Answer to Question #203636 in Linear Algebra for khosa

Question

Consider the following augmented matrix [12]   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.

Answer 1

Question

consider the following augmented matrix

$\begin{bmatrix} \def\arraystretch{1.5} \begin{array}{ ccc:} 1&-1&2&1 \\ 3&-1&5&-2 \\ -4&2&x^2-8&x+2\\ \end{array} \end{bmatrix}$

Determine the values of x for which the system has

(i)no solution

(ii)exactly one solution

(iii)infinitely many solutions.

solution

Here

$[A|B]=\begin{bmatrix} \def\arraystretch{1.5} \begin{array}{ ccc:} 1&-1&2&1 \\ 3&-1&5&-2 \\ -4&2&x^2-8&x+2\\ \end{array} \end{bmatrix}\\\&\space A=\begin{bmatrix} \def\arraystretch{1.5} \begin{array}{ ccc} 1&-1&2& \\ 3&-1&5& \\ -4&2&x^2-8&\\ \end{array} \end{bmatrix}$

$\begin{bmatrix} \def\arraystretch{1.5} \begin{array}{ ccc:} 1&-1&2&1 \\ 3&-1&5&-2 \\ -4&2&x^2-8&x+2\\ \end{array} \end{bmatrix}$

step 1:

multiply 3 to row 1 and subtract from row 2

$\begin{bmatrix} \def\arraystretch{1.5} \begin{array}{ ccc:} 1&-1&2&1 \\ 0&2&-1&-5 \\ -4&2&x^2-8&x+2\\ \end{array} \end{bmatrix}$

step2:

multiply 4 to row 1 and add to row 3

$\begin{bmatrix} \def\arraystretch{1.5} \begin{array}{ ccc:} 1&-1&2&1 \\ 0&2&-1&-5 \\ 0&-2&x^2&x+6\\ \end{array} \end{bmatrix}$

step3:

row 2 add to row 3

$\begin{bmatrix} \def\arraystretch{1.5} \begin{array}{ ccc:} 1&-1&2&1 \\ 0&2&-1&-5 \\ 0&0&x^2-1&x+1\\ \end{array} \end{bmatrix}$

now we want system has

(i)no solution

"If rank(A) < rank(A|B), then the system is no solution"

for no solution

$rank(A)=2 \implies x^2-1=0\implies x=+1 \space ,-1\\ rank(A|B)=3\implies x+1\neq0\implies x\neq-1\\$

hence both condition give value of x=+1

(ii)exactly one solution

"If rank(A) = rank(A|B) = the number of rows in x, then the system has exactly one solution. "

for exactly one solution

$rank(A) =number\space of \space rows=3\implies x^2-1\neq0\implies x\neq+1,-1\\ \\ rank(A|B) =number\space of \space rows=3\implies x+1\neq0 \implies x\neq-1$

hence both condition give value of $x\isin R ; x\neq +1,-1$

(iii)infinitely many solutions.

If rank(A) = rank(A|B) < the number of rows in x, then the system has infinitely -many solutions.

for infinitely -many solutions.

$rank(A) = rank(A|B) < \space number \space of \space rows \space hence\\ rank(A) =2\implies x^2-1=0 \implies x=+1,-1\\ rank(A|B)=2\implies x+1=0 \implies x=-1$

hence both condition give value of x=-1