# Find the Rank of a Matrix with a Parameter ## Problem 103

Find the rank of the following real matrix.
$\begin{bmatrix} a & 1 & 2 \\ 1 &1 &1 \\ -1 & 1 & 1-a \end{bmatrix},$ where $a$ is a real number.

(Kyoto University, Linear Algebra Exam) Add to solve later

## Solution.

The rank is the number of nonzero rows of a (reduced) row echelon form matrix of the given matrix.

We apply elementary row operations as follows.
\begin{align*}
&\begin{bmatrix}
a & 1 & 2 \\
1 &1 &1 \\
-1 & 1 & 1-a
\end{bmatrix}
\xrightarrow{R_1 \leftrightarrow R_2}
\begin{bmatrix}
1 &1 &1 \\
a & 1 & 2 \\
-1 & 1 & 1-a
\end{bmatrix}
\xrightarrow[R_3+R_1]{R_2-aR_1}
\begin{bmatrix}
1 &1 &1 \\
0 & 1-a & 2-a \\
0 & 2 & 2-a
\end{bmatrix}\\[8pt] &\xrightarrow{R_2\leftrightarrow R_3}
\begin{bmatrix}
1 &1 &1 \\
0 & 2 & 2-a\\
0 & 1-a & 2-a
\end{bmatrix}
\xrightarrow{R_3-\frac{1-a}{2}R_2}
\begin{bmatrix}
1 &1 &1 \\
0 & 2 & 2-a\\
0 & 0 & (2-a) (a+1)/2
\end{bmatrix}.
\end{align*}

The last matrix is in row echelon form.
Therefore, if $a \neq -1, 2$, then $(3, 3)$-entry of the last matrix is not zero. From this we see that the rank is $3$ when $a \neq -1, 2$.

On the other hand, when $a=-1$ or $a=2$ the third row is a zero row, hence the rank is $2$. Add to solve later

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