$\left|A_1\right|=3$ and $\left|A_2\right|=\left|A\right|=3$ . Neither of these is zero, so $A$ does have an $LU$ decomposition.

Answer

Let us swap the second and third rows and consider

$B=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 7\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill -2\hfill & \hfill 6\hfill & \hfill 1\hfill \end{array}\right]$

the leading submatrices are

$B_1=1,B_2=\left[\begin{array}{cc}\hfill 1\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 3\hfill \end{array}\right],B_3=B$

which have determinants $1$ , $3$ and $45$ respectively. All of these are non-zero and we conclude that $B$ does indeed have an $LU$ decomposition.

1. 1. We let

      $\left[\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill -4& \hfill -6\end{array}\right]=LU=\left[\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill L_{21}& \hfill 1\end{array}\right]\left[\begin{array}{cc}\hfill U_{11}& \hfill U_{12}\\ \hfill 0& \hfill U_{22}\end{array}\right]=\left[\begin{array}{cc}\hfill U_{11}& \hfill U_{12}\\ \hfill L_{21}{U}_{11}& \hfill L_{21}{U}_{12}+U_{22}\end{array}\right].$

      Comparing the left-hand and right-hand sides row by row gives us that $U_{11}=2$ , $U_{12}=1$ , $L_{21}{U}_{11}=-4$ which implies that $L_{21}=-2$ and, finally, $L_{21}{U}_{12}+U_{22}=-6$ from which we see that $U_{22}=-4$ . Hence

      $\left[\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill -4& \hfill -6\end{array}\right]=\left[\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill -2& \hfill 1\end{array}\right]\left[\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill 0& \hfill -4\end{array}\right]$

      is an $LU$ decomposition of the given matrix.

   2. We let

      $\left[\begin{array}{ccc}\hfill 2& \hfill 1& \hfill -4\\ \hfill 2& \hfill 2& \hfill -2\\ \hfill 6& \hfill 3& \hfill -11\end{array}\right]=LU=\left[\begin{array}{ccc}\hfill U_{11}& \hfill U_{12}& \hfill U_{13}\\ \hfill L_{21}{U}_{11}& \hfill L_{21}{U}_{12}+U_{22}& \hfill L_{21}{U}_{13}+U_{23}\\ \hfill L_{31}{U}_{11}& \hfill L_{31}{U}_{12}+L_{32}{U}_{22}& \hfill L_{31}{U}_{13}+L_{32}{U}_{23}+U_{33}\end{array}\right].$

      Looking at the top row we see that $U_{11}=2$ , $U_{12}=1$ and $U_{13}=-4$ . Now, from the second row, $L_{21}=1$ , $U_{22}=1$ and $U_{23}=2$ . The last three unknowns come from the bottom row: $L_{31}=3$ , $L_{32}=0$ and $U_{33}=1$ . Hence

      $\left[\begin{array}{ccc}\hfill 2& \hfill 1& \hfill -4\\ \hfill 2& \hfill 2& \hfill -2\\ \hfill 6& \hfill 3& \hfill -11\\ \hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill 1& \hfill 1& \hfill 0\\ \hfill 3& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 2& \hfill 1& \hfill -4\\ \hfill 0& \hfill 1& \hfill 2\\ \hfill 0& \hfill 0& \hfill 1\\ \hfill \end{array}\right]$

      is an $LU$ decomposition of the given matrix.

   3. We let

      $\left[\begin{array}{ccc}\hfill 1& \hfill 3& \hfill 2\\ \hfill 2& \hfill 8& \hfill 5\\ \hfill 1& \hfill 11& \hfill 4\end{array}\right]=LU=\left[\begin{array}{ccc}\hfill U_{11}& \hfill U_{12}& \hfill U_{13}\\ \hfill L_{21}{U}_{11}& \hfill L_{21}{U}_{12}+U_{22}& \hfill L_{21}{U}_{13}+U_{23}\\ \hfill L_{31}{U}_{11}& \hfill L_{31}{U}_{12}+L_{32}{U}_{22}& \hfill L_{31}{U}_{13}+L_{32}{U}_{23}+U_{33}\end{array}\right].$

      Looking at the top row we see that $U_{11}=1$ , $U_{12}=3$ and $U_{13}=2$ . Now, from the second row, $L_{21}=2$ , $U_{22}=2$ and $U_{23}=1$ . The last three unknowns come from the bottom row: $L_{31}=1$ , $L_{32}=4$ and $U_{33}=-2$ . Hence

      $\left[\begin{array}{ccc}\hfill 1& \hfill 3& \hfill 2\\ \hfill 2& \hfill 8& \hfill 5\\ \hfill 1& \hfill 11& \hfill 4\\ \hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill 2& \hfill 1& \hfill 0\\ \hfill 1& \hfill 4& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1& \hfill 3& \hfill 2\\ \hfill 0& \hfill 2& \hfill 1\\ \hfill 0& \hfill 0& \hfill -2\\ \hfill \end{array}\right]$

      is an $LU$ decomposition of the given matrix.

2. Direct multiplication provides the necessary check.
3. 1. We begin by solving

      $\left[\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill -2& \hfill 1\end{array}\right]\left[\begin{array}{c}\hfill y_1\\ \hfill y_2\end{array}\right]=\left[\begin{array}{c}\hfill 1\\ \hfill 2\end{array}\right]$

      Clearly $y_1=1$ and therefore $y_2=4$ . The values $y_1$ and $y_2$ appear on the right-hand side of the second system we need to solve:

      $\left[\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill 0& \hfill -4\end{array}\right]\left[\begin{array}{c}\hfill x_1\\ \hfill x_2\end{array}\right]=\left[\begin{array}{c}\hfill 1\\ \hfill 4\end{array}\right]$

      The second equation implies that $x_2=-1$ and therefore, from the first equation, $x_1=1$ .

   2. We begin by solving the system

      $\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill 1& \hfill 1& \hfill 0\\ \hfill 3& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill y_1\\ \hfill y_2\\ \hfill y_3\end{array}\right]=\left[\begin{array}{c}\hfill 4\\ \hfill 0\\ \hfill 11\end{array}\right].$

      Starting with the top equation we see that $y_1=4$ . The second equation then implies that $y_2=-4$ and then, from the third equation, $y_3=-1$ . These values now appear on the right-hand side of the second system

      $\left[\begin{array}{ccc}\hfill 2& \hfill 1& \hfill -4\\ \hfill 0& \hfill 1& \hfill 2\\ \hfill 0& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\\ \hfill x_2\\ \hfill x_3\end{array}\right]=\left[\begin{array}{c}\hfill 4\\ \hfill -4\\ \hfill -1\end{array}\right].$

      The bottom equation shows us that $x_3=-1$ . Moving up to the middle equation we obtain $x_2=-2$ . The top equation yields $x_1=1$ .

   3. We begin by solving the system

      $\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill 2& \hfill 1& \hfill 0\\ \hfill 1& \hfill 4& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill y_1\\ \hfill y_2\\ \hfill y_3\end{array}\right]=\left[\begin{array}{c}\hfill 2\\ \hfill 3\\ \hfill 0\end{array}\right].$

      Starting with the top equation we see that $y_1=2$ . The second equation then implies that $y_2=-1$ and then, from the third equation, $y_3=2$ . These values now appear on the right-hand side of the second system

      $\left[\begin{array}{ccc}\hfill 1& \hfill 3& \hfill 2\\ \hfill 0& \hfill 2& \hfill 1\\ \hfill 0& \hfill 0& \hfill -2\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\\ \hfill x_2\\ \hfill x_3\end{array}\right]=\left[\begin{array}{c}\hfill 2\\ \hfill -1\\ \hfill 2\end{array}\right].$

      The bottom equation shows us that $x_3=-1$ . Moving up to the middle equation we obtain $x_2=0$ . The top equation yields $x_1=4$ .

4. 1. The second leading submatrix has determinant $1\times 12-6\times 2=0$ and this implies that $A$ has no $LU$ decomposition.
   2. Swapping the second and third rows gives $\left[\begin{array}{ccc}\hfill 1& \hfill 6& \hfill 2\\ \hfill -1& \hfill -3& \hfill -1\\ \hfill 2& \hfill 12& \hfill 5\end{array}\right].$ We let

      $\left[\begin{array}{ccc}\hfill 1& \hfill 6& \hfill 2\\ \hfill -1& \hfill -3& \hfill -1\\ \hfill 2& \hfill 12& \hfill 5\end{array}\right]=LU=\left[\begin{array}{ccc}\hfill U_{11}& \hfill U_{12}& \hfill U_{13}\\ \hfill L_{21}{U}_{11}& \hfill L_{21}{U}_{12}+U_{22}& \hfill L_{21}{U}_{13}+U_{23}\\ \hfill L_{31}{U}_{11}& \hfill L_{31}{U}_{12}+L_{32}{U}_{22}& \hfill L_{31}{U}_{13}+L_{32}{U}_{23}+U_{33}\end{array}\right].$

      Looking at the top row we see that $U_{11}=1$ , $U_{12}=6$ and $U_{13}=2$ . Now, from the second row, $L_{21}=-1$ , $U_{22}=3$ and $U_{23}=1$ . The last three unknowns come from the bottom row: $L_{31}=2$ , $L_{32}=0$ and $U_{33}=1$ . Hence

      $\left[\begin{array}{ccc}\hfill 1& \hfill 6& \hfill 2\\ \hfill -1& \hfill -3& \hfill -1\\ \hfill 2& \hfill 12& \hfill 5\\ \hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1& \hfill 6& \hfill 2\\ \hfill 0& \hfill 3& \hfill 1\\ \hfill 0& \hfill 0& \hfill 1\\ \hfill \end{array}\right]$

      is an $LU$ decomposition of the given matrix.

   3. We begin by solving the system

      $\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill y_1\\ \hfill y_2\\ \hfill y_3\end{array}\right]=\left[\begin{array}{c}\hfill 9\\ \hfill 17\\ \hfill -4\end{array}\right].$

      (Note that the second and third rows of the right-hand side vector have been swapped too.) Starting with the top equation we see that $y_1=9$ . The second equation then implies that $y_2=26$ and then, from the third equation, $y_3=-22$ . These values now appear on the right-hand side of the second system

      $\left[\begin{array}{ccc}\hfill 1& \hfill 6& \hfill 2\\ \hfill 0& \hfill 3& \hfill 1\\ \hfill 0& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\\ \hfill x_2\\ \hfill x_3\end{array}\right]=\left[\begin{array}{c}\hfill 9\\ \hfill 26\\ \hfill -22\end{array}\right].$

      The bottom equation shows us that $x_3=-22$ . Moving up to the middle equation we obtain $x_2=16$ . The top equation yields $x_1=-43$ .