If $k$ cannot equal $0$ and $A$ is as given below, what is $A$ inverse? $$
A = 
\pmatrix{
1 & 0 & 2k\\
0 & 1 & k\\
0 & 0 & k
}
$$
I don't know what method to use to solve this problem as I haven't encountered and variable before when solving for the inverse of a matrix. I've tried to use the identity method to solve it. It didn't seem right. I also, tried solving for each row and setting them equal to 0, this also didn't seem right. 
 A: Using the row-reduction method, we try to bring the augmented block matrix $(A|I)$ to the form  $(I|A^{-1})$. In this case, that is fairly simple:
$$
\left(\begin{array}{ccc|ccc}1&0&2k&1&0&0\\0&1&k&0&1&0\\0&0&k&0&0&1\end{array}\right) \to \left(\begin{array}{ccc|ccc}1&0&0&1&0&-2\\0&1&0&0&1&-1\\0&0&1&0&0&\tfrac{1}{k}\end{array}\right)
$$
So that your inverse should be 
$$
A^{-1} = \left(\begin{array}{ccc}1&0&-2\\0&1&-1\\0&0&\tfrac{1}{k}\end{array}\right)
$$
A: Hint: We can write the matrix as the product of elementary matrices:
$$
\begin{bmatrix}
1 & 0 & 2k \\
0 & 1 & k \\
0 & 0 & k \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & k \\
\end{bmatrix}.
$$
Perhaps you'll find it easier to find $$
\left(\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & k \\
\end{bmatrix}\right)^{-1}=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & k \\
\end{bmatrix}^{-1}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix}^{-1}
\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}^{-1}.
$$
Their inverses are described on the linked wikipedia page.
A: Apply Gauss Method:
$\left[\begin{matrix}
  1 & 0 & 2k & 1 & 0 & 0\\
  0 & 1 & k  & 0 & 1 & 0\\
  0 & 0 & k  & 0 & 0 & 1
 \end{matrix}\right]$
$\left[\begin{matrix}
  1 & 0 & 0  & 1 & 0 & -2\\
  0 & 1 & 0  & 0 & 1 & -1\\
  0 & 0 & k  & 0 & 0 & 1
 \end{matrix}\right]$
$\left[\begin{matrix}
  1 & 0 & 2k & 1 & 0 & -2\\
  0 & 1 & k  & 0 & 1 & -1\\
  0 & 0 & 1  & 0 & 0 & \frac 1k
 \end{matrix}\right]$
so the inverse is
$$\left[\begin{matrix}
 1 & 0 & -2\\
 0 & 1 & -1\\
 0 & 0 & \frac 1k
 \end{matrix}\right]$$
A: The first two columns of $A$ are already the columns of an identity matrix, so the first two columns of $A^{-1}$ will be these as well. To finish finding $A^{-1}$ you need to find the last column, which can be done by solving $Ax=e_3$ for $x$, where $e_3$ is the third column of the identity matrix. Once you have constructed this, verify that $AA^{-1}$ gives you the identity by direct computation.
A: Try solving $Ax_k = e_k$ for $x_k$, where $e_1 = (1,0,0)^T$, etc. Then form the matrix $\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}$.
For example, we see that $x_3 = (-2,-1,{1 \over k})^T$.
A: Using the formula $A^{-1}=$$1\over detA)$$(A^t)^*$
$det (A)=k$
$A^t=\matrix {1 &0& k \\ 0&1&0\\2 k&k &k}$
Now we calculate the adjunted matrix $|A_{11}|=|\matrix{1&0\\ k&k}|=k$ ,...,$|A_{33}|=|\matrix{1&0\\ 0&1}|=1$ 
So, $(A^t)^*= \matrix{k&0&-2k\\ 0&k&-k\\0&0&1}$
Finally, $A^{-1}= \matrix{1&0&-2\\ 0&1&-1\\0&0&1/k}$
