# What are the conditions under which the system of equations: x+ y + z = 1 and x + 2y - z = b and 5x + 7y + a z = b ^ { 2}, (a) have only one solution, (b) no solution, and (c) infinite solution?

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Feb 10, 2018

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$\setminus$

1. $a = 1 \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \Leftrightarrow \setminus \quad \text{unique solution.}$
2. $a = 1 , \setminus b \setminus \ne - 1 , 3 \setminus \quad \Leftrightarrow \setminus \quad \text{no solution.}$
3. $a = 1 , \setminus b = - 1 , 3 \setminus \quad \Leftrightarrow \setminus \quad \text{infinitely-many solutions.}$

#### Explanation:

$\setminus$

$\text{Here's the idea: reduce the system to reduced row-echelon}$
$\text{form; examine the coefficient matrix for conditions}$
$\text{that make it different from the identity matrix,}$
$\text{if any, and connect these differences, if any, with the entries}$
$\text{in the vector of constants} .$

$1. \text{Given system:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad x + y + z \setminus = \setminus 1$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad x + 2 y - z \setminus = \setminus b$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 5 x + 7 y + a z \setminus = \setminus {b}^{2.}$

$2. \text{Convert to matrix/vector format:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus q \quad \setminus \quad \setminus \quad \left(\begin{matrix}1 & 1 & 1 \\ 1 & 2 & - 1 \\ 5 & 7 & a\end{matrix}\right) = \left(\begin{matrix}1 \\ b \\ {b}^{2}\end{matrix}\right)$

$3. \text{Begin to reduce it to reduced row-echelon form:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus q \quad \setminus \quad \setminus \quad \left(\begin{matrix}1 & 1 & 1 \\ 1 & 2 & - 1 \\ 5 & 7 & a\end{matrix}\right) = \left(\begin{matrix}1 \\ b \\ {b}^{2}\end{matrix}\right)$

$\text{Subtract Row 1 from Row 2; Subtract 5 * (Row 1) from Row 2.}$
$\text{Obtain the following:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus q \quad \setminus \quad \setminus \quad \left(\begin{matrix}1 & 1 & 1 \\ 0 & 1 & - 2 \\ 0 & 2 & a - 5\end{matrix}\right) = \left(\begin{matrix}1 \\ b - 1 \\ {b}^{2} - 5\end{matrix}\right)$

$\text{Subtract 2 * (Row 1) from Row3. Obtain the following:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus q \quad \setminus \quad \setminus \quad \left(\begin{matrix}1 & 0 & 3 \\ 0 & 1 & - 2 \\ 0 & 0 & \textcolor{red}{a - 1}\end{matrix}\right) = \left(\begin{matrix}2 - b \\ b - 1 \\ {b}^{2} - 2 b - 3\end{matrix}\right)$

$\text{We do not need to reduce any further -- we can complete our}$
$\text{analysis clearly from the matrix in its current state.}$

$4. \text{Analysis of the reduced matrix:}$

$\setminus \text{The key position now, is the" \ \ color(red){a-1} \ \ "entry in the lower right}$
$\text{corner,}$

$\text{i) Case 1:} \setminus \setminus \textcolor{red}{a - 1 \setminus \ne 0.}$

$\text{If" \ \ color(red){a-1 \ne 0,} \ "then the matrix can be further reduced to}$
$\text{all 1's on the diagonal, and there will be a unique solution.}$

$\text{So we have our first conclusion:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad a = 1 \setminus \quad \Rightarrow \setminus \quad \text{unique solution.}$

$\text{ii) Case 2:} \setminus \setminus \textcolor{red}{a - 1 = 0.}$

$\text{If" \ \ color(red){a-1 \ = 0 }, \ "then the matrix has a row of 0's, and so the}$
$\text{system is either inconsistent [no solutions] or dependent }$
$\text{[infinitely-many solutions]. Which of these two situations}$
$\text{is true, depends on the value of the vector entry opposite the" \ \ color(red){a-1} \ \ "entry in the matrix -- it depends on the value of:}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \textcolor{b l u e}{{b}^{2} - 2 b - 3} .$

$\text{If" \ \ a = 1, \ "and" \ \ color(blue){ b^2 - 2b - 3 \ne 0 } \quad rArr}$
$\text{bottom row of matrix is all 0's & bottom entry of vector is} \setminus \ne 0$
$\setminus q \quad \setminus q \quad \setminus \quad \Rightarrow \setminus \quad \text{system is inconsistent; no solution.}$

$\text{If" \ \ a = 1, \ "and" \ \ color(blue){ b^2 - 2b - 3 = 0 } \quad rArr}$
$\text{bottom row of matrix is all 0's & bottom entry of vector is 0}$
$\setminus q \quad \setminus q \quad \setminus \quad \Rightarrow \setminus \quad \text{system is dependent; infinitely-many solutions.}$

$\text{To finish, we need to solve:} \setminus q \quad {b}^{2} - 2 b - 3 = 0.$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {b}^{2} - 2 b - 3 = 0$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left(b - 3\right) \left(b + 1\right) = 0$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad b = - 1 , 3.$

$\text{So we can state the conclusions of this subsection, compactly:}$

$a = 1 , \setminus b \setminus \ne - 1 , 3 \setminus \quad \Rightarrow \setminus \quad \text{no solution; inconsistent.}$

$a = 1 , \setminus b = - 1 , 3 \setminus \quad \Rightarrow \setminus \quad \text{infinitely-many solutions; dependent.}$

$\text{iii) Summary: As these cases cover all the possible situations}$
$\text{for the matrix and vector, we can summarize our results here:}$

1. $a = 1 \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \Leftrightarrow \setminus \setminus \text{unique solution.}$
2. $a = 1 , \setminus b \setminus \ne - 1 , 3 \setminus \quad \Leftrightarrow \setminus \quad \text{no solution.}$
3. $a = 1 , \setminus b = - 1 , 3 \setminus \quad \Leftrightarrow \setminus \quad \text{infinitely-many solutions.}$

$\setminus$

$\text{Summary of Conclusions:}$

1. $a = 1 \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \Leftrightarrow \setminus \quad \text{unique solution.}$
2. $a = 1 , \setminus b \setminus \ne - 1 , 3 \setminus \quad \Leftrightarrow \setminus \quad \text{no solution.}$
3. $a = 1 , \setminus b = - 1 , 3 \setminus \quad \Leftrightarrow \setminus \quad \text{infinitely-many solutions.}$
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