Answer to Question #295998 in Linear Algebra for Sandeep

Question

Consider the following systemof linear equations:

x+2y+2z=1,x+ay+3z=3,x+11y+az=b.

For which values of a does the system have a unique solution. and for which pairs of values (a,b) does the system have more than on solution?

Answer 1

$$x+2y+2z=1$$ $$x+ay+3z=3$$ $$x+11y+az=b$$

Augmented matrix

$$\begin{bmatrix} 1 & 2 & 2 && 1 \\ 1 & a & 3 && 3 \\ 1 & 11 & a && b \\ \end{bmatrix}$$

$R_2=R_2-R_1$

$$\begin{bmatrix} 1 & 2 & 2 && 1 \\ 0 & a-2 & 1 && 2 \\ 1 & 11 & a && b \\ \end{bmatrix}$$

$R_3=R_3-R_1$

$$\begin{bmatrix} 1 & 2 & 2 && 1 \\ 0 & a-2 & 1 && 2 \\ 0 & 9 & a -2&& b-1 \\ \end{bmatrix}$$

If $a-2=0$

$$\begin{bmatrix} 1 & 2 & 2 && 1 \\ 0 & 0 & 1 && 2 \\ 0 & 9 & 0&& b-1 \\ \end{bmatrix}$$

$$y=\dfrac{b-1}{9}, z=2, x=1-\dfrac{2(b-1)}{9}-4$$

If $a=2,$ we have the unique solution

$$(-\dfrac{2b+25}{9}, \dfrac{b-1}{9}, 2)$$

If $a\not=2$

$R_2=\dfrac{R_2}{a-2}$

$$\begin{bmatrix} 1 & 2 & 2 && 1 \\ \\ 0 & 1 & \dfrac{1}{a-2} && \dfrac{2}{a-2} \\ \\ 0 & 9 & a -2&& b-1 \\ \end{bmatrix}$$

$R_1=R_1-2R_2$

$$\begin{bmatrix} 1 & 0 & 2- \dfrac{2}{a-2} && 1-\dfrac{4}{a-2} \\ \\ 0 & 1 & \dfrac{1}{a-2} && \dfrac{2}{a-2} \\ \\ 0 & 9 & a -2&& b-1 \\ \end{bmatrix}$$

$R_3=R_3-9R_2$

$$\begin{bmatrix} 1 & 0 & \dfrac{2a-6}{a-2} && \dfrac{a-6}{a-2} \\ \\ 0 & 1 & \dfrac{1}{a-2} && \dfrac{2}{a-2} \\ \\ 0 & 0 & \dfrac{(a-2)^2-9}{a-2} && \dfrac{(b-1)(a-2)-18}{a-2} \\ \end{bmatrix}$$

If $\dfrac{(a-2)^2-9}{a-2}=0$

$$a_1=-1, a_2=5$$

$a=-1$

$$\begin{bmatrix} 1 & 0 & 8/3 && 7/3 \\ 0 & 1 & -1/3 && -2/3 \\ 0 & 0 & 0 && b+5 \\ \end{bmatrix}$$

If $b\not=-5,$ the system has no solution.

If $b=-5,$ the system has more than one solution.

$a=5$

$$\begin{bmatrix} 1 & 0 & 4/3 && -1/3 \\ 0 & 1 & 1/3 && 2/3 \\ 0 & 0 & 0 && b-7 \\ \end{bmatrix}$$

If $b\not=7,$ the system has no solution.

If $b=7,$ the system has more than one solution.

If $a\not=2, a\not=-1, a\not=5$

$R_3=(\dfrac{a-2}{(a-2)^2-9})R_3$

$$\begin{bmatrix} 1 & 0 & \dfrac{2a-6}{a-2} && \dfrac{a-6}{a-2} \\ \\ 0 & 1 & \dfrac{1}{a-2} && \dfrac{2}{a-2} \\ \\ 0 & 0 & 1 && \dfrac{(b-1)(a-2)-18}{(a-2)^2-9} \\ \end{bmatrix}$$

$R_1=R_1-(\dfrac{2a-6}{a-2})R_3$

$$\begin{bmatrix} 1 & 0 & 0 && \dfrac{a^2-6a+33-2(a-3)b}{(a+1)(a-5)} \\ \\ 0 & 1 & \dfrac{1}{a-2} && \dfrac{2}{a-2} \\ \\ 0 & 0 & 1 && \dfrac{(b-1)(a-2)-18}{(a-2)^2-9} \\ \end{bmatrix}$$

$R_2=R_2-\dfrac{R_3}{a-2}$

$$\begin{bmatrix} 1 & 0 & 0 && \dfrac{a^2-6a+33-2(a-3)b}{(a+1)(a-5)} \\ \\ 0 & 1 & 0 && \dfrac{2a-3-b}{(a+1)(a-5)} \\ \\ 0 & 0 & 1 && \dfrac{(b-1)(a-2)-18}{(a+1)(a-5)} \\ \end{bmatrix}$$

In this case the system has the unique solution.

i) The system has the unique solution for $a\not=-1, a\not=5.$

ii) The system has more than one solution for $(a, b)=(-1, -5)$ or $(a, b)=(5, 7).$