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Proof Jumble 10/2, We will use proof by contradiction to show that if \(ad…
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We will use proof by contradiction to show that if \(ad-bc=0\), then the matrix \(A\) is not invertible.
Let \(A\vec{x}=\vec{0}\), where \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\) and \(\vec{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\)
We first want to construct an augmented matrix that has the matrix \(A\) and \(\vec{0}\), like the following:
\(\begin{bmatrix} a & b & 0 \\ c & d & 0 \end{bmatrix}\).
We can row reduce the matrix to echelon form, under the assumption that \(ad-bc=0\).
We compute: \(\begin{bmatrix} a & b & 0 \\ c & d & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & \frac{b}{a} & 0 \\ 0 & 0 & 0 \end{bmatrix}\)
When the bottom row of a matrix is all zeros, it means that there is a free variable in the solution.
Free variables indicate that there is more than one solution present. In this case, the free variable is \(x_2\).
We know this because the first row of the reduced matrix would be equal to \(x_1+\frac{b}{a}x_2=0\) and it can be simplified to \(x_1=-\frac{b}{a}x_2\).
Therefore, \(\vec{x} =\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\) is one solution, where \(x_2=a\).
Next, let \(ad-bc=0\) and \(A\vec{x}=\vec{0}\), where \(A\vec{x}\) is equal to the \(A\) matrix multiplied by a vector \(\vec{x}\).
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Proof: we will show that the distance between 2 points is d(x,y)=d(y,x)
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we know that when we square a positive number, we will get a positive number.
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