Let A = [1 1 0 4 3 -5 1 1 5]. (1) Solve Ax∗ ^→= b∗^→ where x ^→= (x1, x2, x3) ∈ R3 and b^→ = (b1, b2, b3). (2) Determine if A is invertible. If yes, compute A−1 and express A and A−1 as products of elementary matrices.(3) Compute detA and detA−1. (4) Find LU and LDU decompositions of A. (5) Compute

Question

Let

[latex]A=\left[\begin{matrix} 1 & 1 & 0 \ 4 &3 & -5 \ 1 & 1 & 5 \end{matrix} ight].[/latex]

(1) Solve A [latex]\overrightarrow{x^{*} }=\overrightarrow{b^{*} }[/latex] where [latex]\overrightarrow{x }=\left(x_{1} ,x_{2},x_{3} ight) \in R^{3}[/latex] and [latex]\overrightarrow{b }=\left(b_{1} ,b_{2},b_{3} ight).[/latex]

(2) Determine if A is invertible. If yes, compute [latex]A^{-1}[/latex] and express A and [latex]A^{-1}[/latex] as products of elementary matrices.

(3) Compute det A and det [latex]A^{-1}.[/latex]

(4) Find LU and LDU decompositions of A.

(5) Compute Im(A), Ker(A) and [latex] Im\left(A^{*} ight), Ker\left(A^{*} ight)[/latex] (see (3.7.32)).

(6) Investigate the geometric mapping properties of A.

Accepted Answer

Perform elementary row operations to

[latex] \left[A\mid \overrightarrow{b^{*} }\mid I_{3} ight] = \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} ight][/latex]

[latex]\underset{\begin{matrix} E_{\left(2 ight)-4\left(1 ight) } \ E_{\left(3 ight)-\left(1 ight)} \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} ight] \left(*_{1} ight)[/latex]

[latex]\underset{\begin{matrix} E_{-\left(2 ight)} \ E_{\frac{1}{5}\left(3 ight)} \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} ight) & -\frac{1}{5} & 0 & \frac{1}{5} \end{array} ight] [/latex]

[latex]\underset{\begin{matrix} E_{\left(1 ight)-\left(2 ight)} \ E_{\left(2 ight)-5\left(3 ight)} \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} ight) & -\frac{1}{5} & 0 & \frac{1}{5} \end{array} ight][/latex]

[latex]\underset{ ext{}E_{\left(1 ight)+5\left(3 ight) } }{\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} ight) & -\frac{1}{5} & 0 & \frac{1}{5} \end{array} ight]. \left(*_{2} ight)[/latex]

Stop at [latex]\left(*_{1} ight)[/latex]:

[latex] \begin{matrix} x_{1} & x_{2} & x_{3} \end{matrix}[/latex]

[latex]\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} ight]\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} ight.[/latex]

[latex]\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} ight). \end{matrix} ight.[/latex]

This is the solution of the equations [latex]A\overrightarrow{x^{*} }=\overrightarrow{b^{*} }.[/latex] On the other hand,

[latex]E_{\left(3 ight)-\left(1 ight) } E_{\left(2 ight)-4\left(1 ight)} A=\left[\begin{matrix} 1 & 1 & 0 \ 0 & -1 & -5 \ 0 & 0 & 5 \end{matrix} ight][/latex]

[latex]\Rightarrow A=E^{-1}_{\left(2 ight)-4\left(1 ight)} E^{-1}_{\left(3 ight)-\left(1 ight)} \left[\begin{matrix} 1 & 1 & 0 \ 0 & -1 & -5 \ 0 & 0 & 5 \end{matrix} ight]= E_{\left(2 ight)+4\left(1 ight)} E_{\left(3 ight)+\left(1 ight)}\left[\begin{matrix} 1 & 1 & 0 \ 0 & -1 & -5 \ 0 & 0 & 5 \end{matrix} ight][/latex]

[latex]=\left[\begin{matrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight]\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{matrix} ight]\left[\begin{matrix} 1 & 1 & 0 \ 0 & -1 & -5 \ 0 & 0 & 5 \end{matrix} ight][/latex]

[latex]=\left[\begin{matrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 1 & 0 & 1 \end{matrix} ight]\left[\begin{matrix} 1 & 1 & 0 \ 0 & -1 & -5 \ 0 & 0 & 5 \end{matrix} ight][/latex] (LU-decomposition)

[latex]=\left[\begin{matrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 1 & 0 & 1 \end{matrix} ight]\left[\begin{matrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 5 \end{matrix} ight]\left[\begin{matrix} 1 & 1 & 0 \ 0 & 1 & 5 \ 0 & 0 & 1 \end{matrix} ight][/latex] (LDU-decomposition).

From here, it is easily seen that

det A = the product of the pivots 1,−1 and 5=−5.

[latex]det A^{-1} =\left(det A ight) ^{-1}=-\frac{1}{5}.[/latex]

Also, the four subspaces (see (3.7.32)) are:

[latex] Im \left(A ight)=\ll \left(1,1,0 ight),\left(0,-1,-5 ight),\left(0,0,5 ight) \gg =R^{3} ;[/latex]

[latex]Ker \left(A ight)=\left\{\overrightarrow{0} ight\} ;[/latex]

[latex]Im \left(A^{*} ight)=\ll \left(1,0,0 ight),\left(1,-1,0 ight),\left(0,-5,5 ight) \gg =R^{3} ;[/latex]

[latex]Ker\left(A^{*} ight)=\left\{\overrightarrow{0} ight\} ;[/latex]

which can also be obtained from [latex]\left(*_{2} ight)[/latex] (see Application 8 in Sec. B.5).

Stop at [latex]\left(*_{2} ight)[/latex]:

A is invertible, since

[latex]E_{\left(1 ight)+5\left(3 ight)}E_{\left(2 ight)-5\left(3 ight)}E_{\left(1 ight)-\left(2 ight)}E_{\frac{1}{5}\left(3 ight)}E_{-\left(2 ight)}E_{\left(3 ight)-\left(1 ight)}E_{\left(2 ight)-4\left(1 ight)}A=I_{3}.[/latex]

Therefore,

[latex]A^{-1}=\left[\begin{matrix} -4 & 1 & 1 \ 5 & -1 & -1 \ -\frac{1}{5} & 0 & \frac{1}{5} \end{matrix} ight][/latex]

[latex]=E_{\left(1 ight)+5\left(3 ight)}E_{\left(2 ight)-5\left(3 ight)}E_{\left(1 ight)-\left(2 ight)}E_{\frac{1}{5}\left(3 ight)}E_{-\left(2 ight)}E_{\left(3 ight)-\left(1 ight)}E_{\left(2 ight)-4\left(1 ight)}[/latex]

[latex]\Rightarrow det A^{-1} =\frac{1}{5}\cdot \left(-1 ight)=-\frac{1}{5};[/latex]

and

[latex]A=E^{-1}_{\left(2 ight)-4\left(1 ight)}E^{-1}_{\left(3 ight)-\left(1 ight)}E^{-1}_{-\left(2 ight)}E^{-1}_{\frac{1}{5}\left(3 ight)}E^{-1}_{\left(1 ight)-\left(2 ight)}E^{-1}_{\left(2 ight)-5\left(3 ight)}E^{-1}_{\left(1 ight)+5\left(3 ight)}[/latex]

[latex]=E_{\left(2 ight)+4\left(1 ight)}E_{\left(3 ight)+\left(1 ight)}E_{-\left(2 ight)}E_{5\left(3 ight)}E_{\left(1 ight)+\left(2 ight)}E_{\left(2 ight)+5\left(3 ight)}E_{\left(1 ight)-5\left(3 ight)}[/latex]

[latex]=\left[\begin{matrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight]\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{matrix} ight]\left[\begin{matrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{matrix} ight]\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 5 \end{matrix} ight][/latex]

[latex]\left[\begin{matrix} 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight]\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 5 \ 0 & 0 & 1 \end{matrix} ight]\left[\begin{matrix} 1 & 0 & -5 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight][/latex]

⇒ det A = (−1) ·5 = −5.

From the elementary matrix factorization of A, we can recapture the LDU and hence LU decomposition, since

[latex]E_{\left(2 ight)+4\left(1 ight)}E_{\left(3 ight)+\left(1 ight)}=\left[\begin{matrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 1 & 0 & 1 \end{matrix} ight]=L,[/latex]

[latex]E_{-\left(2 ight)}E_{5\left(3 ight)}=\left[\begin{matrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 5 \end{matrix} ight]=D,[/latex]

[latex]E_{\left(1 ight)+\left(2 ight)}E_{\left(2 ight)+5\left(3 ight)}E_{\left(1 ight)-5\left(3 ight)}=\left[\begin{matrix} 1 & 1 & 0 \ 0 & 1 & 5 \ 0 & 0 & 1 \end{matrix} ight]=U.[/latex]

This is within our reasonable expectation, because in the process of obtaining [latex]\left(*_{2} ight),[/latex] we use [latex]E_{\left(2 ight)-4\left(1 ight)}[/latex] and [latex]E_{\left(3 ight)-\left(1 ight)}[/latex] to transform the original A into an upper triangle as shown in [latex]\left(*_{1} ight),[/latex] and the lower triangle L should be

[latex]\left(E_{\left(3 ight)-\left(1 ight)}E_{\left(2 ight)-4\left(1 ight)} ight) ^{-1}= E_{\left(2 ight)+4\left(1 ight)}E_{\left(3 ight)+\left(1 ight)}.[/latex]

The LU-decomposition can help solving [latex]A\overrightarrow{x^{*} } =\overrightarrow{b^{*} }. [/latex] Notice that

[latex]A\overrightarrow{x^{*} } =\overrightarrow{b^{*} }[/latex]

[latex]\Leftrightarrow \left[\begin{matrix} 1 & 1 & 0 \ 0 & -1 & -5 \ 0 & 0 & 5 \end{matrix} ight] \overrightarrow{x^{*} }=\overrightarrow{y^{*} } [/latex] and [latex]\left[\begin{matrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 1 & 0 & 1 \end{matrix} ight] \overrightarrow{y^{*} }=\overrightarrow{b^{*} } [/latex]

[latex]\Leftrightarrow \overrightarrow{x^{*} }=\left[\begin{matrix} 1 & 1 & 0 \ 0 & -1 & -5 \ 0 & 0 & 5 \end{matrix} ight]^{-1} \left[\begin{matrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 1 & 0 & 1 \end{matrix} ight]^{-1} \overrightarrow{b^{*} }[/latex]

[latex]\Leftrightarrow \overrightarrow{x^{*} }=A^{-1} \overrightarrow{b^{*} }.[/latex]

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 [latex]E_{\left(2 ight)+4\left(1 ight)}[/latex] to the cube followed by [latex]E_{\left(3 ight)+\left(1 ight)}[/latex] · · · then by [latex]E_{\left(1 ight)-5\left(3 ight)}.[/latex]

Also (see Sec. 5.3),

the signed volume of the parallelepiped [latex]= det \left[\begin{matrix} 1 & 1 & 0 \ 4 & 3 & -5 \ 1 & 1 & 5 \end{matrix} ight]=-5[/latex]

[latex]\Rightarrow \frac{ the signed volume of the parallelepiped}{the volume of the unit cube} = det A.[/latex]

Since det A = −5 < 0, so A reverses the orientations in R³.

Question 3.6: Let A = [1 1 0 4 3 -5 1 1 5]. (1) Solve Ax∗ ^→= b∗^→ where x...

Related Answered Questions

Verified Answer:

Give an affine transformation T(x^→ ) = x0^→ + x^→ A, where A =[1/3 2/3 2/3 2/3 1/3 -2/3 2/3 -2/3 1/3]. Determine these x0^→ so that each such T is an orthogonal reflection, and the direction and the plane of invariant points. ...

Verified Answer:

Determine the relative positions of (a) two straight lines in R4,(b) two (two dimensional) planes in R4, and (b) two (two-dimensional) planes in R4, and (c) one (two-dimensional) plane and one (three-dimensional) hyperplane in R4. ...

Verified Answer:

Determine the relative positions of two planes S21 = x0^→ + S1 and S22 = y0^→ + S2 in R3(see Fig. 3.16). ...

Verified Answer:

Determine the relative positions of two lines S11 = x0^→ + S1 and S12 = y0^→ + S2 in R3 (see Fig.3.13). ...

Verified Answer:

Give an affine transformation T(x^→ ) = x0^→ + x^→A, where A = [-1/3 2/3 -2/3 2/3 2/3 1/3 2/3 -1/3 -2/3].Try to determine these x0^→ so that T is a rotation. In this case, determine the axis and the angle of the rotation, also the rotational plane. ...

Verified Answer:

Let a0^→ = (1, 1, 0),a1^→ = (2, 0,−1) and a2^→ = (0,−1, 1). Try to construct a shearing with coefficient k ≠ 0 in the direction a1^→ − a0^→ with a0^→ +<>⊥ as the plane of invariant points. Note that (a1^→ − a0^→ )⊥ (a2^→ − a0^→ ). ...

Verified Answer:

Analyze the rational canonical form ...

Verified Answer:

(a) Find the reflection of R3 along the direction v3= (−1, 1,−1) with respect to the plane (2,−2, 3) + <>. (b) Show that T(x ^→) = x0^→ + xA^→, where x0^→ = (0,−2,−4) and A = [1 0 0 0 5/3 4/3 0 -4/3 -5/3] is a reflection. Determine its direction and plane of invariant points ...

Verified Answer:

Give an affine transformation T(x ) = x0 + x A, where A =[-1 6 -4 2 4 -5 2 6 -7].Try to determine x0 so that T is a two-way stretch. In this case, determine the line of invariant points and the invariant plane. ...

Verified Answer: