Perform elementary row operations to
[ A ∣ b ∗ → ∣ I 3 ] = [ 1 1 0 b 1 1 0 0 4 3 − 5 b 2 0 1 0 1 1 5 b 3 0 0 1 ] \left[A\mid \overrightarrow{b^{*} }\mid I_{3} \right] = \left[\begin{array}{ccc:c:ccc} 1 & 1 & 0 & b_{1} & 1 & 0 & 0 \\ 4 & 3 & -5 & b_{2} & 0 & 1 & 0 \\ 1 & 1 & 5 & b_{3} & 0 & 0 & 1 \end{array} \right] [ A ∣ b ∗ ∣ I 3 ] = ⎣ ⎢ ⎡ 1 4 1 1 3 1 0 − 5 5 b 1 b 2 b 3 1 0 0 0 1 0 0 0 1 ⎦ ⎥ ⎤
⟶ E ( 2 ) − 4 ( 1 ) E ( 3 ) − ( 1 ) [ 1 1 0 b 1 1 0 0 0 − 1 − 5 b 2 − 4 b 1 − 4 1 0 0 0 5 b 3 − b 1 − 1 0 1 ] ( ∗ 1 ) \underset{\begin{matrix} E_{\left(2\right)-4\left(1\right) } \\ E_{\left(3\right)-\left(1\right)} \end{matrix} } {\longrightarrow} \left[\begin{array}{ccc:c:ccc} 1 & 1 & 0 & b_{1} & 1 & 0 & 0 \\ 0 & -1 & -5 & b_{2}-4b_{1} & -4 & 1 & 0 \\ 0 & 0 & 5 & b_{3}-b_{1} & -1 & 0 & 1 \end{array} \right] \left(*_{1} \right) E ( 2 ) − 4 ( 1 ) E ( 3 ) − ( 1 ) ⟶ ⎣ ⎢ ⎡ 1 0 0 1 − 1 0 0 − 5 5 b 1 b 2 − 4 b 1 b 3 − b 1 1 − 4 − 1 0 1 0 0 0 1 ⎦ ⎥ ⎤ ( ∗ 1 )
⟶ E − ( 2 ) E 1 5 ( 3 ) [ 1 1 0 b 1 1 0 0 0 1 5 4 b 1 − b 2 4 − 1 0 0 0 1 1 5 ( b 3 − b 1 ) − 1 5 0 1 5 ] \underset{\begin{matrix} E_{-\left(2\right)} \\ E_{\frac{1}{5}\left(3\right)} \end{matrix} }{\longrightarrow} \left[\begin{array}{ccc:c:ccc} 1 & 1 & 0 & b_{1} & 1 & 0 & 0 \\ 0 & 1 & 5 & 4b_{1}-b_{2} & 4 & -1 & 0 \\ 0 & 0 & 1 &\frac{1}{5}\left(b_{3}-b_{1}\right) & -\frac{1}{5} & 0 & \frac{1}{5} \end{array} \right] E − ( 2 ) E 5 1 ( 3 ) ⟶ ⎣ ⎢ ⎡ 1 0 0 1 1 0 0 5 1 b 1 4 b 1 − b 2 5 1 ( b 3 − b 1 ) 1 4 − 5 1 0 − 1 0 0 0 5 1 ⎦ ⎥ ⎤
⟶ E ( 1 ) − ( 2 ) E ( 2 ) − 5 ( 3 ) [ 1 0 − 5 − 3 b 1 + b 2 − 3 1 0 0 1 0 5 b 1 − b 2 − b 3 5 − 1 − 1 0 0 1 1 5 ( b 3 − b 1 ) − 1 5 0 1 5 ] \underset{\begin{matrix} E_{\left(1\right)-\left(2\right)} \\ E_{\left(2\right)-5\left(3\right)} \end{matrix} }{\longrightarrow} \left[\begin{array}{ccc:c:ccc} 1 & 0 & -5 & -3b_{1}+b_{2} & -3 & 1 & 0 \\ 0 & 1 & 0 & 5b_{1}-b_{2}-b_{3} & 5 & -1 & -1 \\ 0 & 0 & 1 &\frac{1}{5}\left(b_{3}-b_{1}\right) & -\frac{1}{5} & 0 & \frac{1}{5} \end{array} \right] E ( 1 ) − ( 2 ) E ( 2 ) − 5 ( 3 ) ⟶ ⎣ ⎢ ⎡ 1 0 0 0 1 0 − 5 0 1 − 3 b 1 + b 2 5 b 1 − b 2 − b 3 5 1 ( b 3 − b 1 ) − 3 5 − 5 1 1 − 1 0 0 − 1 5 1 ⎦ ⎥ ⎤
⟶ E ( 1 ) + 5 ( 3 ) [ 1 0 0 − 4 b 1 + b 2 + b 3 − 4 1 1 0 1 0 5 b 1 − b 2 − b 3 5 − 1 − 1 0 0 1 1 5 ( b 3 − b 1 ) − 1 5 0 1 5 ] . ( ∗ 2 ) \underset{\text{}E_{\left(1\right)+5\left(3\right) } }{\longrightarrow} \left[\begin{array}{ccc:c:ccc} 1 & 0 & 0 & -4b_{1}+b_{2}+b_{3} & -4 & 1 & 1 \\ 0 & 1 & 0 & 5b_{1}-b_{2}-b_{3} & 5 & -1 & -1 \\ 0 & 0 & 1 &\frac{1}{5}\left(b_{3}-b_{1}\right) & -\frac{1}{5} & 0 & \frac{1}{5} \end{array} \right]. \left(*_{2} \right) E ( 1 ) + 5 ( 3 ) ⟶ ⎣ ⎢ ⎡ 1 0 0 0 1 0 0 0 1 − 4 b 1 + b 2 + b 3 5 b 1 − b 2 − b 3 5 1 ( b 3 − b 1 ) − 4 5 − 5 1 1 − 1 0 1 − 1 5 1 ⎦ ⎥ ⎤ . ( ∗ 2 )
Stop at ( ∗ 1 ) \left(*_{1} \right) ( ∗ 1 )
x 1 x 2 x 3 \begin{matrix} x_{1} & x_{2} & x_{3} \end{matrix} x 1 x 2 x 3
[ 1 1 0 b 1 0 − 1 − 5 b 2 − 4 b 1 0 0 5 b 3 − b 1 ] ⇒ { x 1 + x 2 = b 1 − x 2 − 5 x 3 = b 2 − 4 b 1 5 x 3 = b 3 − b 1 \left[\begin{array}{ccc:c} 1 & 1 & 0 & b_{1} \\ 0 & -1 & -5 & b_{2}-4b_{1} \\ 0 & 0 & 5 & b_{3}-b_{1} \end{array} \right]\Rightarrow \left\{\begin{matrix} x_{1}+x_{2}=b_{1} \\ -x_{2}-5x_{3}=b_{2}-4b_{1} \\ 5x_{3}=b_{3}-b_{1} \end{matrix} \right. ⎣ ⎢ ⎡ 1 0 0 1 − 1 0 0 − 5 5 b 1 b 2 − 4 b 1 b 3 − b 1 ⎦ ⎥ ⎤ ⇒ ⎩ ⎪ ⎨ ⎪ ⎧ x 1 + x 2 = b 1 − x 2 − 5 x 3 = b 2 − 4 b 1 5 x 3 = b 3 − b 1
⇒ { x 1 = − 4 b 1 + b 2 + b 3 x 2 = 5 b 1 − b 2 − b 3 x 3 = 1 5 ( b 3 − b 1 ) . \Rightarrow \left\{\begin{matrix} x_{1}=-4b_{1}+b_{2}+b_{3} \\ x_{2}=5b_{1}-b_{2}-b_{3} \\ x_{3}=\frac{1}{5} \left(b_{3}-b_{1}\right). \end{matrix} \right. ⇒ ⎩ ⎪ ⎨ ⎪ ⎧ x 1 = − 4 b 1 + b 2 + b 3 x 2 = 5 b 1 − b 2 − b 3 x 3 = 5 1 ( b 3 − b 1 ) .
This is the solution of the equations A x ∗ → = b ∗ → . A\overrightarrow{x^{*} }=\overrightarrow{b^{*} }. A x ∗ = b ∗ .
E ( 3 ) − ( 1 ) E ( 2 ) − 4 ( 1 ) A = [ 1 1 0 0 − 1 − 5 0 0 5 ] E_{\left(3\right)-\left(1\right) } E_{\left(2\right)-4\left(1\right)} A=\left[\begin{matrix} 1 & 1 & 0 \\ 0 & -1 & -5 \\ 0 & 0 & 5 \end{matrix} \right] E ( 3 ) − ( 1 ) E ( 2 ) − 4 ( 1 ) A = ⎣ ⎢ ⎡ 1 0 0 1 − 1 0 0 − 5 5 ⎦ ⎥ ⎤
⇒ A = E ( 2 ) − 4 ( 1 ) − 1 E ( 3 ) − ( 1 ) − 1 [ 1 1 0 0 − 1 − 5 0 0 5 ] = E ( 2 ) + 4 ( 1 ) E ( 3 ) + ( 1 ) [ 1 1 0 0 − 1 − 5 0 0 5 ] \Rightarrow A=E^{-1}_{\left(2\right)-4\left(1\right)} E^{-1}_{\left(3\right)-\left(1\right)} \left[\begin{matrix} 1 & 1 & 0 \\ 0 & -1 & -5 \\ 0 & 0 & 5 \end{matrix} \right]= E_{\left(2\right)+4\left(1\right)} E_{\left(3\right)+\left(1\right)}\left[\begin{matrix} 1 & 1 & 0 \\ 0 & -1 & -5 \\ 0 & 0 & 5 \end{matrix} \right] ⇒ A = E ( 2 ) − 4 ( 1 ) − 1 E ( 3 ) − ( 1 ) − 1 ⎣ ⎢ ⎡ 1 0 0 1 − 1 0 0 − 5 5 ⎦ ⎥ ⎤ = E ( 2 ) + 4 ( 1 ) E ( 3 ) + ( 1 ) ⎣ ⎢ ⎡ 1 0 0 1 − 1 0 0 − 5 5 ⎦ ⎥ ⎤
= [ 1 0 0 4 1 0 0 0 1 ] [ 1 0 0 0 1 0 1 0 1 ] [ 1 1 0 0 − 1 − 5 0 0 5 ] =\left[\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} 1 & 1 & 0 \\ 0 & -1 & -5 \\ 0 & 0 & 5 \end{matrix} \right] = ⎣ ⎢ ⎡ 1 4 0 0 1 0 0 0 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 1 0 1 0 0 0 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 0 1 − 1 0 0 − 5 5 ⎦ ⎥ ⎤
= [ 1 0 0 4 1 0 1 0 1 ] [ 1 1 0 0 − 1 − 5 0 0 5 ] =\left[\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} 1 & 1 & 0 \\ 0 & -1 & -5 \\ 0 & 0 & 5 \end{matrix} \right] = ⎣ ⎢ ⎡ 1 4 1 0 1 0 0 0 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 0 1 − 1 0 0 − 5 5 ⎦ ⎥ ⎤
= [ 1 0 0 4 1 0 1 0 1 ] [ 1 0 0 0 − 1 0 0 0 5 ] [ 1 1 0 0 1 5 0 0 1 ] =\left[\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{matrix} \right]\left[\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{matrix} \right] = ⎣ ⎢ ⎡ 1 4 1 0 1 0 0 0 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 0 0 − 1 0 0 0 5 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 0 1 1 0 0 5 1 ⎦ ⎥ ⎤
From here, it is easily seen that
det A = the product of the pivots 1,−1 and 5=−5.
d e t A − 1 = ( d e t A ) − 1 = − 1 5 . det A^{-1} =\left(det A\right) ^{-1}=-\frac{1}{5}. d e t A − 1 = ( d e t A ) − 1 = − 5 1 .
Also, the four subspaces (see (3.7.32)) are:
I m ( A ) = ≪ ( 1 , 1 , 0 ) , ( 0 , − 1 , − 5 ) , ( 0 , 0 , 5 ) ≫ = R 3 ; Im \left(A\right)=\ll \left(1,1,0\right),\left(0,-1,-5\right),\left(0,0,5\right) \gg =R^{3} ; I m ( A ) = ≪ ( 1 , 1 , 0 ) , ( 0 , − 1 , − 5 ) , ( 0 , 0 , 5 ) ≫ = R 3 ;
K e r ( A ) = { 0 → } ; Ker \left(A\right)=\left\{\overrightarrow{0} \right\} ; K e r ( A ) = { 0 } ;
I m ( A ∗ ) = ≪ ( 1 , 0 , 0 ) , ( 1 , − 1 , 0 ) , ( 0 , − 5 , 5 ) ≫ = R 3 ; Im \left(A^{*} \right)=\ll \left(1,0,0\right),\left(1,-1,0\right),\left(0,-5,5\right) \gg =R^{3} ; I m ( A ∗ ) = ≪ ( 1 , 0 , 0 ) , ( 1 , − 1 , 0 ) , ( 0 , − 5 , 5 ) ≫ = R 3 ;
K e r ( A ∗ ) = { 0 → } ; Ker\left(A^{*} \right)=\left\{\overrightarrow{0} \right\} ; K e r ( A ∗ ) = { 0 } ;
which can also be obtained from ( ∗ 2 ) \left(*_{2} \right) ( ∗ 2 )
Stop at ( ∗ 2 ) \left(*_{2} \right) ( ∗ 2 )
A is invertible, since
E ( 1 ) + 5 ( 3 ) E ( 2 ) − 5 ( 3 ) E ( 1 ) − ( 2 ) E 1 5 ( 3 ) E − ( 2 ) E ( 3 ) − ( 1 ) E ( 2 ) − 4 ( 1 ) A = I 3 . E_{\left(1\right)+5\left(3\right)}E_{\left(2\right)-5\left(3\right)}E_{\left(1\right)-\left(2\right)}E_{\frac{1}{5}\left(3\right)}E_{-\left(2\right)}E_{\left(3\right)-\left(1\right)}E_{\left(2\right)-4\left(1\right)}A=I_{3}. E ( 1 ) + 5 ( 3 ) E ( 2 ) − 5 ( 3 ) E ( 1 ) − ( 2 ) E 5 1 ( 3 ) E − ( 2 ) E ( 3 ) − ( 1 ) E ( 2 ) − 4 ( 1 ) A = I 3 .
Therefore,
A − 1 = [ − 4 1 1 5 − 1 − 1 − 1 5 0 1 5 ] A^{-1}=\left[\begin{matrix} -4 & 1 & 1 \\ 5 & -1 & -1 \\ -\frac{1}{5} & 0 & \frac{1}{5} \end{matrix} \right] A − 1 = ⎣ ⎢ ⎡ − 4 5 − 5 1 1 − 1 0 1 − 1 5 1 ⎦ ⎥ ⎤
= E ( 1 ) + 5 ( 3 ) E ( 2 ) − 5 ( 3 ) E ( 1 ) − ( 2 ) E 1 5 ( 3 ) E − ( 2 ) E ( 3 ) − ( 1 ) E ( 2 ) − 4 ( 1 ) =E_{\left(1\right)+5\left(3\right)}E_{\left(2\right)-5\left(3\right)}E_{\left(1\right)-\left(2\right)}E_{\frac{1}{5}\left(3\right)}E_{-\left(2\right)}E_{\left(3\right)-\left(1\right)}E_{\left(2\right)-4\left(1\right)} = E ( 1 ) + 5 ( 3 ) E ( 2 ) − 5 ( 3 ) E ( 1 ) − ( 2 ) E 5 1 ( 3 ) E − ( 2 ) E ( 3 ) − ( 1 ) E ( 2 ) − 4 ( 1 )
⇒ d e t A − 1 = 1 5 ⋅ ( − 1 ) = − 1 5 ; \Rightarrow det A^{-1} =\frac{1}{5}\cdot \left(-1\right)=-\frac{1}{5}; ⇒ d e t A − 1 = 5 1 ⋅ ( − 1 ) = − 5 1 ;
and
A = E ( 2 ) − 4 ( 1 ) − 1 E ( 3 ) − ( 1 ) − 1 E − ( 2 ) − 1 E 1 5 ( 3 ) − 1 E ( 1 ) − ( 2 ) − 1 E ( 2 ) − 5 ( 3 ) − 1 E ( 1 ) + 5 ( 3 ) − 1 A=E^{-1}_{\left(2\right)-4\left(1\right)}E^{-1}_{\left(3\right)-\left(1\right)}E^{-1}_{-\left(2\right)}E^{-1}_{\frac{1}{5}\left(3\right)}E^{-1}_{\left(1\right)-\left(2\right)}E^{-1}_{\left(2\right)-5\left(3\right)}E^{-1}_{\left(1\right)+5\left(3\right)} A = E ( 2 ) − 4 ( 1 ) − 1 E ( 3 ) − ( 1 ) − 1 E − ( 2 ) − 1 E 5 1 ( 3 ) − 1 E ( 1 ) − ( 2 ) − 1 E ( 2 ) − 5 ( 3 ) − 1 E ( 1 ) + 5 ( 3 ) − 1
= E ( 2 ) + 4 ( 1 ) E ( 3 ) + ( 1 ) E − ( 2 ) E 5 ( 3 ) E ( 1 ) + ( 2 ) E ( 2 ) + 5 ( 3 ) E ( 1 ) − 5 ( 3 ) =E_{\left(2\right)+4\left(1\right)}E_{\left(3\right)+\left(1\right)}E_{-\left(2\right)}E_{5\left(3\right)}E_{\left(1\right)+\left(2\right)}E_{\left(2\right)+5\left(3\right)}E_{\left(1\right)-5\left(3\right)} = E ( 2 ) + 4 ( 1 ) E ( 3 ) + ( 1 ) E − ( 2 ) E 5 ( 3 ) E ( 1 ) + ( 2 ) E ( 2 ) + 5 ( 3 ) E ( 1 ) − 5 ( 3 )
= [ 1 0 0 4 1 0 0 0 1 ] [ 1 0 0 0 1 0 1 0 1 ] [ 1 0 0 0 − 1 0 0 0 1 ] [ 1 0 0 0 1 0 0 0 5 ] =\left[\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \end{matrix} \right] = ⎣ ⎢ ⎡ 1 4 0 0 1 0 0 0 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 1 0 1 0 0 0 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 0 0 − 1 0 0 0 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 0 0 1 0 0 0 5 ⎦ ⎥ ⎤
[ 1 1 0 0 1 0 0 0 1 ] [ 1 0 0 0 1 5 0 0 1 ] [ 1 0 − 5 0 1 0 0 0 1 ] \left[\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} 1 & 0 & -5 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] ⎣ ⎢ ⎡ 1 0 0 1 1 0 0 0 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 0 0 1 0 0 5 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 0 0 1 0 − 5 0 1 ⎦ ⎥ ⎤
⇒ det A = (−1) ·5 = −5.
From the elementary matrix factorization of A, we can recapture the LDU and hence LU decomposition, since
E ( 2 ) + 4 ( 1 ) E ( 3 ) + ( 1 ) = [ 1 0 0 4 1 0 1 0 1 ] = L , E_{\left(2\right)+4\left(1\right)}E_{\left(3\right)+\left(1\right)}=\left[\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} \right]=L, E ( 2 ) + 4 ( 1 ) E ( 3 ) + ( 1 ) = ⎣ ⎢ ⎡ 1 4 1 0 1 0 0 0 1 ⎦ ⎥ ⎤ = L ,
E − ( 2 ) E 5 ( 3 ) = [ 1 0 0 0 − 1 0 0 0 5 ] = D , E_{-\left(2\right)}E_{5\left(3\right)}=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{matrix} \right]=D, E − ( 2 ) E 5 ( 3 ) = ⎣ ⎢ ⎡ 1 0 0 0 − 1 0 0 0 5 ⎦ ⎥ ⎤ = D ,
E ( 1 ) + ( 2 ) E ( 2 ) + 5 ( 3 ) E ( 1 ) − 5 ( 3 ) = [ 1 1 0 0 1 5 0 0 1 ] = U . E_{\left(1\right)+\left(2\right)}E_{\left(2\right)+5\left(3\right)}E_{\left(1\right)-5\left(3\right)}=\left[\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{matrix} \right]=U. E ( 1 ) + ( 2 ) E ( 2 ) + 5 ( 3 ) E ( 1 ) − 5 ( 3 ) = ⎣ ⎢ ⎡ 1 0 0 1 1 0 0 5 1 ⎦ ⎥ ⎤ = U .
This is within our reasonable expectation, because in the process of obtaining ( ∗ 2 ) , \left(*_{2} \right), ( ∗ 2 ) , E ( 2 ) − 4 ( 1 ) E_{\left(2\right)-4\left(1\right)} E ( 2 ) − 4 ( 1 ) E ( 3 ) − ( 1 ) E_{\left(3\right)-\left(1\right)} E ( 3 ) − ( 1 ) ( ∗ 1 ) , \left(*_{1} \right), ( ∗ 1 ) ,
( E ( 3 ) − ( 1 ) E ( 2 ) − 4 ( 1 ) ) − 1 = E ( 2 ) + 4 ( 1 ) E ( 3 ) + ( 1 ) . \left(E_{\left(3\right)-\left(1\right)}E_{\left(2\right)-4\left(1\right)}\right) ^{-1}= E_{\left(2\right)+4\left(1\right)}E_{\left(3\right)+\left(1\right)}. ( E ( 3 ) − ( 1 ) E ( 2 ) − 4 ( 1 ) ) − 1 = E ( 2 ) + 4 ( 1 ) E ( 3 ) + ( 1 ) .
The LU-decomposition can help solving A x ∗ → = b ∗ → . A\overrightarrow{x^{*} } =\overrightarrow{b^{*} }. A x ∗ = b ∗ .
A x ∗ → = b ∗ → A\overrightarrow{x^{*} } =\overrightarrow{b^{*} } A x ∗ = b ∗
⇔ [ 1 1 0 0 − 1 − 5 0 0 5 ] x ∗ → = y ∗ → \Leftrightarrow \left[\begin{matrix} 1 & 1 & 0 \\ 0 & -1 & -5 \\ 0 & 0 & 5 \end{matrix} \right] \overrightarrow{x^{*} }=\overrightarrow{y^{*} } ⇔ ⎣ ⎢ ⎡ 1 0 0 1 − 1 0 0 − 5 5 ⎦ ⎥ ⎤ x ∗ = y ∗ [ 1 0 0 4 1 0 1 0 1 ] y ∗ → = b ∗ → \left[\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} \right] \overrightarrow{y^{*} }=\overrightarrow{b^{*} } ⎣ ⎢ ⎡ 1 4 1 0 1 0 0 0 1 ⎦ ⎥ ⎤ y ∗ = b ∗
⇔ x ∗ → = [ 1 1 0 0 − 1 − 5 0 0 5 ] − 1 [ 1 0 0 4 1 0 1 0 1 ] − 1 b ∗ → \Leftrightarrow \overrightarrow{x^{*} }=\left[\begin{matrix} 1 & 1 & 0 \\ 0 & -1 & -5 \\ 0 & 0 & 5 \end{matrix} \right]^{-1} \left[\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} \right]^{-1} \overrightarrow{b^{*} } ⇔ x ∗ = ⎣ ⎢ ⎡ 1 0 0 1 − 1 0 0 − 5 5 ⎦ ⎥ ⎤ − 1 ⎣ ⎢ ⎡ 1 4 1 0 1 0 0 0 1 ⎦ ⎥ ⎤ − 1 b ∗
⇔ x ∗ → = A − 1 b ∗ → . \Leftrightarrow \overrightarrow{x^{*} }=A^{-1} \overrightarrow{b^{*} }. ⇔ x ∗ = A − 1 b ∗ .
Readers are urged to carry out actual computations to solve out the solution.
The elementary matrices, LU and LDU decompositions can be used to help investigating geometric mapping properties of A, better using GSP. For example, the image of the unit cube under A is the parallelepiped as shown in Fig. 3.50. This parallelepiped can be obtained by performing successive mappings E ( 2 ) + 4 ( 1 ) E_{\left(2\right)+4\left(1\right)} E ( 2 ) + 4 ( 1 ) E ( 3 ) + ( 1 ) E_{\left(3\right)+\left(1\right)} E ( 3 ) + ( 1 ) E ( 1 ) − 5 ( 3 ) . E_{\left(1\right)-5\left(3\right)}. E ( 1 ) − 5 ( 3 ) .
Also (see Sec. 5.3),
the signed volume of the parallelepiped = d e t [ 1 1 0 4 3 − 5 1 1 5 ] = − 5 = det \left[\begin{matrix} 1 & 1 & 0 \\ 4 & 3 & -5 \\ 1 & 1 & 5 \end{matrix} \right]=-5 = d e t ⎣ ⎢ ⎡ 1 4 1 1 3 1 0 − 5 5 ⎦ ⎥ ⎤ = − 5
⇒ t h e s i g n e d v o l u m e o f t h e p a r a l l e l e p i p e d t h e v o l u m e o f t h e u n i t c u b e = d e t A . \Rightarrow \frac{ the signed volume of the parallelepiped}{the volume of the unit cube} = det A. ⇒ t h e v o l u m e o f t h e u n i t c u b e t h e s i g n e d v o l u m e o f t h e p a r a l l e l e p i p e d = d e t A .
Since det A = −5 < 0, so A reverses the orientations in R³.
