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Question of 146

Q.Solve the following system of equations by matrix method :
3x−2y+3z=83x - 2y + 3z = 8
2x+y−z=12x + y - z = 1
4x−3y+2z=44x - 3y + 2z = 4.

Karnataka PUCKarnataka II PUC Board 2022Subjective· 5mImportance★★★★★
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Writing the system as AX=BAX=B and using X=A−1BX=A^{-1}B gives x=1, y=2, z=3x=1,\ y=2,\ z=3.

  1. Write the system in matrix form AX=BAX=B:

A=[3−2321−14−32],X=[xyz],B=[814].A=\begin{bmatrix}3&-2&3\\2&1&-1\\4&-3&2\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}8\\1\\4\end{bmatrix}.

  1. Find ∣A∣|A| (expand along Row 1):

∣A∣=3(1⋅2−(−1)(−3))−(−2)(2⋅2−(−1)⋅4)+3(2⋅(−3)−1⋅4)|A|=3(1\cdot2-(-1)(-3))-(-2)(2\cdot2-(-1)\cdot4)+3(2\cdot(-3)-1\cdot4)

=3(2−3)+2(4+4)+3(−6−4)=3(−1)+2(8)+3(−10)=−3+16−30=−17.=3(2-3)+2(4+4)+3(-6-4)=3(-1)+2(8)+3(-10)=-3+16-30=-17.

Since ∣A∣=−17≠0|A|=-17\neq0, A−1A^{-1} exists and the system has a unique solution.

  1. Find the cofactors CijC_{ij} of AA:

C11=−1, C12=−8, C13=−10,C_{11}=-1,\ C_{12}=-8,\ C_{13}=-10,

C21=−5, C22=−6, C23=1,C_{21}=-5,\ C_{22}=-6,\ C_{23}=1,

C31=−1, C32=9, C33=7.C_{31}=-1,\ C_{32}=9,\ C_{33}=7.

  1. Adjoint = transpose of the cofactor matrix:

adj⁡A=[−1−5−1−8−69−1017].\operatorname{adj}A=\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}.

  1. Inverse: A−1=1∣A∣adj⁡A=1−17[−1−5−1−8−69−1017].A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A=\dfrac{1}{-17}\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}. …

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