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Q.Solve the following equations by matrix method : x+2y−3z=4x + 2y - 3z = 4, 2x+4y−5z=122x + 4y - 5z = 12, 3x−y+z=33x - y + z = 3.

Odisha ChseOdisha CHSE +2 Science Board Exam 2023Subjective· 6mImportance★★★★★
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Writing the system as AX=BAX=B, computing A−1A^{-1}, and finding X=A−1BX=A^{-1}B gives x=2, y=7, z=4x=2,\ y=7,\ z=4.

The system is x+2y−3z=4x+2y-3z=4, 2x+4y−5z=122x+4y-5z=12, 3x−y+z=33x-y+z=3, i.e. AX=BAX=B with

A=[12−324−53−11]A=\begin{bmatrix}1&2&-3\\2&4&-5\\3&-1&1\end{bmatrix}, B=[4123]B=\begin{bmatrix}4\\12\\3\end{bmatrix}

∣A∣=1(4⋅1−(−5)(−1))−2(2⋅1−(−5)(3))+(−3)(2(−1)−4⋅3)|A| = 1(4\cdot1-(-5)(-1)) - 2(2\cdot1-(-5)(3)) + (-3)(2(-1)-4\cdot3)

=1(4−5)−2(2+15)−3(−2−12)=−1−34+42=7= 1(4-5) - 2(2+15) - 3(-2-12) = -1-34+42 = 7

Since ∣A∣=7≠0|A|=7\ne0, a unique solution exists.

Cofactors: C11=−1, C12=−17, C13=−14, C21=1, C22=10, C23=7, C31=2, C32=−1, C33=0C_{11}=-1,\,C_{12}=-17,\,C_{13}=-14,\,C_{21}=1,\,C_{22}=10,\,C_{23}=7,\,C_{31}=2,\,C_{32}=-1,\,C_{33}=0

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