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Q.Solve the system of linear equations x+y+2z=0x + y + 2z = 0, x+2y−z=9x + 2y - z = 9, x−3y+3z=−14x - 3y + 3z = -14 by using matrix method.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2019Subjective· 3mImportance★★★★★
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Write the system as AX=BAX=B, find A−1A^{-1} via the adjoint, then X=A−1BX=A^{-1}B gives x=1,y=3,z=−2x=1,y=3,z=-2.

Write the system as AX=BAX=B with A=[11212−11−33], X=[xyz], B=[09−14]A=\begin{bmatrix}1&1&2\\1&2&-1\\1&-3&3\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ B=\begin{bmatrix}0\\9\\-14\end{bmatrix}

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

=1(6−3)−1(3+1)+2(−3−2)=3−4−10=−11≠0=1(6-3)-1(3+1)+2(-3-2) = 3-4-10 = -11\ne 0

Cofactors: C11=3, C12=−4, C13=−5, C21=−9, C22=1, C23=4, C31=−5, C32=3, C33=1C_{11}=3,\ C_{12}=-4,\ C_{13}=-5,\ C_{21}=-9,\ C_{22}=1,\ C_{23}=4,\ C_{31}=-5,\ C_{32}=3,\ C_{33}=1

adj(A)=[3−9−5−413−541],A−1=1−11[3−9−5−413−541]adj(A)=\begin{bmatrix}3&-9&-5\\-4&1&3\\-5&4&1\end{bmatrix},\qquad A^{-1}=\dfrac{1}{-11}\begin{bmatrix}3&-9&-5\\-4&1&3\\-5&4&1\end{bmatrix}

X=A−1BX=A^{-1}B:

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