$\square$  Summary: System with six equations, nine variables. Consistent. Null space of coefficient matrix has dimension 5.

$\square$  A system of linear equations (Definition SLE).\begin{align*} x_1 +2x_ 2-2x_3 +9x_4 +3x_5 -5x_6-2x_7 +x_8 +27x_9 &= -5 \\ 2x_1 +4x_2 +3x_3 +4x_4 -x_5 +4x_6 +10x_7 +2x_8 -23x_9 &=18 \\ x_1 +2x_2 +x_3 + 3x_4 +x_5 +x_6 +5x_7 +2x_8 -7x_9 &=6 \\ 2x_1 +4x_2 +3x_3 + 4x_4 -7x_5 +2x_6 +4x_7 -11x_ 9&=20 \\ x_1 +2x_2 + 5x_4 +2x_5 -4x_6 +3x_7 +8x_8 +13x_9 &= -4 \\ -3x_1 -6x_2 -x_3 -13x_4 +2x_5 -5x_6 -4x_7 +13x_ 8+10x_ 9&=-29 \end{align*}

$\square$  Some solutions to the system of linear equations, not necessarily exhaustive (Definition SSLE):

$x_1=6$, $x_2= 0$, $x_3= -1$, $x_4= 0$, $x_5= -1$, $x_6= 2$, $x_7= 0$, $x_8= 0$, $x_9= 0$
$x_1=4$, $x_2=1$, $x_3=-1$, $x_4=0$, $x_5=-1$, $x_6=2$, $x_7=0$, $x_8=0$, $x_9= 0$
$x_1=-17$, $x_2=7$, $x_3=3$, $x_4=2$, $x_5=-1$, $x_6=14$, $x_7=-1$, $x_8=3$, $x_9=2$
$x_1=-11$, $x_2=-6$, $x_3=1$, $x_4=5$, $x_5=-4$, $x_6=7$, $x_7=3$, $x_8=1$, $x_9=1$

$\square$  Augmented matrix of the linear system of equations (Definition AM):\begin{bmatrix} 1 & 2 & -2 & 9 & 3 & -5 & -2 & 1 & 27 & -5 \\ 2 & 4 & 3 & 4 & -1 & 4 & 10 & 2 & -23 & 18 \\ 1 & 2 & 1 & 3 & 1 & 1 & 5 & 2 & -7 & 6 \\ 2 & 4 & 3 & 4 & -7 & 2 & 4 & 0 & -11 & 20 \\ 1 & 2 & 0 & 5 & 2 & -4 & 3 & 8 & 13 & -4 \\ -3 & -6 & -1 & -13 & 2 & -5 & -4 & 13 & 10 & -29 \end{bmatrix}

$\square$  Matrix in reduced row-echelon form, row-equivalent to the augmented matrix. (Definition RREF)\begin{bmatrix} \leading{1} & 2 & 0 & 5 & 0 & 0 & 1 & -2 & 3 & 6 \\ 0 & 0 & \leading{1} & -2 & 0 & 0 & 3 & 5 & -6 & -1 \\ 0 & 0 & 0 & 0 & \leading{1} & 0 & 1 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 & 0 & \leading{1} & 0 & -2 & -3 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

$\square$  Analysis of the augmented matrix (Definition RREF).\begin{align*}r&=4&D&=\set{1,\,3,\,5,\,6}&F&=\set{2,\,4,\,7,\,8,\,9,\,10}\end{align*}

$\square$  Vector form of the solution set to the system of equations (Theorem VFSLS). Notice the relationship between the free variables and the set $F$ above. Also, notice the pattern of 0's and 1's in the entries of the vectors corresponding to elements of the set $F$ in the larger examples.

$\colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\\x_8\\x_9}= \colvector{6\\0\\-1\\0\\-1\\2\\0\\0\\0} + x_2\colvector{-2\\1\\0\\0\\0\\0\\0\\0\\0}+ x_4\colvector{-5\\0\\2\\1\\0\\0\\0\\0\\0}+ x_7\colvector{-1\\0\\-3\\0\\-1\\0\\1\\0\\0}+ x_8\colvector{2\\0\\-5\\0\\-1\\2\\0\\1\\0}+ x_9\colvector{-3\\0\\6\\0\\1\\3\\0\\0\\1}$

$\square$  Given a system of equations we can always build a new, related, homogeneous system (Definition HS) by converting the constant terms to zeros and retaining the coefficients of the variables. Properties of this new system will have precise relationships with various properties of the original system.\begin{align*} x_1 +2x_ 2-2x_3 +9x_4 +3x_5 -5x_6-2x_7 +x_8 +27x_9 &= 0 \\ 2x_1 +4x_2 +3x_3 +4x_4 -x_5 +4x_6 +10x_7 +2x_8 -23x_9 &=0 \\ x_1 +2x_2 +x_3 + 3x_4 +x_5 +x_6 +5x_7 +2x_8 -7x_9 &=0 \\ 2x_1 +4x_2 +3x_3 + 4x_4 -7x_5 +2x_6 +4x_7 -11x_ 9&=0 \\ x_1 +2x_2 + + 5x_4 +2x_5 -4x_6 +3x_7 +8x_8 +13x_9 &= 0\\ -3x_1 -6x_2 -x_3 -13x_4 +2x_5 -5x_6 -4x_7 +13x_ 8+10x_ 9&=0 \end{align*}

$\square$  Some solutions to the associated homogenous system of linear equations, not necessarily exhaustive (Definition SSLE). Review Theorem HSC as you consider these solutions.

$x_1=0$, $x_2=0$, $x_3=0$, $x_4=0$, $x_5=0$, $x_6=0$, $x_7=0$, $x_8=0$, $x_9=0$
$x_1=-2$, $x_2=1$, $x_3=0$, $x_4=0$, $x_5=0$, $x_6=0$, $x_7=0$, $x_8=0$, $x_9=0$
$x_1=-23$, $x_2=7$, $x_3=4$, $x_4=2$, $x_5=0$, $x_6=12$, $x_7=-1$, $x_8=3$, $x_9=2$
$x_1=-17$, $x_2=-6$, $x_3=2$, $x_4=5$, $x_5=-3$, $x_6=5$, $x_7=3$, $x_8=1$, $x_9=1$

$\square$  Form the augmented matrix of the homogenous linear system, and use row operations to convert to reduced row-echelon form. Notice how the entries of the final column remain zeros.\begin{bmatrix} \leading{1} & 2 & 0 & 5 & 0 & 0 & 1 & -2 & 3 & 0 \\ 0 & 0 & \leading{1} & -2 & 0 & 0 & 3 & 5 & -6 & 0 \\ 0 & 0 & 0 & 0 & \leading{1} & 0 & 1 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & \leading{1} & 0 & -2 & -3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

$\square$  Analysis of the augmented matrix for the homogenous system (Definition RREF). Compare this with the same analysis of the original system, especially in the case where the original system is inconsistent (Theorem RCLS).\begin{align*}r&=4&D&=\set{1,\,3,\,5,\,6}&F&=\set{2,\,4,\,7,\,8,\,9,\,10}\end{align*}

$\square$  For any system of equations we can isolate the coefficient matrix, which will be identical to the coefficient matrix of the associated homogenous system. For the remainder of the discussion of this system of equations, we will analyze just the coefficient matrix.\begin{bmatrix} 1 & 2 & -2 & 9 & 3 & -5 & -2 & 1 & 27 \\ 2 & 4 & 3 & 4 & -1 & 4 & 10 & 2 & -23 \\ 1 & 2 & 1 & 3 & 1 & 1 & 5 & 2 & -7 \\ 2 & 4 & 3 & 4 & -7 & 2 & 4 & 0 & -11 \\ 1 & 2 & 0 & 5 & 2 & -4 & 3 & 8 & 13 \\ -3 & -6 & -1 & -13 & 2 & -5 & -4 & 13 & 10 \end{bmatrix}

$\square$  Row-equivalent matrix in reduced row-echelon form (Definition RREF).\begin{bmatrix} \leading{1} & 2 & 0 & 5 & 0 & 0 & 1 & -2 & 3 \\ 0 & 0 & \leading{1} & -2 & 0 & 0 & 3 & 5 & -6 \\ 0 & 0 & 0 & 0 & \leading{1} & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & \leading{1} & 0 & -2 & -3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

$\square$  Analysis of the reduced row-echelon form of the matrix (Definition RREF). For archetypes begin as systems of equations, compare this analysis with the analysis for the coefficient matrices of the original system, and of the associated homogeneous system.\begin{align*}r&=4&D&=\set{1,\,3,\,5,\,6}&F&=\set{2,\,4,\,7,\,8,\,9}\end{align*}

$\square$  Is the matrix nonsingular or singular? (Consider Theorem NMRRI. At the same time, examine the sizes of the sets $D$ and $F$ for the analysis of the reduced row-echelon version of the matrix.)

Since the matrix is not square, the question does not apply.

$\square$  The null space of the matrix. The set of vectors used in the span construction is a linearly independent set of column vectors that spans the null space of the matrix (Theorem SSNS, Theorem BNS). Solve a homogenous system with this matrix as the coefficient matrix and write the solutions in vector form (Theorem VFSLS) to see these vectors arise. Compare the entries of these vectors for indices in $D$ versus entries for indices in $F$.\begin{align*}\spn{\set{ \colvector{-2\\1\\0\\0\\0\\0\\0\\0\\0},\, \colvector{-5\\0\\2\\1\\0\\0\\0\\0\\0},\, \colvector{-1\\0\\-3\\0\\-1\\0\\1\\0\\0},\, \colvector{2\\0\\-5\\0\\-1\\2\\0\\1\\0},\, \colvector{-3\\0\\6\\0\\1\\3\\0\\0\\1} } }\end{align*}

$\square$  The column space of the matrix, expressed as the span of a set of linearly independent vectors that are also columns of the matrix. These columns have indices that form the set $D$ above (Theorem BCS).\begin{align*}\spn{\set{\colvector{1\\2\\1\\2\\1\\-3},\,\colvector{-2\\3\\1\\3\\0\\-1},\,\colvector{3\\-1\\1\\-7\\2\\2},\,\colvector{-5\\4\\1\\2\\-4\\-5}} }\end{align*}

$\square$  The column space of the matrix, as it arises from the extended echelon form of the matrix. The matrix $L$ is computed as described in Definition EEF. This is followed by the column space described as the span of a set of linearly independent vectors that equals the null space of $L$, computed as according to Theorem FS and Theorem BNS. When $r=m$, the matrix $L$ has no rows and the column space is all of $\complex{m}$.\begin{align*}L&=\begin{bmatrix}1&0&\frac{186}{131}&\frac{51}{131}&-\frac{188}{131}&\frac{77}{131}\\0&1&-\frac{272}{131}&-\frac{45}{131}&\frac{58}{131}&-\frac{14}{131}\end{bmatrix}\end{align*}\begin{align*}\spn{\set{ \colvector{-\frac{77}{131}\\\frac{14}{131}\\0\\0\\0\\1},\, \colvector{\frac{188}{131}\\-\frac{58}{131}\\0\\0\\1\\0},\, \colvector{-\frac{51}{131}\\\frac{45}{131}\\0\\1\\0\\0},\, \colvector{-\frac{186}{131}\\\frac{272}{131}\\1\\0\\0\\0} } }\end{align*}

$\square$  The column space of the matrix, expressed as the span of a set of linearly independent vectors. These vectors are computed by bringing the transpose of the matrix into reduced row-echelon form, tossing out the zero rows, and writing the remaining nonzero rows as column vectors. By Theorem CSRST and Theorem BRS, and in the style of Example CSROI, this yields a linearly independent set of vectors that span the column space.\begin{align*}\spn{\set{\colvector{1\\0\\0\\0\\-1\\-\frac{29}{7}},\,\colvector{0\\1\\0\\0\\-\frac{11}{2}\\-\frac{94}{7}},\,\colvector{0\\0\\1\\0\\10\\22},\,\colvector{0\\0\\0\\1\\\frac{3}{2}\\3}} }\end{align*}

$\square$  Row space of the matrix, expressed as a span of a set of linearly independent vectors, obtained from the nonzero rows of the row-equivalent matrix in reduced row-echelon form. (Theorem BRS)\begin{align*}\spn{\set{\colvector{1\\2\\0\\5\\0\\0\\1\\-2\\3},\,\colvector{0\\0\\1\\-2\\0\\0\\3\\5\\-6},\, \colvector{0\\0\\0\\0\\1\\0\\1\\1\\-1},\,\colvector{0\\0\\0\\0\\0\\1\\0\\-2\\-3}} }\end{align*}

$\square$  Subspace dimensions associated with the matrix (Definition ROM, Definition NOM). Verify Theorem RPNC.\begin{align*}\text{Rank: }4&&\text{Nullity: }5&&\text{Matrix columns: }9&\end{align*}

$\square$  Determinant of the matrix. The matrix is nonsingular if and only if the determinant is nonzero (Theorem SMZD).The determinant is not defined for matrices that are not square.