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Q.Solve the following system of Linear 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 2023Subjective· 5mImportance★★★★★
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Cast the system as AX=BAX=B; since det⁡A=−17≠0\det A=-17\neq0, A−1A^{-1} exists and X=A−1BX=A^{-1}B gives x=1, y=2, z=3x=1,\,y=2,\,z=3.

Step 1 — Matrix form.

A=[3−2321−14−32],X=[xyz],B=[814],AX=B.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},\qquad AX=B.

Step 2 — Determinant of AA.

det⁡A=3(1⋅2−(−1)(−3))−(−2)(2⋅2−(−1)⋅4)+3(2(−3)−1⋅4)\det A=3\big(1\cdot2-(-1)(-3)\big)-(-2)\big(2\cdot2-(-1)\cdot4\big)+3\big(2(-3)-1\cdot4\big)

=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 3 — Cofactors and adjoint.

The cofactors AijA_{ij} are

A11=−1, A12=−8, A13=−10,A_{11}=-1,\ A_{12}=-8,\ A_{13}=-10,

A21=−5, A22=−6, A23=1,A_{21}=-5,\ A_{22}=-6,\ A_{23}=1,

A31=−1, A32=9, A33=7.A_{31}=-1,\ A_{32}=9,\ A_{33}=7.

Taking the 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}.

Step 4 — Inverse and solution. …

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