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Prove that the set of invertible real $2 \times 2$ matrices is dense in the set of all real $2 \times 2$ matrices.
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To prove that the set of invertible real $2 \times 2$ matrices is dense in the set of all real $2 \times 2$ matrices, we will show that for any real $2 \times 2$ matrix $A$, there exists a sequence of invertible real $2 \times 2$ matrices $B_n$ such that the sequence $B_n$ converges to $A$ as $n$ goes to infinity.

To construct the sequence $B_n$, we will use the fact that the determinant is a continuous function on the set of all $2 \times 2$ matrices. Specifically, let det be the determinant function, which maps a $2 \times 2$ matrix to its determinant. Then for any real $2 \times 2$ matrix $A$, we can define a sequence $B_n$ as follows:

$B_n = (1 - \frac{1}{n}) A + \frac{1}{n} I$

where $I$ is the $2 \times 2$ identity matrix. Then

$det(B_n) = (1 - \frac{1}{n})$

$det(A) + \frac{1}{n} det(I)$

$= 1 - \frac{1}{n} + \frac{1}{n} = 1$

for all n, so $B_n$ is invertible for all $n$.

Now, we will show that the sequence $B_n$ converges to $A$ as $n$ goes to infinity. To do this, we will use the fact that the matrix norm is a continuous function on the set of all $2 \times 2$ matrices. Specifically, let || || be a matrix norm, which maps a $2 \times 2$ matrix to a nonnegative real number. Then

$||B_n - A|| = ||(1 - 1/n) A - (1 - 1/n) A + (1/n) I||$

$= ||(1/n) I|| = 1/n$

for all $n$, so the sequence $B_n$ converges to $A$ as $n$ goes to infinity. This completes the proof.
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