##### Practice Midterm SOLUTION: #5
abbyto1607
Question five asks for the image and kernel of the matrix $\left( \begin{array}{ccc} 1&2&1 \\ 3&-2&4 \\ 1&-6&2 \end{array} \right)$. Put this into RREF $\left( \begin{array}{ccc} 1&0&5/4 \\ 0&1&-1/8\\ 0&0&0\end{array} \right)$. Since the rank of this matrix is 2, we know the dimension of the image is also 2. Rows without a leading one in the RREF correspond to a redundant vector in the original matrix. Excluding the redundant vector gives the image, where im(A) is spanned by $\left( \begin{array}{ccc} 1\\ 3 \\ 1 \end{array} \right)$ and $\left( \begin{array}{ccc} 2\\ -2 \\ -6\end{array} \right)$. To find the kernel, we need to find the vectors that map to zero when multiplied by the matrix. This amounts to solving for the matrix $\left( \begin{array}{ccc} 1&2&1 \\ 3&-2&4 \\ 1&-6&2 \end{array} \right)$ $\left( \begin{array}{ccc} x_{1}\\ x_{2} \\ x_{3}\end{array} \right)$ = $\left( \begin{array}{ccc} 0\\ 0 \\ 0 \end{array} \right)$ To simplify this, use the RREF found earlier. $\left( \begin{array}{ccc} 1&0&5/4 \\ 0&1&-1/8\\ 0&0&0\end{array} \right)$ Thus, $\left( \begin{array}{ccc} x_{1}\\ x_{2} \\ x_{3}\end{array} \right)$ = $\left( \begin{array}{ccc} -5/4x_{3}\\ 1/8x_{3} \\ 0\end{array} \right)$ where $x_{3}$ is arbitrary. ker(A) is then spanned by$\left( \begin{array}{ccc} -5/4\\ 1/8 \\ 0\end{array} \right)$
henry_lee: May 5, 2015, 6:37 p.m.
If x3 is arbitrary, then isn't x3 set to t, and therefore the ker(A) is spanned by (-5/4, 1/8, 1)?
DonnaBranchevsky: May 5, 2015, 7:49 p.m.
I also got that it should be spanned by (-5/4, 1/8, 1)
emmyralds: May 5, 2015, 9:58 p.m.
Agreed with the ones above, should be (-5/4, 1/8, 1) because x3 = x3, ergo the coefficient of x3 is 1.
abbyto1607: May 5, 2015, 10:29 p.m.
Yep, you guys are right. My $x_{3}$ is $"t"$. The zero in the last row should be a 1. That was a mistake copied from an earlier draft. Thanks!