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Q.Solve the following system of linear equations by matrix method 4x+3y+2z=604x+3y+2z=60, 2x+4y+6z=902x+4y+6z=90, 6x+2y+3z=706x+2y+3z=70.

Karnataka PUCKarnataka II PUC Board 2025Subjective· 5mImportance★★★★★
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Matrix method X=A−1BX=A^{-1}B for the 3×33\times3 system gives x=5, y=8, z=8x=5,\ y=8,\ z=8.

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

A=[432246623],X=[xyz],B=[609070].A=\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}60\\90\\70\end{bmatrix}.

Step 1 — Determinant.

∣A∣=4(4⋅3−6⋅2)−3(2⋅3−6⋅6)+2(2⋅2−4⋅6)=4(0)−3(−30)+2(−20)=0+90−40=50.|A|=4(4\cdot3-6\cdot2)-3(2\cdot3-6\cdot6)+2(2\cdot2-4\cdot6)=4(0)-3(-30)+2(-20)=0+90-40=50.

Since ∣A∣=50≠0|A|=50\neq0, A−1A^{-1} exists and the system has a unique solution.

Step 2 — Cofactors.

A11=0, A12=30, A13=−20,A_{11}=0,\ A_{12}=30,\ A_{13}=-20,

A21=−5, A22=0, A23=10,A_{21}=-5,\ A_{22}=0,\ A_{23}=10,

A31=10, A32=−20, A33=10.A_{31}=10,\ A_{32}=-20,\ A_{33}=10.

Step 3 — Adjoint (transpose of the cofactor matrix):

adj A=[0−510300−20−201010].\text{adj}\,A=\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}.

Step 4 — Solve X=A−1B=1∣A∣(adj A)BX=A^{-1}B=\dfrac{1}{|A|}(\text{adj}\,A)B: …

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