Question #350968

Solve the system of linear equations using matrix inverse method.

x-y+z=1

2y-z=1

2x+3y=1


1
Expert's answer
2022-06-16T10:31:09-0400
A=(111021230),X=(xyz),B=(111)A=\begin{pmatrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{pmatrix}, X=\begin{pmatrix} x \\ y\\ z \end{pmatrix}, B=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}

AX=BAX=B

A1AX=A1B=>X=A1BA^{-1}AX=A^{-1}B=>X=A^{-1}B

augment the matrix with the identity matrix:


(111100021010230001)\begin{pmatrix} 1 & -1 & 1 & & 1& 0 & 0 \\ 0 & 2 & -1 & & 0& 1 & 0 \\ 2 & 3 & 0 & & 0 & 0 & 1 \\ \end{pmatrix}

R3=R32R1R_3=R_3-2R_1


(111100021010052201)\begin{pmatrix} 1 & -1 & 1 & & 1& 0 & 0 \\ 0 & 2 & -1 & & 0& 1 & 0 \\ 0 & 5 & -2 & & -2 & 0 & 1 \\ \end{pmatrix}

R2=R2/2R_2=R_2/2


(111100011/201/20052201)\begin{pmatrix} 1 & -1 & 1 & & 1& 0 & 0 \\ 0 & 1 & -1/2 & & 0& 1/2 & 0 \\ 0 & 5 & -2 & & -2 & 0 & 1 \\ \end{pmatrix}

R1=R1+R2R_1=R_1+R_2


(101/211/20011/201/20052201)\begin{pmatrix} 1 & 0 & 1/2 & & 1& 1/2 & 0 \\ 0 & 1 & -1/2 & & 0& 1/2 & 0 \\ 0 & 5 & -2 & & -2 & 0 & 1 \\ \end{pmatrix}

R3=R35R2R_3=R_3-5R_2


(101/211/20011/201/20001/225/21)\begin{pmatrix} 1 & 0 & 1/2 & & 1& 1/2 & 0 \\ 0 & 1 & -1/2 & & 0& 1/2 & 0 \\ 0 & 0 & 1/2 & & -2 & -5/2 & 1 \\ \end{pmatrix}

R3=2R3R_3=2R_3


(101/211/20011/201/20001452)\begin{pmatrix} 1 & 0 & 1/2 & & 1& 1/2 & 0 \\ 0 & 1 & -1/2 & & 0& 1/2 & 0 \\ 0 & 0 & 1 & & -4 & -5 & 2 \\ \end{pmatrix}

R1=R1R3/2R_1=R_1-R_3/2


(100331011/201/20001452)\begin{pmatrix} 1 & 0 & 0 & & 3 & 3 & -1 \\ 0 & 1 & -1/2 & & 0& 1/2 & 0 \\ 0 & 0 & 1 & & -4 & -5 & 2 \\ \end{pmatrix}

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


(100331010221001452)\begin{pmatrix} 1 & 0 & 0 & & 3 & 3 & -1 \\ 0 & 1 & 0 & & -2 & -2 & 1 \\ 0 & 0 & 1 & & -4 & -5 & 2 \\ \end{pmatrix}

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


A1=(331221452)A^{-1}=\begin{pmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{pmatrix}

X=A1BX=A^{-1}B

=(331221452)(111)=\begin{pmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{pmatrix}\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}

=(3+3122+145+2)=(537)=\begin{pmatrix} 3+3-1\\ -2-2+1\\ -4-5+2 \end{pmatrix}=\begin{pmatrix} 5\\ -3\\ -7 \end{pmatrix}

(5,3,7)(5, -3, -7)


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