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Question of 146

Q.Solve the following system of equations using 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 2026Subjective· 5mImportance★★★★★
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Writing AX=BAX=B and using X=A−1BX=A^{-1}B with det⁡A=−17\det A=-17 gives x=1, y=2, z=3x=1,\ y=2,\ z=3.

Write the system as AX=BAX=B where

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}.

Step 1 — Determinant:

det⁡A=3(1⋅2−(−1)(−3))−(−2)(2⋅2−(−1)(4))+3(2(−3)−1⋅4)\det A = 3(1\cdot2-(-1)(-3)) -(-2)(2\cdot2-(-1)(4)) + 3(2(-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 det⁡A=−17≠0\det A=-17\neq0, A−1A^{-1} exists and the system has a unique solution.

Step 2 — Cofactors:

A11=(2−3)=−1,A12=−(4+4)=−8,A13=(−6−4)=−10,A_{11}=(2-3)=-1,\quad A_{12}=-(4+4)=-8,\quad A_{13}=(-6-4)=-10,

A21=−(−4+9)=−5,A22=(6−12)=−6,A23=−(−9+8)=1,A_{21}=-(-4+9)=-5,\quad A_{22}=(6-12)=-6,\quad A_{23}=-(-9+8)=1,

A31=(2−3)=−1,A32=−(−3−6)=9,A33=(3+4)=7.A_{31}=(2-3)=-1,\quad A_{32}=-(-3-6)=9,\quad A_{33}=(3+4)=7.

Step 3 — Adjoint (transpose of 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}.

Step 4 — Solve X=1det⁡A(adj⁡A)BX=\dfrac{1}{\det A}(\operatorname{adj}A)B: …

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