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Q.Solve the following system of equations by matrix method: 3x−2y+3z=83x-2y+3z=8, 2x+y−z=12x+y-z=1 and 4x−3y+2z=44x-3y+2z=4. OR Solve the differential equation (tan⁡−1y−x) dy=(1+y2) dx(\tan^{-1}y-x)\, dy=(1+y^2)\, dx.

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2023Subjective· 8mImportance★★★★★
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Form AX=BAX=B, confirm det⁡A=−17≠0\det A=-17\ne0, and solve X=A−1BX=A^{-1}B to get x=1,y=2,z=3x=1,y=2,z=3. (Answering the main part of the OR.)

Concept. A system AX=BAX=B with det⁡A≠0\det A\ne0 has the unique solution X=A−1BX=A^{-1}B, where A−1=1det⁡A adj AA^{-1}=\dfrac{1}{\det A}\,\text{adj}\,A.

Matrix form.

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\cdot(-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 (transpose of the cofactor matrix):

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=1det⁡A(adj A)BX=\dfrac{1}{\det A}(\text{adj}\,A)B: …

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