Question
Consider a function $D$ from $\mathbb{R}^{2 \times 2}$ to $\mathbb{R}$ that is linear in both columns and alternating on the columns. See Examples 4 and 6 and the subsequent discussions. Assume that $D\left(I_{2}\right)=1$.Using Exercises 62 and 63 as a guide, show that $D(A)=a d-b c=\operatorname{det} A$ for all $2 \times 2$ matrices $A$
Step 1
The function \( D \) is linear in each column and alternating, meaning \( D(A) = -D(A') \) if the columns of \( A \) are swapped to get \( A' \). Also, \( D(I_2) = 1 \) where \( I_2 \) is the \( 2 \times 2 \) identity matrix. Show more…
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Key Concepts
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Consider a function $D$ from $\mathbb{R}^{2 \times 2}$ to $\mathbb{R}$ that is linear in both columns and alternating on the columns. See Examples 4 and 6 and the subsequent discussions. Assume that $D\left(I_{2}\right)=1$. Show that $D(A)=0$ for any $2 \times 2$ matrix $A$ whose two columns are equal.
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Consider a function $D$ from $\mathbb{R}^{2 \times 2}$ to $\mathbb{R}$ that is linear in both columns and alternating on the columns. See Examples 4 and 6 and the subsequent discussions. Assume that $D\left(I_{2}\right)=1$. Show that $D\left[\begin{array}{ll}a & b \\ 0 & d\end{array}\right]=$ ad. Hint: Write $\left[\begin{array}{l}b \\ d\end{array}\right]=$ $\left[\begin{array}{l}b \\ 0\end{array}\right]+\left[\begin{array}{l}0 \\ d\end{array}\right]$ and use linearity in the second column $\operatorname{}: D\left[\begin{array}{cc}a & b \\ 0 & d\end{array}\right]=D\left[\begin{array}{cc}a & b \\ 0 & 0\end{array}\right]+D\left[\begin{array}{cc}a & 0 \\ 0 & d\end{array}\right]=$ $a b D\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]+\ldots$ Use Exercise 62
Find all $2 \times 2$ matrices $A$ such that det $A=1$ and $A=A^{-1}$
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