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{\bf Question}

Write the following system of simultaneous equations in matrix
form and calculate the determinant as a function of $a$ and $b$.
$$ x+y+z=3,\ \ \ \  x+2y+2z=5,\ \ \ \ x+ay+bz=3.$$ For each of the
cases given below decide whether the equations have a unique
solution, no solutions or infinitely many solutions; find the
solutions where possible:

\begin{description}
\item[(i)]
$a=b=1$;
\item[(ii)]
$a=1$ and $b=1$;
\item[(iii)]
$a \ne 1$ and $b=1$;
\item[(iv)]
$a=b \ne 1$;
\item[(v)]
$1 \ne a \ne b \ne 1$.
\end{description}


{\bf Answer}

In matrix form ($A{\bf x}={\bf b}$):

$$\left(\begin{array}{ccc} 1 & 1 & 1\\ 1 & 2 & 2\\ 1 & a & b
\end{array} \right) \left(\begin{array}{c} x\\y\\t \end{array}\right)=
\left(\begin{array}{c} 3\\5\\3 \end{array}\right)$$
$$\det(A)=\left|\begin{array}{ccc} 1 & 1 & 1\\ 1 & 2 & 2\\ 1 & a &
b
\end{array}\right|=\left|\begin{array}{rcc} 1 & 1 & 1\\ -1 & 0 & 0\\ 1 & a &
b
\end{array}\right| \ \ (R_2'=R_2-2R_1)$$
\begin{eqnarray*}
& = & (-1)(-1)^{2+1} \left|\begin{array}{cc} 1 & 1\\ a & b
\end{array} \right| + 0 - 0\\
& {\ \ } & (\rm Expanding\ determinant\ along\ second\ row)\\ & =
& b-a.
\end{eqnarray*}

To solve the equations, take augmented matrix and reduce to upper
triangular form using row operations.
\begin{eqnarray*}
\left(\begin{array}{ccc} 1 & 1 & 1\\ 1 & 2 & 2\\ 1 & a & b
\end{array} \right| \left. \begin{array}{c} 3 \\ 5 \\ 3 \end{array} \right)
& \begin{array}{l} {R_2'=R_2-R_1}\\ {R_3'=R_3-R_1}\\ {-----
\longrightarrow}
\end{array} &
\left(\begin{array}{ccc} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & {a-1} & {b-1}
\end{array} \right| \left. \begin{array}{c} 3 \\ 2 \\ 0 \end{array}
\right)\\ & \begin{array}{l} {R_3'=R_3-(a-1)R_2}\\ {--------
\longrightarrow}
\end{array} & \left(\begin{array}{ccc} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & {b-a}
\end{array} \right| \left. \begin{array}{c} 3 \\ 2 \\ {2-2a} \end{array}
\right)
\end{eqnarray*}

So the equations reduce to:

$\left. \begin{array}{rcl}  {x+y+z} & = & 3\\ {y+z} & = & 2\\
{(b-a)z} & = & {2-2a}
\end{array} \right\}$

\begin{description}
\item[(i)]
$a=b=1$. $\det(A)=b-a=1-1=0$.

So either no solutions or an infinite number of solutions.

Equations become: $\left.
\begin{array}{rcl}  {x+y+z} & = & 3\\ {y+z} & = & 2\\ 0 & =
& 0
\end{array} \right\}$

Let $z=\alpha$, so $y=2-\alpha$ and
$x=3-y-z=3-(2-\alpha)-\alpha=1$. Infinitely many solutions:
$(x,y,z)=(1,2-\alpha,\alpha)$.

\item[(ii)]
$a=1,\ b \ne 1$. $\det(A)=b-1 \ne 0$. So a unique solution.

Equations become: $\left. \begin{array}{rcl}  {x+y+z} & = & 3\\
{y+z} & = & 2\\ {(b-1)z} & = & 0
\end{array} \right\} \begin{array}{l}z=0\ (\rm since\ b \ne 1)\\y=2\\x=2 \end{array}$

Hence the unique solution: $(x,y,z)=(1,2,0)$.

\item[(iii)]
$a \ne 1,\ b = 1$.  $\det(A)=1-a \ne 0$, so a unique solution.

Equations become: $\left. \begin{array}{rcl} {x+y+z} & = & 3\\
{y+z} & = & 2\\ {(1-a)z} & = & {2-2a}
\end{array} \right\} \begin{array}{l}z=\displaystyle \frac{2-2a}{1-a}=2\ (1-a \ne 0)\\y=0\\x=1 \end{array}$
so a unique solution $(x,y,z)=(1,0,2)$.

\item[(iv)]
$a=b \ne 1$. $\det(A)=b-a=0$.

So either no solutions or an infinite number of solutions.

Equations become: $\left. \begin{array}{rcl}  {x+y+z} & = & 3\\
{y+z} & = & 2\\ 0 & = & {2-2a}
\end{array} \right\}$

since $a \ne 1,\ 0=2-2a \ne 0$ which is a contradiction.

So equations are inconsistent and have no solutions.

\item[(v)]
$1 \ne a \ne b \ne 1$.  $\det(A)=b-a \ne 0$, so a unique solution.

Equations become: $\left. \begin{array}{rcl} {x+y+z} & = & 3\\
{y+z} & = & 2\\ {(b-a)z} & = & {2-2a}
\end{array} \right\} \begin{array}{l}z=\displaystyle \frac{2-2a}{b-a}\ (b-a \ne 0)\\
y=2-\displaystyle \frac{2-2a}{b-a}=\displaystyle
\frac{2b-2}{b-a}\\x=1
\end{array}$ so a unique solution $\displaystyle
(x,y,z)=(1,\frac{2b-2}{b-a},\frac{2-2a}{b-a})$.

\end{description}



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