Answer
$det(A)=24$
$\begin{array} \\A^{-1}=\\ \end{array}
\begin{bmatrix} 1&0&0&-\frac{1}{4}\\0&\frac{1}{2}&0&-\frac{1}{4}\\0&0&\frac{1}{3}&-\frac{1}{4} \\0&0&0&\frac{1}{4}\end{bmatrix}$
Work Step by Step
Step 1. List the matrix needed:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&0&0&1\\ 0&2&0&2\\0&0&3&3 \\0&0&0&4\end{bmatrix}$
Step 2. Calculate the determinant using column-1 expansion twice (use the formula from section 10.6):
$|A|=2(3\times4-0)=24$
Step 3. Derive $A^{-1}$ (use the procedures from section 10.5):
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 1&0&0&1 &|&1&0&0&0\\ 0&2&0&2 &|&0&1&0&0\\0&0&3&3 &|&0&0&1&0 \\0&0&0&4 &|&0&0&0&1\end{bmatrix}
\begin{array} \\ \\R_2/2\to R_2\\R_3/3\to R_3 \\R_4/4\to R_4\end{array}$
Step 4. Peform the row operations shown on the right side of the matrix:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 1&0&0&1 &|&1&0&0&0\\ 0&1&0&1 &|&0&1/2&0&0\\0&0&1&1 &|&0&0&1/3&0 \\0&0&0&1 &|&0&0&0&1/4\end{bmatrix}
\begin{array} \\ R_1-R_4\to R_1 \\R_2-R_4\to R_2\\R_3-R_4\to R_3 \\ \\ \end{array}$
Step 5. Peform the row operations shown on the right side of the matrix:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 1&0&0&0 &|&1&0&0&-\frac{1}{4}\\ 0&1&0&0 &|&0&\frac{1}{2}&0&-\frac{1}{4}\\0&0&1&0 &|&0&0&\frac{1}{3}&-\frac{1}{4} \\0&0&0&1 &|&0&0&0&\frac{1}{4}\end{bmatrix}$
Step 6. Write the result as:
$\begin{array} \\A^{-1}=\\ \end{array}
\begin{bmatrix} 1&0&0&-\frac{1}{4}\\0&\frac{1}{2}&0&-\frac{1}{4}\\0&0&\frac{1}{3}&-\frac{1}{4} \\0&0&0&\frac{1}{4}\end{bmatrix}$
