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Solving systems of linear equations using matrix method.

Let's consider a system of n linear equations with n unknowns of the general form.

{a11x1+a12x2+...+a1nxn=b1a21x1+a22x2+...+a2nxn=b2..............................an1x1+an2x2+...+annxn=bn,(1)

or, in matrix form, AX=B, where

A=(a11a12...a1na21a22...a2n............an1an2...ann);

X=(x1x2...xn); B=(b1b2...bn).

If detA≠0, meaning the matrix A has an inverse, then system (1) has a unique solution given by X=A−1B.

Examples:

Solve the following systems using the matrix method:

1. {3x−5y=132x+7y=81 Solution.

The matrix A=(3−527) is non-singular because

detA=|3−527|=21+10=31≠0. Therefore, the system has a unique solution X=A−1B. Let's find the inverse matrix A−1:

Let's find the cofactors of the corresponding elements of matrix A:

A11=(−1)1+1⋅7=7;

A12=(−1)1+2⋅2=−2;

A21=(−1)2+1⋅(−5)=5;

A22=(−1)2+2⋅3=3.

From here, we find the adjugate matrix:

A∗=(75−23).

A−1=1detAA∗=131(75−23)=(731531−231331).

(xy)=A−1B=(731531−231331)(1381)=(731∗13+531∗81−231∗13+331∗81)=

=(91+40531−26+24331)=(167).

Answer: x=16; y=7.

2. {7x+2y+3z=155x−3y+2z=1510x−11y+5z=36

Solution.

The matrix A=(7235−3210−115) is non-singular because

detA=|7235−3210−115|=−105−165+40+90+154−50=−36≠0. Therefore, the system has a unique solution given by

X=A−1B. Let's find the inverse matrix A−1:

Let's find the cofactors of the corresponding elements of matrix A:

A11=(−1)1+1⋅|−32−115|=−15+22=7;

A12=(−1)1+2⋅|52105|=−(25−20)=−5;

A13=(−1)1+3⋅|5−310−11|=−55+30=−25;

A21=(−1)2+1⋅|23−115|=−(10+33)=−43;

A22=(−1)2+2⋅|73105|=35−30=5;

A23=(−1)2+3⋅|7210−11|=−(−77−20)=97;

A31=(−1)3+1⋅|23−32|=4+9=13;

A32=(−1)3+2⋅|7352|=−(14−15)=1;

A33=(−1)3+3⋅|725−3|=−21−10=−31;

From here, we find the adjugate matrix:

A∗=(7−4313−551−2597−31).

A−1=1detAA∗=1−36(7−4313−551−2597−31)=(−7364336−1336536−536−1362536−97363136).

(xyz)=A−1B=(−7364336−1336536−536−1362536−97363136)(151536)=

=(−736∗15+4336∗15−1336∗36536∗15−536∗15−136∗362536∗15−9736∗15+3136∗36)=(2−11).

Answer: x=2; y=−1; z=1.

3. {x+y−2z=62x+3y−7z=165x+2y+z=16

Solution.

The matrix A=(11−223−7521) is non-singular because

detA=|11−223−7521|=3−8−35+30+14−2=2≠0. Therefore, the system has a unique solution

X=A−1B. Let's find the inverse matrix A−1:

Let's find the cofactors of the corresponding elements of the matrix A:

A11=(−1)1+1⋅|3−721|=3+14=17;

A12=(−1)1+2⋅|2−751|=−(2+35)=−37;

A13=(−1)1+3⋅|2352|=4−15=−11;

A21=(−1)2+1⋅|1−221|=−(1+4)=−5;

A22=(−1)2+2⋅|1−251|=1+10=11;

A23=(−1)2+3⋅|1152|=−(2−5)=3;

A31=(−1)3+1⋅|1−23−7|=−7+6=−1;

A32=(−1)3+2⋅|1−22−7|=−(−7+4)=3;

A33=(−1)3+3⋅|1123|=3−2=1;

From here, we find the adjugate matrix:

A∗=(17−5−1−37113−1131).

A−1=1detAA∗=12(17−5−1−37113−1131)=(172−52−12−37211232−1123212).

(xyz)=A−1B=(172−52−12−37211232−1123212)(61616)=

=(172∗6−52∗16−12∗16−372∗6+112∗16+32∗16−112∗6+32∗16+12∗16)=(31−1).

Answer: x=3; y=1; z=−1.

Homework:

Solve the systems of equations using the matrix method.

1. {3x−4y=−63x+4y=18.

Answer: x=2;y=3.

2. {2x+y=5x+3z=165y−z=10.

Answer: x=1;y=3;z=5.

3. {4x1+4x2+5x3+5x4=02x1+3x3−x4=10x1+x2−5x3=−103x2+2x3=1.

Answer: x1=1;x2=−1;x3=2;x4=−2.

Tags: inverse matrix, matrix, matrix method, systems of linear equations

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