Answer
$det(A)=0$, $det(B)=2$
$\begin{array} \\B^{-1}= \\ \end{array}\begin{bmatrix} 1&-2&0\\0&1/2&0\\3&-6&1\end{bmatrix}$
Work Step by Step
Step 1. Evaluate $det(A)$ using row2 expansion:
$\begin{array} \\det(A)= \\ \end{array}\begin{vmatrix} 1&4&1\\0&2&0\\1&0&1\end{vmatrix}
=2(1-1)=0$
Thus matrix A does not have an inverse.
Step 2. Evaluate $det(B)$ using row2 expansion:
$\begin{array} \\det(B)= \\ \end{array}\begin{vmatrix} 1&4&0\\0&2&0\\-3&0&1\end{vmatrix}
=2(1-0)=2$
Thus, matrix B should have an inverse.
Step 3. Derive $B^{-1}$
$\begin{array} \\BI= \\ \end{array}\begin{bmatrix} 1&4&0&|&1&0&0\\0&2&0&|&0&1&0\\-3&0&1&|&0&0&1\end{bmatrix}\begin{array} \\ \\ \\3R_1+R_3\to R_3 \end{array}$
Step 4. Perform the row operations shown on the right side of the matrix:
$\begin{array} \\BI= \\ \end{array}\begin{bmatrix} 1&4&0&|&1&0&0\\0&2&0&|&0&1&0\\0&12&1&|&3&0&1\end{bmatrix}\begin{array} \\ \\ \\R_3-6R_2\to R_3 \end{array}$
Step 5. Perform the row operations shown on the right side of the matrix:
$\begin{array} \\BI= \\ \end{array}\begin{bmatrix} 1&4&0&|&1&0&0\\0&2&0&|&0&1&0\\0&0&1&|&3&-6&1\end{bmatrix}\begin{array} ( R_1-2R_2\to R_1\\R_2/2\to R_2 \\ \\ \end{array}$
Step 6. Perform the row operations shown on the right side of the matrix:
$\begin{array} \\BI= \\ \end{array}\begin{bmatrix} 1&0&0&|&1&-2&0\\0&1&0&|&0&1/2&0\\0&0&1&|&3&-6&1\end{bmatrix}$
Step 7. Write the inverse:
$\begin{array} \\B^{-1}= \\ \end{array}\begin{bmatrix} 1&-2&0\\0&1/2&0\\3&-6&1\end{bmatrix}$
