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Q.Solve the following system of linear equations, using matrix method - x−y+2z=7x - y + 2z = 7 3x+4y−5z=−53x + 4y - 5z = -5 2x−y+3z=122x - y + 3z = 12

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2022Subjective· 6mImportance★★★★★
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Write the system as AX=BAX=B, find A−1A^{-1} via the adjoint (cofactors), then X=A−1BX=A^{-1}B.

x−y+2z=7,3x+4y−5z=−5,2x−y+3z=12x-y+2z=7,\quad 3x+4y-5z=-5,\quad 2x-y+3z=12

A=[1−1234−52−13],X=[xyz],B=[7−512]A=\begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}7\\-5\\12\end{bmatrix}

Determinant:

∣A∣=1[(4)(3)−(−5)(−1)]−(−1)[(3)(3)−(−5)(2)]+2[(3)(−1)−(4)(2)]|A| = 1[(4)(3)-(-5)(-1)] -(-1)[(3)(3)-(-5)(2)] + 2[(3)(-1)-(4)(2)]

=1(12−5)+1(9+10)+2(−3−8)=7+19−22=4e0= 1(12-5) + 1(9+10) + 2(-3-8) = 7+19-22 = 4 e 0

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

Cofactors:

C11=7, C12=−19, C13=−11C_{11}=7,\ C_{12}=-19,\ C_{13}=-11

C21=1, C22=−1, C23=−1C_{21}=1,\ C_{22}=-1,\ C_{23}=-1

C31=−3, C32=11, C33=7C_{31}=-3,\ C_{32}=11,\ C_{33}=7

adj(A)=[71−3−19−111−11−17],A−1=14[71−3−19−111−11−17]\text{adj}(A) = \begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}, \qquad A^{-1} = \frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}

Solve X=A−1BX=A^{-1}B: …

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