Linear Algebra

Find a Row-Equivalent Matrix which is in Reduced Row Echelon Form and Determine the Rank

Problem 643

For each of the following matrices, find a row-equivalent matrix which is in reduced row echelon form. Then determine the rank of each matrix.

(a) $A = \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix}$.

(b) $B = \begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix}$.

(c) $C = \begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix}$.

(d) $D = \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}$.

(e) $E = \begin{bmatrix} -2 & 3 & 1 \end{bmatrix}$.

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Definition (Rank of a Matrix).

The rank of a matrix $A$ is the number of nonzero rows in the reduced row echelon form matrix $\rref(A)$ that is row equivalent to $A$.

Solution.

(a) $A = \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix}$

The matrix $A$ has rank 2, which can be seen by computing
\[ \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix} \xrightarrow{R_2 + 2 R_1} \begin{bmatrix} 1 & 3 \\ 0 & 8 \end{bmatrix} \xrightarrow{\frac{1}{8} R_2 } \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \xrightarrow{R_1 – 3 R_2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \] Because the row-reduced matrix has two non-zero rows, the rank of $A$ is $2$.

(b) $B = \begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix}$

The matrix $B$ has rank 2, which can be seen by computing
\begin{align*}
\begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix} \xrightarrow{\frac{1}{2} R_1 } \begin{bmatrix} 1 & 3 & -1 \\ 3 & -2 & 8 \end{bmatrix} \xrightarrow{R_2 – 3 R_1} \begin{bmatrix} 1 & 3 & -1 \\ 0 & -11 & 11 \end{bmatrix} \\[6pt] \xrightarrow{\frac{-1}{11} R_2 } \begin{bmatrix} 1 & 3 & -1 \\ 0 & 1 & -1 \end{bmatrix} \xrightarrow{R_1 – 3 R_2} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \end{bmatrix}
\end{align*}

(c) $C = \begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix}$

The matrix $C$ has rank 2, which can be seen by computing
\begin{align*}
\begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix} \xrightarrow{\frac{1}{2} R_1 } \begin{bmatrix} 1 & -1 & 2 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix} \xrightarrow[R_3 – 6 R_1]{R_2 – 4 R_1} \begin{bmatrix} 1 & -1 & 2 \\ 0 & 5 & -10 \\ 0 & 5 & -10 \end{bmatrix} \\[6pt] \xrightarrow{R_3 – R_2} \begin{bmatrix} 1 & -1 & 2 \\ 0 & 5 & -10 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{ \frac{1}{5} R_2 } \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{R_1 + R_2} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix} .
\end{align*}

(d) $D = \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}$

The matrix $D$ has rank 1, which can be seen by calculating:
\[\begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix} \xrightarrow{ \frac{-1}{2} R_1 } \begin{bmatrix} 1 \\ 3 \\1 \end{bmatrix} \xrightarrow{R_2 – 3 R_1 , R_3 – R_1} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}.\]

(e) $E = \begin{bmatrix} -2 & 3 & 1 \end{bmatrix}$

The matrix $E$ has rank 1, which can be seen by calculating:
\[\begin{bmatrix} -2 & 3 & 1 \end{bmatrix} \xrightarrow{ \frac{-1}{2} R_1 } \begin{bmatrix} 1 & \frac{-3}{2} & \frac{-1}{2} \end{bmatrix}.\]

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