## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.2 - Inconsistent and Dependent Systems and Their Applications - Exercise Set - Page 904: 45

#### Answer

The given system is inconsistent for $a=1\text{ or }a=3$.

#### Work Step by Step

The augmented matrix for the given system will be, $\left[ \begin{matrix} 1 & 3 & 1 \\ 2 & 5 & 2a \\ 1 & 1 & {{a}^{2}} \\ \end{matrix}\left| \begin{matrix} {{a}^{2}} \\ 0 \\ -9 \\ \end{matrix} \right. \right]$ Now the necessary condition for inconsistency is that the determinant should be zero, So, $\left| \begin{matrix} 1 & 3 & 1 \\ 2 & 5 & 2a \\ 1 & 1 & {{a}^{2}} \\ \end{matrix} \right|=0$ Solve the determinant: Expanding along row 1, \begin{align} & 1\left( 5{{a}^{2}}-2a \right)-3\left( 2{{a}^{2}}-2a \right)+1\left( 2-5 \right)=0 \\ & 5{{a}^{2}}-2a-6{{a}^{2}}+6a-3=0 \\ & -{{a}^{2}}+4a-3=0 \\ & {{a}^{2}}-4a+3=0 \end{align} Solve the above quadratic equation as below: \begin{align} & {{a}^{2}}-3a-a+3=0 \\ & a\left( a-3 \right)-\left( a-3 \right)=0 \\ & \left( a-3 \right)\left( a-1 \right)=0 \\ & a=1,3 \end{align} So, the equation may be inconsistent for $a=1\text{ or }a=3$ When, $a=1$ \begin{align} & x+3y+z=1 \\ & 2x+5y+2z=0 \\ & x+y+z=-9 \\ & \text{Augmented matrix,} \\ \end{align} $\left[ \begin{matrix} 1 & 3 & 1 \\ 2 & 5 & 2 \\ 1 & 1 & 1 \\ \end{matrix}\left| \begin{matrix} 1 \\ 0 \\ -9 \\ \end{matrix} \right. \right]$ Using Gaussian elimination, $\text{Replace row 2 by }{{\text{R}}_{2}}-2{{R}_{1}},$ $\left[ \begin{matrix} 1 & 3 & 1 \\ 0 & -1 & 0 \\ 1 & 1 & 1 \\ \end{matrix}\left| \begin{matrix} 1 \\ -2 \\ -9 \\ \end{matrix} \right. \right]$ \begin{align} & \text{Replace row 2 by -1}{{\text{R}}_{2}},\text{ row 3 by }{{\text{R}}_{3}}-{{R}_{1}} \\ & \left[ \begin{matrix} 1 & 3 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \\ \end{matrix}\left| \begin{matrix} 1 \\ 2 \\ -10 \\ \end{matrix} \right. \right] \\ \end{align} \begin{align} & \text{Replace row 3 by }{{\text{R}}_{3}}-{{R}_{2}}, \\ & \left[ \begin{matrix} 1 & 3 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{matrix}\left| \begin{matrix} 1 \\ 2 \\ -12 \\ \end{matrix} \right. \right] \\ \end{align} The equation will be \begin{align} & x+3y+z=1 \\ & y=2 \\ & 0x+0y+0z=-12 \end{align} So, the last equation can never be a true statement; hence, the system has no solution or the system is inconsistent. When $a=3$ \begin{align} & x+3y+z=9 \\ & 2x+5y+6z=0 \\ & x+y+9z=-9 \end{align} Augmented matrix, $\left[ \begin{matrix} 1 & 3 & 1 \\ 2 & 5 & 6 \\ 1 & 1 & 9 \\ \end{matrix}\left| \begin{matrix} 9 \\ 0 \\ -9 \\ \end{matrix} \right. \right]$ Using Gaussian elimination, \begin{align} & \text{Replace row 2 by }{{\text{R}}_{2}}-2{{R}_{1}}, \\ & \left[ \begin{matrix} 1 & 3 & 1 \\ 0 & -1 & 4 \\ 1 & 1 & 9 \\ \end{matrix}\left| \begin{matrix} 9 \\ -18 \\ -9 \\ \end{matrix} \right. \right] \\ \end{align} \begin{align} & \text{Replace row 2 by -1}{{\text{R}}_{2}},\text{ row 3 by }{{\text{R}}_{3}}-{{R}_{1}} \\ & \left[ \begin{matrix} 1 & 3 & 1 \\ 0 & 1 & -4 \\ 0 & -2 & 8 \\ \end{matrix}\left| \begin{matrix} 9 \\ 18 \\ -18 \\ \end{matrix} \right. \right] \\ \end{align} \begin{align} & \text{Replace row 3 by }{{\text{R}}_{3}}+2{{R}_{2}}, \\ & \left[ \begin{matrix} 1 & 3 & 1 \\ 0 & 1 & -4 \\ 0 & 0 & 0 \\ \end{matrix}\left| \begin{matrix} 9 \\ 18 \\ 18 \\ \end{matrix} \right. \right] \\ \end{align} Therefore, the equation will be, \begin{align} & x+3y+z=-9 \\ & y-4z=18 \\ & 0x+0y+0z=18 \\ \end{align} So, the last equation can never be a true statement; hence, the system has no solution or the system is inconsistent. So, the system is inconsistent for $a=1\text{ or }a=3$

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