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Q.Solve the system of linear equations using matrix method: 2x+3y+3z=52x+3y+3z=5, x−2y+z=−4x-2y+z=-4, 3x−y−2z=33x-y-2z=3.

Jharkhand JacJAC Intermediate Board 2026Subjective· 5mImportance★★★★★
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Write the system as AX=BAX=B, find A−1A^{-1} via the adjoint (since det⁡A≠0\det A\ne0), and compute X=A−1BX=A^{-1}B.

System: 2x+3y+3z=52x+3y+3z=5, x−2y+z=−4x-2y+z=-4, 3x−y−2z=33x-y-2z=3.

Write as AX=BAX=B where A=[2331−213−1−2]A=\begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix}, X=[xyz]X=\begin{bmatrix}x\\y\\z\end{bmatrix}, B=[5−43]B=\begin{bmatrix}5\\-4\\3\end{bmatrix}.

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

=2(4+1)−3(−2−3)+3(−1+6)=2(5)−3(−5)+3(5)=10+15+15=40≠0= 2(4+1) - 3(-2-3) + 3(-1+6) = 2(5)-3(-5)+3(5) = 10+15+15=40\ne 0, so A−1A^{-1} exists.

Cofactors: C11=5, C12=5, C13=5, C21=3, C22=−13, C23=11, C31=9, C32=1, C33=−7C_{11}=5,\ C_{12}=5,\ C_{13}=5,\ C_{21}=3,\ C_{22}=-13,\ C_{23}=11,\ C_{31}=9,\ C_{32}=1,\ C_{33}=-7

Adjoint (transpose of cofactor matrix): adj(A)=[5395−131511−7]\text{adj}(A) = \begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}

A−1=140adj(A)A^{-1} = \dfrac{1}{40}\text{adj}(A)

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