Question #249535

Consider the following system of linear algebraic equations 𝐴π‘₯=𝑏:


[1 1 3 [ X = [2


5 3 1 Y 3


2 3 1] Z] -1]


Solve the system using (you can compare your solution with MATLAB)


1. Cramer’s Rule


2. Gauss elimination without pivoting


3. Gauss-Jordan elimination


4. LU factorization


1
Expert's answer
2022-02-21T16:24:28-0500

A=(113531231),x=(xyz),b=(23βˆ’1)A=\begin{pmatrix} 1 & 1&3 \\ 5&3&1\\ 2&3&1 \end{pmatrix}, x=\begin{pmatrix} x \\ y\\z \end{pmatrix},b=\begin{pmatrix} 2 \\ 3\\-1 \end{pmatrix}

1. Cramer’s Rule

Ξ”=∣113531231∣=3+2+45βˆ’18βˆ’5βˆ’3=24Ξ”x=∣213331βˆ’131∣=6βˆ’1+27+9βˆ’3βˆ’6=32Ξ”y=∣1235312βˆ’11∣=3+4βˆ’15βˆ’18βˆ’10+1=βˆ’35\Delta=\begin{vmatrix} 1 & 1&3 \\ 5&3&1\\ 2&3&1 \end{vmatrix}=3+2+45-18-5-3=24\\ \Delta_x=\begin{vmatrix} 2 & 1&3 \\ 3&3&1\\ -1&3&1 \end{vmatrix}=6-1+27+9-3-6=32\\ \Delta_y=\begin{vmatrix} 1 & 2&3 \\ 5&3&1\\ 2&-1&1 \end{vmatrix}=3+4-15-18-10+1=-35

Ξ”z=∣11253323βˆ’1∣=βˆ’3+6+30βˆ’12+5βˆ’9=17\Delta_z=\begin{vmatrix} 1 & 1&2 \\ 5&3&3\\ 2&3&-1 \end{vmatrix}=-3+6+30-12+5-9=17

x=Ξ”1Ξ”=3224=43y=Ξ”2Ξ”=βˆ’3524=βˆ’3524z=Ξ”3Ξ”=1724x=\frac{\Delta _1}{\Delta}=\frac{32}{24}=\frac{4}{3}\\ y=\frac{\Delta _2}{\Delta}=\frac{-35}{24}=-\frac{35}{24}\\ z=\frac{\Delta _3}{\Delta}=\frac{17}{24}

2. Gauss elimination without pivoting

(113∣2531∣3231βˆ£βˆ’1)β†’\begin{pmatrix} 1 & 1&3 &|&2\\ 5&3&1&|&3\\ 2&3&1&|&-1 \end{pmatrix}\to

2r+1r(-5)

3r+1r(-2)

(113∣20βˆ’2βˆ’14βˆ£βˆ’701βˆ’5βˆ£βˆ’5)β†’\begin{pmatrix} 1 & 1&3 &|&2\\ 0&-2&-14&|&-7\\ 0&1&-5&|&-5 \end{pmatrix}\to

2r↔\leftrightarrow 3r

(113∣201βˆ’5βˆ£βˆ’500βˆ’24βˆ£βˆ’17)\begin{pmatrix} 1 & 1&3 &|&2\\ 0&1&-5&|&-5\\ 0&0&-24&|&-17\\ \end{pmatrix}

x+y+3z=2βˆ’2yβˆ’14z=βˆ’7βˆ’24z=βˆ’17↦z=1724x+y+3z=2\\ -2y-14z=-7\\ -24z=-17\mapsto z=\frac{17}{24}

βˆ’2yβˆ’14β‹…1724=βˆ’7↦x2=βˆ’3524xβˆ’3524+3β‹…1724=2↦x=34-2y-14\cdot\frac{17}{24}=-7\mapsto x_2=-\frac{35}{24}\\ x-\frac{35}{24}+3\cdot\frac{17}{24}=2\mapsto x=\frac{3}{4}


3. Gauss-Jordan elimination

(113∣2531∣3231βˆ£βˆ’1)β†’\begin{pmatrix} 1 & 1&3 &|&2\\ 5&3&1&|&3\\ 2&3&1&|&-1 \end{pmatrix}\to

2r+1r(-5)

3r+1r(-2)

(113∣20βˆ’2βˆ’14βˆ£βˆ’701βˆ’5βˆ£βˆ’5)β†’\begin{pmatrix} 1 & 1&3 &|&2\\ 0&-2&-14&|&-7\\ 0&1&-5&|&-5 \end{pmatrix}\to

2r↔\leftrightarrow 3r

(113∣201βˆ’5βˆ£βˆ’50βˆ’2βˆ’14βˆ£βˆ’7)β†’\begin{pmatrix} 1 & 1&3 &|&2\\ 0&1&-5&|&-5\\ 0&-2&-14&|&-7 \end{pmatrix}\to

3r+2r(2)

(113∣201βˆ’5βˆ£βˆ’500βˆ’24βˆ£βˆ’17)β†’\begin{pmatrix} 1 & 1&3 &|&2\\ 0&1&-5&|&-5\\ 0&0&-24&|&-17\\ \end{pmatrix}\to

3r(βˆ’124\frac{-1}{24} )

(113∣201βˆ’5βˆ£βˆ’5001∣1724)β†’\begin{pmatrix} 1 & 1&3 &|&2\\ 0&1&-5&|&-5\\ 0&0&1&|&\frac{17}{24} \end{pmatrix}\to

2r+3r(5)

1r+3r(-3)

(110βˆ£βˆ’324010βˆ£βˆ’3524001∣1724)β†’\begin{pmatrix} 1 & 1&0 &|&-\frac{3}{24}\\ 0&1&0&|&-\frac{35}{24}\\ 0&0&1&|&\frac{17}{24} \end{pmatrix}\to

1r+2r(-1)

(100∣3224010βˆ£βˆ’3524001∣1724)\begin{pmatrix} 1 & 0&0 &|&\frac{32}{24}\\ 0&1&0&|&-\frac{35}{24}\\ 0&0&1&|&\frac{17}{24} \end{pmatrix}

x=3224=43,y=βˆ’3524,z=1724x=\frac{32}{24}=\frac{4}{3}, y=-\frac{35}{24}, z=\frac{17}{24}


4. LU factorization

L=(1005102βˆ’121)L=\begin{pmatrix} 1 & 0&0 \\ 5 & 1&0\\ 2&-\frac{1}{2}&1 \end{pmatrix}

5: 2:

(1130βˆ’2βˆ’1401βˆ’5)\begin{pmatrix} 1 & 1&3 \\ 0&-2&-14\\ 0&1&-5 \end{pmatrix}

βˆ’12-\frac{1}{2} :

(1130βˆ’2βˆ’1400βˆ’12)=U\begin{pmatrix} 1 & 1&3 \\ 0&-2&-14\\ 0&0&-12 \end{pmatrix}=U

LY=BLY=B\\

(1005102βˆ’121)(y1y2y3)=(23βˆ’1)\begin{pmatrix} 1 & 0&0 \\ 5 & 1&0\\ 2&-\frac{1}{2}&1 \end{pmatrix}\begin{pmatrix} y_1\\ y_2\\ y_3 \end{pmatrix}=\begin{pmatrix} 2\\ 3\\ -1 \end{pmatrix}

y1=25y1+y2=3↦y2=βˆ’72y1βˆ’12y2+y3=βˆ’1↦y3=βˆ’172y_1=2\\ 5y_1+y_2=3\mapsto y_2=-7\\ 2y_1-\frac{1}{2}y_2+y_3=-1\mapsto y_3=-\frac{17}{2}

UX=YUX=Y\\

(1130βˆ’2βˆ’1400βˆ’12)(xyz)=(2βˆ’7βˆ’172)\begin{pmatrix} 1 & 1&3 \\ 0&-2&-14\\ 0&0&-12 \end{pmatrix}\begin{pmatrix} x \\ y\\z \end{pmatrix}=\begin{pmatrix} 2 \\ -7\\-\frac{17}{2} \end{pmatrix}

x+y+3z=2↦x=βˆ’3224βˆ’2yβˆ’14z=βˆ’7↦y=βˆ’3524βˆ’12z=βˆ’172↦z=1724x+y+3z=2\mapsto x=-\frac{32}{24} \\ -2y-14z=-7\mapsto y=-\frac{35}{24}\\ -12z=-\frac{17}{2}\mapsto z=\frac{17}{24}


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