kyle wrote a matrix to represent a system of equations. the matrix is shown below\n\nwhich of the following…

Explanation:

Step1: Find the determinant of the coefficient matrix

The coefficient matrix (A=\begin{bmatrix}1&-3&1\3&2&1\-6&-4&-2\end{bmatrix}) [ \begin{align*} \det(A)&=1\times\begin{vmatrix}2&1\-4&-2\end{vmatrix}-(-3)\times\begin{vmatrix}3&1\-6&-2\end{vmatrix}+1\times\begin{vmatrix}3&2\-6&-4\end{vmatrix}\ &=1\times(2\times(- 2)-1\times(-4))+3\times(3\times(-2)-1\times(-6))+1\times(3\times(-4)-2\times(-6))\ &=1\times(-4 + 4)+3\times(-6 + 6)+1\times(-12+12)\ &=0+0+0\ &=0 \end{align*} ]

Step2: Check the augmented matrix for consistency

The augmented matrix (\left[\begin{array}{ccc|c}1&-3&1&4\3&2&1&3\-6&-4&-2&1\end{array}\right]) Perform row operations: (R_2-3R_1): (\left[\begin{array}{ccc|c}1&-3&1&4\3 - 3\times1&2-3\times(-3)&1-3\times1&3-3\times4\-6&-4&-2&1\end{array}\right]=\left[\begin{array}{ccc|c}1&-3&1&4\0&11&-2&-9\-6&-4&-2&1\end{array}\right]) (R_3 + 6R_1): (\left[\begin{array}{ccc|c}1&-3&1&4\0&11&-2&-9\-6+6\times1&-4+6\times(-3)&-2+6\times1&1+6\times4\end{array}\right]=\left[\begin{array}{ccc|c}1&-3&1&4\0&11&-2&-9\0&-22&4&25\end{array}\right]) (R_3+2R_2): (\left[\begin{array}{ccc|c}1&-3&1&4\0&11&-2&-9\0+2\times0&-22 + 2\times11&4+2\times(-2)&25+2\times(-9)\end{array}\right]=\left[\begin{array}{ccc|c}1&-3&1&4\0&11&-2&-9\0&0&0&7\end{array}\right])

Since we have a row (\left[0\ 0\ 0\big|7\right]), which implies (0x+0y + 0z=7) (a contradiction)

Answer:

inconsistent