Prove Matrix $A$ is Invertible Prove that $A$ is invertible if $a\not=0$ $a \not=b$
$A = \begin{bmatrix} a & b & b \\ a & a & b \\a & a & a \end{bmatrix}$
 A: Compute the determinant.
To do this quickly, subtract row $2$ from row $3$ and develop from row $3$. (Otherwise use the usual Sarrus rule.)
This gives $\Delta=(a-b)(a^2-ab)$.
A: More generally a matrix $M=(M_{i,j})_{i,j=1}^n$, where $M_{i,j}=b$ if $i<j$ and $M_{i,j}=a$ otherwise, is invertible if and only if $a\neq0$ and $a\neq b$. To see this multiply on the left by a lower triangular matrix
$$
  L=\begin{pmatrix}1&0&0&\ldots&0\\ -1&1&0&\ddots&\vdots\\ 0 &-1&1&\ddots&0 \\
 \vdots & \ddots &-1&1& 0\\ 0 &\ldots &0 & -1 & 1
\end{pmatrix}
\qquad \text{(that is $L_{i,j}=\delta_{i,j}-\delta_{i,j+1}$),}
$$
which is clearly invertible, to obtain
$$
  LM=\begin{pmatrix}a&b&b&\ldots&b\\ 0&a-b&0&\ddots&0\\ 0 &0&a-b&\ddots&0 \\
 \vdots & \ddots &0&a-b& 0\\ 0 &\ldots &0 & 0 & a-b
\end{pmatrix},
$$
which being upper triangular is invertible if and only if all its diagonal entries are nonzero.
A: Denote such a $n\times n$-Matrix by $A_{a,b,n}$.
If we subtract row $1$ from all rows $2,3 \dotsc, n$ we end up at $$\det A_{a,b,n} = \det \begin{pmatrix}a & b\dotsc  b \\ 0 & A_{a-b,0,n-1}\end{pmatrix}$$
Since $a-b \neq 0$, this inductively shows the assertion for all $n$, not only the $3 \times 3$ case.
