Answer
No. see explanations.
Work Step by Step
Step 1. Write the new matrix together with an identify as:
$\begin{array}( \\A|I= \\ \\ \end{array}
\begin{bmatrix} 3 & 1&4 &| &1 &0&0\\4 & 2&6 &| &0 &1&0\\3 & 2&5 &| &0 &0&1 \end{bmatrix} \begin{array}( 4R_1\\3R_2 \\4R_3 \\\end{array}$
Step 2. Use row operations to transform the left half into a reduced row-echelon:
$\begin{bmatrix} 12 & 4&16 &| &4 &0&0\\12& 6&18 &| &0 &3&0\\12 &8&20 &| &0 &0&4 \end{bmatrix} \begin{array}( \\R_2-R_1 \\R_3-R_2 \\\end{array}$
$\begin{bmatrix} 12 & 4&16 &| &4 &0&0\\0& 2&2 &| &-4 &3&0\\0 &2&2 &| &0 &-3&4 \end{bmatrix} $
Since the last two rows of the left half matrix are the same, it is not possible to find an inverse of the orginal matrix.
Step 3. Without the inverse matrix, it will not be possible to use matrix inversion to solve parts (b), (c), and (d) of Exercise 61.
