Question #196382

Consider the following augmented matrix [ 1 -1 2 1]

[3 -1 5 -2]

[-4 2 x^2-8 x+2]


Determine the values of x for which the system has

I) no solution,

Ii) exactly one solution,

Iii) infinitely many solutions


Expert's answer

Augmented matrix 


A=[1βˆ’1213βˆ’15βˆ’2βˆ’42x2βˆ’8x+2]A=\begin{bmatrix} 1 & -1 & 2 & 1 \\ 3 & -1 & 5 & -2 \\ -4 & 2 & x^2-8 & x+2 \end{bmatrix}



R2=R2βˆ’3R1R_2=R_2-3R_1


[1βˆ’12102βˆ’1βˆ’5βˆ’42x2βˆ’8x+2]\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 2 & -1 & -5 \\ -4 & 2 & x^2-8 & x+2 \end{bmatrix}

R3=R3+4R1R_3=R_3+4R_1


[1βˆ’12102βˆ’1βˆ’50βˆ’2x2x+6]\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 2 & -1 & -5 \\ 0 & -2 & x^2 & x+6 \end{bmatrix}

R2=R2/2R_2=R_2/2


[1βˆ’12101βˆ’1/2βˆ’5/20βˆ’2x2x+6]\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & -1/2 & -5/2 \\ 0 & -2 & x^2 & x+6 \end{bmatrix}

R1=R1+R2R_1=R_1+R_2


[103/2βˆ’3/201βˆ’1/2βˆ’5/20βˆ’2x2x+6]\begin{bmatrix} 1 & 0 & 3/2 & -3/2 \\ 0 & 1 & -1/2 & -5/2 \\ 0 & -2 & x^2 & x+6 \end{bmatrix}

R3=R3+2R2R_3=R_3+2R_2


[103/2βˆ’3/201βˆ’1/2βˆ’5/200x2βˆ’1x+1]\begin{bmatrix} 1 & 0 & 3/2 & -3/2 \\ 0 & 1 & -1/2 & -5/2 \\ 0 & 0 & x^2-1 & x+1 \end{bmatrix}

I) no solution


x2βˆ’1=0x+1=ΜΈ0=>x=1\begin{matrix} x^2-1=0 \\ x+1\not=0 \end{matrix}=>x=1

Ii) exactly one solution


x2βˆ’1=ΜΈ0=>x=ΜΈΒ±1x^2-1\not=0=>x\not=\pm1

Iii) infinitely many solutions


x2βˆ’1=0x+1=0=>x=βˆ’1\begin{matrix} x^2-1=0 \\ x+1=0 \end{matrix}=>x=-1

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