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Question of 146

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

Karnataka PUCKarnataka II PUC Board 2019Subjective· 5mImportance★★★★★
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det⁡A=−17e0\det A=-17 e0, so X=A−1BX=A^{-1}B gives x=1, y=2, z=3x=1,\,y=2,\,z=3.

Concept. For AX=BAX=B with det⁡Ae0\det A e0, the unique solution is X=A−1B=1det⁡A(adj A)BX=A^{-1}B=\dfrac{1}{\det A}(\text{adj }A)B.

Set-up.

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}.

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)\cdot4)+3(2(-3)-1\cdot4)

=3(2−3)+2(4+4)+3(−6−4)=−3+16−30=−17 (e0).=3(2-3)+2(4+4)+3(-6-4)=-3+16-30=-17\ ( e0).

Adjoint. The cofactors give

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

Solve X=1−17(adj A)BX=\dfrac{1}{-17}(\text{adj }A)B. …

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