Solve $$$\begin{cases} a + c = 0 \\ a + b + 2 c + d = 0 \\ a + c + 2 d = 2 \\ a + b + d = 0 \end{cases}$$$ for $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$

The calculator will solve the system of linear equations $$$\begin{cases} a + c = 0 \\ a + b + 2 c + d = 0 \\ a + c + 2 d = 2 \\ a + b + d = 0 \end{cases}$$$ for $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$, with steps shown.

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Solve $$$\begin{cases} a + c = 0 \\ a + b + 2 c + d = 0 \\ a + c + 2 d = 2 \\ a + b + d = 0 \end{cases}$$$ for $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ using the Gauss-Jordan Elimination method.

Solution

Write down the augmented matrix: $$$\left[\begin{array}{cccc|c}1 & 0 & 1 & 0 & 0\\1 & 1 & 2 & 1 & 0\\1 & 0 & 1 & 2 & 2\\1 & 1 & 0 & 1 & 0\end{array}\right]$$$.

Perform the Gauss-Jordan elimination (for steps, see Gauss-Jordan elimination calculator): $$$\left[\begin{array}{cccc|c}1 & 0 & 1 & 0 & 0\\0 & 1 & 1 & 1 & 0\\0 & 0 & -2 & 0 & 0\\0 & 0 & 0 & 2 & 2\end{array}\right]$$$.

Back-substitute:

$$$d = \frac{2}{2} = 1$$$

$$$c = \frac{0 - \left(0\right) \left(1\right)}{-2} = 0$$$

$$$b = 0 - \left(1\right)^{2} - \left(0\right) \left(1\right) = -1$$$

$$$a = 0 - \left(0\right) \left(1\right) - \left(0\right) \left(1\right) - \left(-1\right) \left(0\right) = 0$$$

Answer

$$$a = 0$$$A

$$$b = -1$$$A

$$$c = 0$$$A

$$$d = 1$$$A