Tento p��klad lze �e�it samoz�ejm� r�zn�, nap��klad opakovan�m rozvojem
podle sloupce.
Jedna z mo�nost� v�hodn�ch �prav:
\vskip-20pt  $$
    \multline
    \det A=\cr
=[\text{vytkni } b \text{ ze t�et�ho sloupce; sloupcov� �pravy: }
 I.-V. @>>> I., II.-IV.  @>>> II]=\cr
 =
b\cdot \left|  \matrix
     -1    &    0    &   1    &    1     &   a+1     \cr
      0    &   -1    &   1    &   a+1    &     1     \cr
      0    &    0    &   1    &    -1    &     1     \cr
      0    &     1   &   1    &    a     &     1     \cr
       1   &    0    &   1    &      1   &     a     \cr
   \endmatrix\right|
=\cr
=[\text{��dkov� �pravy: } V.+I.- 2\cdot III.@>>> V., IV.+II. @>>> IV.
]=\cr
=b\cdot \left|  \matrix
     -1    &    0    &   1    &    1     &   a+1     \cr
      0    &   -1    &   1    &   a+1    &     1     \cr
      0    &    0    &   1    &    -1    &     1     \cr
      0    &    0    &   2    &   2a+1   &     2     \cr
      0    &    0    &   0    &    4     &    2a-1   \cr
   \endmatrix\right|
=\cr
    =[IV.-2\cdot III.@>>> IV.]=\cr
=b\cdot \left|  \matrix
     -1    &    0    &   1    &    1     &   a+1     \cr
      0    &   -1    &   1    &   a+1    &     1     \cr
      0    &    0    &   1    &    -1    &     1     \cr
      0    &    0    &   0    &   2a+3   &     0     \cr
      0    &    0    &   0    &    4     &    2a-1   \cr
   \endmatrix\right|
=\cr
    =[IV.\leftrightarrow V.]=\cr
=-b\cdot \left|  \matrix
     -1    &    0    &   1    &    1     &   a+1     \cr
      0    &   -1    &   1    &   a+1    &     1     \cr
      0    &    0    &   1    &    -1    &     1     \cr
      0    &    0    &   0    &    4     &    2a-1   \cr
      0    &    0    &   0    &   2a+3   &     0     \cr
   \endmatrix\right|
=\cr
=[V.-\frac{2a+3}{4}\cdot IV.@>>> V.]=\cr
=-b\cdot \left|  \matrix
     -1    &    0    &   1    &    1     &   a+1     \cr
      0    &   -1    &   1    &   a+1    &     1     \cr
      0    &    0    &   1    &    -1    &     1     \cr
      0    &    0    &   0    &    4     &    2a-1   \cr
      0    &    0    &   0    &     0    &   \frac{(2a+3)(2a-1)}{4}     \cr
   \endmatrix\right|
=\cr
=   (2a+3)(2a-1)b.
    \endmultline
$$
