a+b =7
b+c =10
c+a=13
a+b+c=?
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MathBot Answer:
The system of linear equations has exactly one solution.
The substitutions are \( a = 5 \), \( b = 2 \), \( c = 8 \).
\[a + b + c = 15\]
Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}1 & 1 & 0 &\bigm |& 7\\0 & 1 & 1 &\bigm |& 10\\1 & 0 & 1 &\bigm |& 13\end{bmatrix}$$
Then use a series of elementary row operations to convert the matrix into reduced-row echelon form. The three elementary row operations are:
1. Swap the positions of any two rows.
2. Multiply any row by a nonzero scalar.
3. Multiply a row by a nonzero scalar and add it to any other row.
First, switch the rows in the matrix such that the row with the leftmost non-zero entry with the greatest magnitude is at the top.
$$\begin{bmatrix}1 & 1 & 0 &\bigm |& 7\\0 & 1 & 1 &\bigm |& 10\\1 & 0 & 1 &\bigm |& 13\end{bmatrix}$$
The leading term of row \(1\) is already \(1\) so this row does not need to be multiplied by a scalar.
$$\begin{bmatrix}1 & 1 & 0 &\bigm |& 7\\0 & 1 & 1 &\bigm |& 10\\1 & 0 & 1 &\bigm |& 13\end{bmatrix}$$
Multiply row \(1\) by scalar \(-1\) and add it to row \(3\).
$$\begin{bmatrix}1 & 1 & 0 &\bigm |& 7\\0 & 1 & 1 &\bigm |& 10\\0 & -1 & 1 &\bigm |& 6\end{bmatrix}$$
Switch the rows in the matrix such that the row with the next leftmost non-zero entry with the next greatest magnitude is the next row from the top.
$$\begin{bmatrix}1 & 1 & 0 &\bigm |& 7\\0 & 1 & 1 &\bigm |& 10\\0 & -1 & 1 &\bigm |& 6\end{bmatrix}$$
The leading term of row \(2\) is already \(1\) so this row does not need to be multiplied by a scalar.
$$\begin{bmatrix}1 & 1 & 0 &\bigm |& 7\\0 & 1 & 1 &\bigm |& 10\\0 & -1 & 1 &\bigm |& 6\end{bmatrix}$$
Multiply row \(2\) by scalar \(-1\) and add it to row \(1\).
$$\begin{bmatrix}1 & 0 & -1 &\bigm |& -3\\0 & 1 & 1 &\bigm |& 10\\0 & -1 & 1 &\bigm |& 6\end{bmatrix}$$
Multiply row \(2\) by scalar \(1\) and add it to row \(3\).
$$\begin{bmatrix}1 & 0 & -1 &\bigm |& -3\\0 & 1 & 1 &\bigm |& 10\\0 & 0 & 2 &\bigm |& 16\end{bmatrix}$$
Multiply row \(3\) by scalar \(\frac{1}{2}\) to make the leading term \(1\).
$$\begin{bmatrix}1 & 0 & -1 &\bigm |& -3\\0 & 1 & 1 &\bigm |& 10\\0 & 0 & 1 &\bigm |& 8\end{bmatrix}$$
Multiply row \(3\) by scalar \(1\) and add it to row \(1\).
$$\begin{bmatrix}1 & 0 & 0 &\bigm |& 5\\0 & 1 & 1 &\bigm |& 10\\0 & 0 & 1 &\bigm |& 8\end{bmatrix}$$
Multiply row \(3\) by scalar \(-1\) and add it to row \(2\).
$$\begin{bmatrix}1 & 0 & 0 &\bigm |& 5\\0 & 1 & 0 &\bigm |& 2\\0 & 0 & 1 &\bigm |& 8\end{bmatrix}$$
Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot a+ 0 \cdot b+ 0 \cdot c = 5 \\ a = 5\end{aligned}$$$$\begin{aligned}0 \cdot a+ 1 \cdot b+ 0 \cdot c = 2 \\ b = 2\end{aligned}$$$$\begin{aligned}0 \cdot a+ 0 \cdot b+ 1 \cdot c = 8 \\ c = 8\end{aligned}$$
By substituting these values into the expression, we find that:$$\begin{aligned} a + b + c &= 2 + 5 + 8\\ &=15\end{aligned}$$