Question #295998

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?

1
Expert's answer
2022-02-10T17:38:29-0500

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

Augmented matrix


[12211a33111ab]\begin{bmatrix} 1 & 2 & 2 && 1 \\ 1 & a & 3 && 3 \\ 1 & 11 & a && b \\ \end{bmatrix}

R2=R2R1R_2=R_2-R_1


[12210a212111ab]\begin{bmatrix} 1 & 2 & 2 && 1 \\ 0 & a-2 & 1 && 2 \\ 1 & 11 & a && b \\ \end{bmatrix}

R3=R3R1R_3=R_3-R_1


[12210a21209a2b1]\begin{bmatrix} 1 & 2 & 2 && 1 \\ 0 & a-2 & 1 && 2 \\ 0 & 9 & a -2&& b-1 \\ \end{bmatrix}

If a2=0a-2=0


[12210012090b1]\begin{bmatrix} 1 & 2 & 2 && 1 \\ 0 & 0 & 1 && 2 \\ 0 & 9 & 0&& b-1 \\ \end{bmatrix}

y=b19,z=2,x=12(b1)94y=\dfrac{b-1}{9}, z=2, x=1-\dfrac{2(b-1)}{9}-4

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


(2b+259,b19,2)(-\dfrac{2b+25}{9}, \dfrac{b-1}{9}, 2)

If a2a\not=2

R2=R2a2R_2=\dfrac{R_2}{a-2}


[1221011a22a209a2b1]\begin{bmatrix} 1 & 2 & 2 && 1 \\ \\ 0 & 1 & \dfrac{1}{a-2} && \dfrac{2}{a-2} \\ \\ 0 & 9 & a -2&& b-1 \\ \end{bmatrix}

R1=R12R2R_1=R_1-2R_2


[1022a214a2011a22a209a2b1]\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}

R3=R39R2R_3=R_3-9R_2


[102a6a2a6a2011a22a200(a2)29a2(b1)(a2)18a2]\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 (a2)29a2=0\dfrac{(a-2)^2-9}{a-2}=0


a1=1,a2=5a_1=-1, a_2=5

a=1a=-1


[108/37/3011/32/3000b+5]\begin{bmatrix} 1 & 0 & 8/3 && 7/3 \\ 0 & 1 & -1/3 && -2/3 \\ 0 & 0 & 0 && b+5 \\ \end{bmatrix}

If b5,b\not=-5, the system has no solution.

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


a=5a=5


[104/31/3011/32/3000b7]\begin{bmatrix} 1 & 0 & 4/3 && -1/3 \\ 0 & 1 & 1/3 && 2/3 \\ 0 & 0 & 0 && b-7 \\ \end{bmatrix}

If b7,b\not=7, the system has no solution.

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


If a2,a1,a5a\not=2, a\not=-1, a\not=5

R3=(a2(a2)29)R3R_3=(\dfrac{a-2}{(a-2)^2-9})R_3


[102a6a2a6a2011a22a2001(b1)(a2)18(a2)29]\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}

R1=R1(2a6a2)R3R_1=R_1-(\dfrac{2a-6}{a-2})R_3


[100a26a+332(a3)b(a+1)(a5)011a22a2001(b1)(a2)18(a2)29]\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}

R2=R2R3a2R_2=R_2-\dfrac{R_3}{a-2}


[100a26a+332(a3)b(a+1)(a5)0102a3b(a+1)(a5)001(b1)(a2)18(a+1)(a5)]\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 a1,a5.a\not=-1, a\not=5.


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


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