Let a,b,c real numbers. Show that the system of linear equations has only one solution

Question

$$x_1 + x_2 + x_3 = a$$

$$2x_1 - x_2 + 3x_3 = b$$

$$4x_1 + x_2 + 9x_3 = c$$

Now, looking at the solution of this, the author does the following:

The matrix of the coefficients is invertible because $det(E) \ne 0$ (where E is the matrix). Therefore the linear system has only one solution.

Can someone elaborate of the above solution? I do not understand this deduction.

Answer 1

See the invertible matrix theorem. http://mathworld.wolfram.com/InvertibleMatrixTheorem.html

Answer 2

If you write the system as a matrix equation

\begin{align*} \left( \begin{matrix} 1 &1 &1\\ 2 &-1 &3\\ 4 &1 &9 \end{matrix} \right) \left( \begin{matrix} x_1\\ x_2\\ x_3 \end{matrix}\right) = \left( \begin{matrix} a\\ b\\ c \end{matrix}\right), \end{align*}

then the determinant of your coefficient matrix is non-zero, so there is a unique solution, which is the right hand side of your equation times the inverse of the coefficient matrix:.

Answer 3

When the matrix is not singular, i.e. $\det(A)\neq 0$, the columns of $A$ are linearly independent and thus, since $A\vec x$ represents a combination of column vectors which are a basis and span $R^3$, we have that $$A\vec x=b$$

has always one and only one solution.

Answer 4

Multiplying the first equation by $-2$ and adding to the second: $$-3x_2+x_3=b-2a$$ Multiplying the first by $-4$ and adding to the third we get $$-3x_2+5x_3=c-4a$$ From this equations we get $$4x_3=-2a-b+c$$

Answer 5

Theorem: If $A$ is an invertible $n\times n$ matrix, then $\forall b\in\Bbb R^n$, the system $Ax=b$ has a unique solution.

You must know this theorem by heart!

Proof: Indeed, if $x_0$ is a solution, then $$Ax_0 = b\Rightarrow A^{-1}(Ax_0) = A^{-1}b \Rightarrow (A^{-1}A)x_0 = Ix_0 = x_0 =A^{-1}b$$ so this proves that the only possible solution is $A^{-1}b$.

Now, $A^{-1}b$ is really a solution since $A(A^{-1}b) = (AA^{-1})b = Ib = b$.

Hence $Ax=b$ has one and only one solution if $A$ is invertible.$\;\;\blacksquare$

Remark: Of course, this is only a possible proof, and other answers above are also good.