Answer
No inverse.
Work Step by Step
Step 1. Write the original matrix together with an identity matrix.
$ \begin{array}( \\B=A|I= \\ \\ \end{array} \begin{bmatrix} 1 & 2 & 3 & | &1 & 0 & 0\\4 & 5 & -1 & | &0 & 1 & 0\\1 & -1 & -10 & | &0 & 0 & 1 \end{bmatrix} \begin{array}( \\4R_1-R_2\to R_2 \\R_1-R_3\to R_3 \\ \end{array}$
Step 2. Use row operations on the right side of the matrix to transform the left side into reduced row-echelon form.
$ \begin{array}( \\B= \\ \\ \end{array} \begin{bmatrix} 1 & 2 & 3 & | &1 & 0 & 0\\0 & 3 & 13 & | &4 & -1 & 0\\0 & 3 & 13 & | &1 & 0 & -1 \end{bmatrix} \begin{array}( \\ \\R_2-R_3\to R_3 \\ \end{array}$
Step 3. Perform the operations given on the right side of the matrix.
$ \begin{array}( \\B= \\ \\ \end{array} \begin{bmatrix} 1 & 2 & 3 & | &1 & 0 & 0\\0 & 3 & 13 & | &4 & -1 & 0\\0 & 0 & 0 & | &1 & -1 & 1 \end{bmatrix} \begin{array}( \\ \\ \\ \end{array}$
Step 4. The last row of the left half matrix are all zeros. As the left half of the matrix can not be transformed into a reduced row-echelon form, we conclude that the original matrix does not have an inverse.
