Question #350138

Solve the following using matrix algebra

2X+Y-Z=5

3X-2Y+2Z=-3

X-3Y-3Z=-2


1
Expert's answer
2022-06-13T15:56:39-0400
A=(211322133),X=(xyz),B=(532)A=\begin{pmatrix} 2 & 1 & -1 \\ 3 & -2 & 2 \\ 1 & -3 & -3 \\ \end{pmatrix}, X=\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix}, B=\begin{pmatrix} 5 \\ -3 \\ -2 \\ \end{pmatrix}

AX=BAX=B

A1AX=A1BA^{-1}AX=A^{-1}B

X=A1BX=A^{-1}B

Augment the matrix with the identity matrix


(211100322010133001)\begin{pmatrix} 2 & 1 & -1& & 1 & 0 & 0 \\ 3 & -2 & 2 & & 0 & 1 & 0 \\ 1 & -3 & -3 & & 0 & 0 & 1 \\ \end{pmatrix}

R1=R1/2R_1=R_1/2


(11/21/21/200322010133001)\begin{pmatrix} 1 & 1/2 & -1/2& & 1/2 & 0 & 0 \\ 3 & -2 & 2 & & 0 & 1 & 0 \\ 1 & -3 & -3 & & 0 & 0 & 1 \\ \end{pmatrix}

R2=R23R1R_2=R_2-3R_1


(11/21/21/20007/27/23/210133001)\begin{pmatrix} 1 & 1/2 & -1/2& & 1/2 & 0 & 0 \\ 0 & -7/2 & 7/2 & & -3/2 & 1 & 0 \\ 1 & -3 & -3 & & 0 & 0 & 1 \\ \end{pmatrix}

R3=R3R1R_3=R_3-R_1


(11/21/21/20007/27/23/21007/25/21/201)\begin{pmatrix} 1 & 1/2 & -1/2& & 1/2 & 0 & 0 \\ 0 & -7/2 & 7/2 & & -3/2 & 1 & 0 \\ 0 & -7/2 & -5/2 & & -1/2 & 0 & 1 \\ \end{pmatrix}

R2=2R2/7R_2=-2R_2/7


(11/21/21/2000113/72/7007/25/21/201)\begin{pmatrix} 1 & 1/2 & -1/2& & 1/2 & 0 & 0 \\ 0 & 1 & -1 & & 3/7 & -2/7 & 0 \\ 0 & -7/2 & -5/2 & & -1/2 & 0 & 1 \\ \end{pmatrix}

R1=R1R2/2R_1=R_1-R_2/2


(1002/71/700113/72/7007/25/21/201)\begin{pmatrix} 1 & 0& 0 & & 2/7 & 1/7 & 0 \\ 0 & 1 & -1 & & 3/7 & -2/7 & 0 \\ 0 & -7/2 & -5/2 & & -1/2 & 0 & 1 \\ \end{pmatrix}

R3=R3+7R2/2R_3=R_3+7R_2/2


(1002/71/700113/72/70006111)\begin{pmatrix} 1 & 0& 0 & & 2/7 & 1/7 & 0 \\ 0 & 1 & -1 & & 3/7 & -2/7 & 0 \\ 0 & 0 & -6 & & 1 & -1 & 1 \\ \end{pmatrix}

R3=R3/6R_3=-R_3/6


(1002/71/700113/72/700011/61/61/6)\begin{pmatrix} 1 & 0& 0 & & 2/7 & 1/7 & 0 \\ 0 & 1 & -1 & & 3/7 & -2/7 & 0 \\ 0 & 0 & 1 & & -1/6 & 1/6 & -1/6 \\ \end{pmatrix}

R2=R2+R3R_2=R_2+R_3


(1002/71/7001011/425/421/60011/61/61/6)\begin{pmatrix} 1 & 0& 0 & & 2/7 & 1/7 & 0 \\ 0 & 1 & 0 & & 11/42 & -5/42 & -1/6 \\ 0 & 0 & 1 & & -1/6 & 1/6 & -1/6 \\ \end{pmatrix}

We are done. On the left is the identity matrix. On the right is the inverse matrix.


A1=(2/71/7011/425/421/61/61/61/6)A^{-1}=\begin{pmatrix} 2/7 & 1/7 & 0 \\ 11/42 & -5/42 & -1/6 \\ -1/6 & 1/6 & -1/6 \\ \end{pmatrix}

X=(xyz)=(2/71/7011/425/421/61/61/61/6)(532)X=\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix}=\begin{pmatrix} 2/7 & 1/7 & 0 \\ 11/42 & -5/42 & -1/6 \\ -1/6 & 1/6 & -1/6 \\ \end{pmatrix}\begin{pmatrix} 5\\ -3 \\ -2 \\ \end{pmatrix}

=(10/73/755/42+15/42+14/425/63/6+2/6)=(121)=\begin{pmatrix} 10/7-3/7\\ 55/42+15/42+14/42 \\ -5/6-3/6+2/6 \\ \end{pmatrix}=\begin{pmatrix} 1\\ 2\\ -1 \\ \end{pmatrix}

(1,2,1)(1, 2, -1)


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