I am very confused about how to solve this system of linear equation

Question

I need to solve a system of 3 equations:

$$ \begin{align} a + b + c &= 7 \tag{i}\\ 9a + 3b + c &= 13 \tag{ii}\\ 81a + 9b + c &= 7 \tag{iii} \end{align} $$

This is how I solved.

Subtract $(i)$ from $(ii)$:

$ \begin{align} && 9a + 3b + c - a - b - c &= 13 - 7\\ \Rightarrow && 8a + 2b &= 6\\ \Rightarrow && 4a + b &= 3\\ \Rightarrow && b &= 4a - 3\tag{iv} \end{align} $

Substitute $(iv)$ in $(i)$:

$ \begin{align} && a + (4a - 3) + c &= 7\\ \Rightarrow && 5a - 3 + c &= 7\\ \Rightarrow && 5a + c &= 7 + 3\\ \Rightarrow && 5a + c &= 10\\ \Rightarrow && c &= 10 - 5a \tag{v} \end{align} $

Substitute $(iv)$ and $(v)$ in $(iii)$:

$ \begin{align} && 81a + 9(4a - 3) + (10 - 5a) &= 7\\ \Rightarrow && 81a + 36a - 27 + 10 - 5a &= 7\\ \Rightarrow && 117a - 5a &= 27 + 7 - 10\\ \Rightarrow && 112a &= 24\\ \Rightarrow && a &= \frac{3}{14} \tag{vi} \end{align} $

But according to the official solution in the book, the correct value for $a$ is $\frac{1}{2}$.

What am I doing wrong?

Answer 1

Your calculation would be more easier ,using your method in another way.

First,(ii)-(i) gives , \begin{equation} \label{iv} 8a+2b =6 \Rightarrow 4a+b =3 \end{equation}

(iii)-(i) gives \begin{equation} 80a+8b=0 \Rightarrow10a+b=0 \end{equation}

Now , Subtract 2nd one from 1st one, $$ -6a=3\Rightarrow a=-1/2$$

Answer 2

Another approach to compare with.

The coefficient matrix is the transpose from a Vandermonde's matrix. Precisely, one has: \begin{align*} \det \begin{bmatrix} 1 & 1 & 1\\ 9 & 3 & 1\\ 81 & 9 & 1 \end{bmatrix} = \det \begin{bmatrix} 1 & 9 & 81\\ 1 & 3 & 9\\ 1 & 1 & 1 \end{bmatrix} = (1 - 9)(1 - 3)(3 - 9) = -96 \end{align*}

According to the Cramer's rule, we need the following determinant: \begin{align*} \det \begin{bmatrix} 7 & 1 & 1\\ 13 & 3 & 1\\ 7 & 9 & 1 \end{bmatrix} = 21 + 7 + 107 - 21 - 13 - 63 = 48 \end{align*}

as well as the determinant: \begin{align*} \det \begin{bmatrix} 1 & 7 & 1\\ 9 & 13 & 1\\ 81 & 7 & 1 \end{bmatrix} = 13 + 567 + 63 - 1053 - 63 - 7 = -480 \end{align*} and, finally, the next determinant too:

\begin{align*} \det \begin{bmatrix} 1 & 1 & 7\\ 9 & 3 & 13\\ 81 & 9 & 7 \end{bmatrix} = 21 + 1053 + 567 - 1701 - 63 - 117 = -240 \end{align*}

Consequently, the solution set is given by the triple: \begin{align*} \left(a,b,c\right) = \left(-\frac{48}{96},\frac{480}{96},\frac{240}{96}\right) = \left(-\frac{1}{2},5,\frac{5}{2}\right) \end{align*}

Hopefully this helps!