Question

For each of the systems of equations that follow, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in row echelon form. Indicate whether the system is consistent. If the system is consistent and involves no free variables, use back substitution to find the unique solution. If the system is consistent and there are free variables, transform it to reduced row echelon form and find all solutions. b) 2x1 - 3x2 = 5 -4x1 + 6x2 = 8 d) 3x1 + 2x2 - 3x3 = 4 x1 -2x2 +2x3 = 1 11x1 + 2x2 + x3 = 14 f) x1 - x2 + 2x3 = 4 2x1 + 3x2 - x3 = 1 7x1 + 3x2 + 4x3 = 7 j) x1 + 2x2 - 3x3 + x4 = 1 -x1 - x2 + 4x3 - x4 =6 -2x1 - 4x2 + 7x3 - x4 = 1

Answer 1

Answer:

b) inconsistent

d) inconsistent

f) inconsistent

j) X_1 = 2 - 6a , X_2 = 4 + a , X_3 = 3 - a, X_4 = a (free variable)

Step-by-step explanation:

b)

Given: Augmented matrix

$\left[\begin{array}{ccc}2&-3&5\\-4&6&8\end{array}\right]$

Row operation : R_2 + 2*R_1

$\left[\begin{array}{ccc}2&-3&5\\0&0&18\end{array}\right]$

Looking at R_2 : 0 not equal to 18 ; Hence, inconsistent

c)

Given: Augmented matrix

$\left[\begin{array}{cccc}3&2&-3&4\\1&-2&2&1\\11&2&1&14\end{array}\right]$

Row operation : R_2 <---> R_1

$\left[\begin{array}{cccc}1&-2&2&1\\3&2&-3&4\\11&2&1&14\end{array}\right]$

Row operations : R_2 - 3*R_1   &   R_3 - 11*R_1

$\left[\begin{array}{cccc}1&-2&2&1\\0&8&-9&1\\0&24&-21&3\end{array}\right]$

Row operations : R_3 - 3*R_2  

$\left[\begin{array}{cccc}1&-2&2&1\\0&8&-9&1\\0&0&6&0\end{array}\right]$

Looking at R_3 : 0 not equal to 6 ; Hence, inconsistent

f)

Given: Augmented matrix

$\left[\begin{array}{cccc}1&-1&2&4\\2&3&-1&1\\7&3&4&7\end{array}\right]$

Row operations : R_2 - 2*R_1   &   R_3 - 7*R_1

$\left[\begin{array}{cccc}1&-1&2&4\\0&5&-5&-7\\0&10&-10&-21\end{array}\right]$

Row operations : R_2 --> ( 1 / 5 )* R_2  & R_3 --> ( 1 / 10 )* R_3

$\left[\begin{array}{cccc}1&-1&2&4\\0&1&-1&-1.4\\0&1&-1&-2.1\end{array}\right]$

Row operations : R_3 - R_1

$\left[\begin{array}{cccc}1&-1&2&4\\0&1&-1&-1.4\\0&0&0&-0.7\end{array}\right]$

Looking at R_3 : 0 not equal to -0.7 ; Hence, inconsistent

j)

Given: Augmented matrix

$\left[\begin{array}{ccccc}1&2&-3&1&1\\-1&-1&4&-1&6\\-2&-4&7&-1&1\end{array}\right]$

Row operations : R_2 + R_1   &   R_3 + 2*R_1

$\left[\begin{array}{ccccc}1&2&-3&1&1\\0&1&1&0&7\\0&0&1&1&3\end{array}\right]$

corresponding equations are:

X_1 + 2*X_2 - 3*X_3 + X_4 = 1

X_2 + X_3  = 7

X_3 + X_4 = 3

Free parameter : X_4 = a , then:

X_3 = 3 - a

X_2 = 4 + a

X_1 = 1 - a + 3*(3 - a) - 2*(4 + a) = 2 - 6 a