System of linear equations: condition for uniqueness

Question

Show that the system $$bx + ay =c \\ cx + az = b \\ cy + bz = a $$ has a unique solution if and only if $ abc \neq 0 $. In this case, fix the system.

I gathered the relationships and got $x = y = z = 1/2$, but I couldn't solve the system. Did I miscalculate?

Answer 1

You're trying to solve the equation

$$\begin{pmatrix} b & a & 0\\ c & 0 & a\\ 0 & c & b \end{pmatrix} \begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}c \\ b \\ a\end{pmatrix},$$ which has a unique solution if and only if the matrix is invertible. So calculating its determinant you get $-2abc$, which is nonzero if and only if $abc \neq 0$. This solves the first part of your question. For the second part, invert the matrix and you get the inverse

$$\frac{1}{-2abc}\begin{pmatrix} -ac & -ab & a^2\\ -bc & b^2 & -ab\\ c^2 & -bc & -ac \end{pmatrix}.$$

So now you just have to multiply by the vector $$\begin{pmatrix}c \\ b \\ a\end{pmatrix}.$$ So the unique solution is

$$\frac{1}{-2abc}\begin{pmatrix} -ac & -ab & a^2\\ -bc & b^2 & -ab\\ c^2 & -bc & -ac \end{pmatrix} \begin{pmatrix}c \\ b \\ a\end{pmatrix} = \frac{1}{-2abc}\begin{pmatrix} -ac^2-ab^2+a^3 \\ -bc^2 + b^3 - a^2b \\ c^3 - b^2c - a^2c\end{pmatrix}.$$

Answer 2

The hint.

The determinant of this system is equal to $-2abc$

Answer 3

This system is equivalent to solving

$$\begin{pmatrix} b & a & 0\\ c & 0 & a\\ 0 & c & b \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} c \\ b \\ a \end{pmatrix}$$

The determinant of this matrix is $-2abc$. Thus, the matrix is invertible if and only if $abc\neq 0$. If the matrix is invertible, then a unique solution exists for the system. If not, the system has zero or infinite solutions.

Answer 4

Just solve it.

If $abc\ne 0$ then $a\ne 0; b\ne 0; c\ne 0$

So $bx + ay = c$ means $x = \frac {c-ay}b=\frac cb - \frac aby$.

And $cx + az=b$ means $z = \frac {b-cx}a =\frac ba -\frac cax=\frac ba - \frac ca(\frac cb - \frac aby)=$

$\frac ba - \frac {c^2}{ab}+\frac cby$

And $cy + bz=a$ means $y = \frac {a-bz}c = \frac ac -\frac bc(z)=\frac ac -\frac bc(\frac ba - \frac {c^2}{ab}+\frac cby)=$

$\frac ac - \frac {b^2}{ac} +\frac ca - y$

So we can solve for $y= \frac a{2c} - \frac {b^2}{2ac} +\frac c{2a}$

And plug in $y$ to get $x=\frac cb-\frac ab(\frac a{2c} - \frac {b^2}{2ac} +\frac c{2a})=$

$\frac cb-\frac {a^2}{2bc}+\frac b{2c}-\frac c{2b}=\frac b{2c}-\frac {a^2}{2bc}-\frac c{2b}$

And $z= \frac ba -\frac cax = \frac ba -\frac ca(\frac b{2c}-\frac {a^2}{2bc}-\frac c{2b})=$

$\frac b{2a} +\frac a{2b} +\frac {c^2}{2ab}$

.....

And if $abc = 0$ then one of $a,b$ or $c = 0$

If, say $a=0$ then we have $bx = c$, $cx = b$ and $cy=-bz$.

If $b\ne 0$ then $x = \frac cb$ and we can only have solutions if $c^2=b^2$ and if so than $y=-\frac bc z$ can have infinitely many solutions.

If $b = 0$ then we must also have $c=0$ and $x,y, z$ can be any values.

Similarly sets of infinite solutions occur if $b$ or $c=0$.