(\det(\mathbf{A})=0.\)

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Let \(\mathbf{A}\in M_{n,n}(\mathbb{K})\) be such that \(\det(\mathbf{A})=0,\) then at least one of its minors \(\mathbf{A}^{(k,l)}\in M_{n-1,n-1}(\mathbb{K}), 1\leqslant k,l\leqslant n\) must have vanishing determinant.

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• False