Solution. \begin {align*} \begin {pmatrix} -1&0&2\\1&-2&3\\2&0&4\\ \end {pmatrix} &\hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2+r_1\\ \longrightarrow \\ r_3\rightarrow r_3+2r_1\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&0&2\\0&-2&5\\0&0&8\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_3\rightarrow \frac {1}{8}r_3\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&0&2\\0&-2&5\\0&0&1\\ \end {pmatrix}\\\\ &\hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2-5r_3\\ \longrightarrow \\ r_1\rightarrow r_1-2r_3\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&0&0\\0&-2&0\\0&0&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_1\rightarrow -r_1\\ \longrightarrow \\ r_2\rightarrow -\frac {1}{2}r_2\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0\\0&1&0\\0&0&1\\ \end {pmatrix}\\ \end {align*}

Therefore the normal form of \( \begin {pmatrix} -1&0&2\\1&-2&3\\2&0&4\\ \end {pmatrix} \) is \(I_3\).

Solution. \[ \begin {pmatrix} 0&5&13&6\\-4&2&3&1\\0&0&3&-1\\0&-2&-3&6\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_1\leftrightarrow r_2\\ \longrightarrow \\ r_3\leftrightarrow r_4\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -4&2&3&1\\0&5&13&6\\0&-2&-3&6\\0&0&3&-1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2+2r_3\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -4&2&3&1\\0&1&7&18\\0&-2&-3&6\\0&0&3&-1\\ \end {pmatrix} \] \begin {align*} & \begin {matrix} r_3\rightarrow r_3+2r_2\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -4&2&3&1\\0&1&7&18\\0&0&11&42\\0&0&3&-1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} c_1\rightarrow -\frac {1}{4}c_1\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&2&3&1\\0&1&7&18\\0&0&11&42\\0&0&3&-1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} c_2\rightarrow c_2-2c_1\\c_3\rightarrow c_3-3c_1\\ \longrightarrow \\ c_4\rightarrow c_4-c_1\\ \end {matrix}\\\\ & \begin {pmatrix} 1&0&0&0\\0&1&7&18\\0&0&11&42\\0&0&3&-1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} c_3\rightarrow c_3-7c_2\\ \longrightarrow \\ c_4\rightarrow c_4-18c_2\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0&0\\0&1&0&0\\0&0&11&42\\0&0&3&-1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} c_4\rightarrow c_4-4c_3\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0&0\\0&1&0&0\\0&0&11&-2\\0&0&3&-13\\ \end {pmatrix}\\\\ & \begin {matrix} r_4\rightarrow r_4-3/11r_3\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0&0\\0&1&0&0\\0&0&11&-2\\0&0&0&-137/11\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_4\rightarrow -\frac {11}{137}r_4\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0&0\\0&1&0&0\\0&0&11&-2\\0&0&0&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_3\rightarrow r_3+2r_4\\ \longrightarrow \\ \end {matrix}\\\\ & \begin {pmatrix} 1&0&0&0\\0&1&0&0\\0&0&11&0\\0&0&0&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_3\rightarrow \frac {1}{11}\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\ \end {pmatrix} \hspace {0.3cm} = I_4\\ \end {align*}

Proof. By definition of the normal form, \(A\) can be carried to \(N\) by a finite sequence of elementary row operations followed by a finite sequence of elementary column operations.

By Theorem 2.6.3(1), performing the row operations is the same as multiplying on the left by elementary \(m\) by \(m\) matrices \(E_1,\dots ,E_k\), in the order applied. The corresponding statement for columns is proved in the same way, with the roles of rows and columns exchanged: applying an elementary column operation to \(A\) is multiplication on the right by the \(n\) by \(n\) elementary matrix obtained by applying that operation to \(I_n\). Let \(F_1,\dots ,F_l\) be those matrices. Then \[E_k\cdots E_1\,A\,F_1\cdots F_l=N.\] Put \(P=E_k\cdots E_1\) and \(Q=F_1\cdots F_l\), so that \(PAQ=N\). Each \(E_t\) and each \(F_t\) is invertible by Theorem 2.6.3(2), and a product of invertible matrices is invertible, so \(P\) and \(Q\) are invertible — of sizes \(m\) by \(m\) and \(n\) by \(n\) respectively, as required. □

Solution. \begin {align*} & \begin {pmatrix} -1&0&2&&|&1&0\\1&-2&3&&|&0&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2+r_1\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&0&2&&|&1&0\\0&-2&5&&|&1&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_1\rightarrow -r_1\\ \longrightarrow \\ \end {matrix}\\\\ & \begin {pmatrix} 1&0&-2&&|&-1&0\\0&-2&5&&|&1&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\rightarrow -\frac {1}{2}r_2\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&-2&&|&-1&0\\0&1&-5/2&&|&-1/2&-1/2\\ \end {pmatrix}\\ \end {align*}

Take \(P= \begin {pmatrix} -1&0\\\\-1/2&-1/2\\ \end {pmatrix} \)

\begin {align*} &\text {Now}\hspace {0.3cm} \begin {pmatrix} 1&0&-2&&|&1&0&0\\0&1&-5/2&&|&0&1&0\\&&&&|&0&0&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} c_3\rightarrow c_3+2c_1\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0&&|&1&0&2\\0&1&-5/2&&|&0&1&0\\&&&&|&0&0&1\\ \end {pmatrix}\\\\ & \begin {matrix} c_3\rightarrow c_3+2c_2\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0&&|&1&0&2\\0&1&0&&|&0&1&5/2\\&&&&|&0&0&1\\ \end {pmatrix}\\ \end {align*}

\[\text {Take}\hspace {0.4cm}Q= \begin {pmatrix} 1&0&2\\0&1&5/2\\0&0&1\\ \end {pmatrix} \]

\[\text {Then}\hspace {0.3cm} PAQ= \begin {pmatrix} -1&0\\-1/2&-1/2\\ \end {pmatrix} \begin {pmatrix} -1&0&2\\1&-2&3\\ \end {pmatrix} \begin {pmatrix} 1&0&2\\0&1&5/2\\0&0&1\\ \end {pmatrix} = \begin {pmatrix} 1&0&0\\0&1&0\\ \end {pmatrix} \]