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Q.Solve the system of the linear equations, using matrix method: 3x−y+z=53x - y + z = 5, 2x−2y+3z=72x - 2y + 3z = 7, x+y−z=−1x + y - z = -1.

Jharkhand JacJAC Intermediate Board 2023Subjective· 5mImportance★★★★★
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Write the system as AX=BAX=B, verify det⁡A≠0\det A\ne0, then solve using Cramer's rule (equivalent to the matrix-inverse method).

System: 3x−y+z=53x-y+z=5; 2x−2y+3z=72x-2y+3z=7; x+y−z=−1x+y-z=-1.

A=[3−112−2311−1]A=\begin{bmatrix}3&-1&1\\2&-2&3\\1&1&-1\end{bmatrix}, B=[57−1]B=\begin{bmatrix}5\\7\\-1\end{bmatrix}.

det⁡A=3[(−2)(−1)−3(1)]−(−1)[2(−1)−3(1)]+1[2(1)−(−2)(1)]\det A = 3[(-2)(-1)-3(1)] -(-1)[2(-1)-3(1)] +1[2(1)-(-2)(1)]

=3(2−3)+1(−2−3)+1(2+2)=−3−5+4=−4= 3(2-3) + 1(-2-3) + 1(2+2) = -3-5+4 = -4.

Since det⁡A=−4≠0\det A = -4\ne 0, a unique solution exists. Using Cramer's rule:

Dx=∣5−117−23−11−1∣=5(2−3)+1(−7+3)+1(7−2)=−5−4+5=−4D_x = \begin{vmatrix}5&-1&1\\7&-2&3\\-1&1&-1\end{vmatrix} = 5(2-3)+1(-7+3)+1(7-2) = -5-4+5=-4, so x=Dxdet⁡A=−4−4=1x=\dfrac{D_x}{\det A}=\dfrac{-4}{-4}=1.

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