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Q.Solve the following system of linear equations, using matrix method- x+y+z=6x+y+z=6, x+3z=11x+3z=11, x−2y+z=0x-2y+z=0

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2023Subjective· 5mImportance★★★★★
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Write the system as AX=BAX=B, compute ∣A∣|A| and adj(A)\text{adj}(A), then X=A−1B=1∣A∣adj(A) BX=A^{-1}B=\dfrac{1}{|A|}\text{adj}(A)\,B.

System: x+y+z=6x+y+z=6, x+0y+3z=11x+0y+3z=11, x−2y+z=0x-2y+z=0.

A=[1111031−21]A=\begin{bmatrix}1&1&1\\1&0&3\\1&-2&1\end{bmatrix}, X=[xyz]X=\begin{bmatrix}x\\y\\z\end{bmatrix}, B=[6110]B=\begin{bmatrix}6\\11\\0\end{bmatrix}.

∣A∣=1(0⋅1−3⋅(−2))−1(1⋅1−3⋅1)+1(1⋅(−2)−0⋅1)=1(6)−1(−2)+1(−2)=6+2−2=6≠0|A|=1(0\cdot1-3\cdot(-2))-1(1\cdot1-3\cdot1)+1(1\cdot(-2)-0\cdot1)=1(6)-1(-2)+1(-2)=6+2-2=6\neq0.

Since ∣A∣≠0|A|\neq0, AA is invertible and a unique solution exists.

Cofactors: C11=6, C12=2, C13=−2, C21=−3, C22=0, C23=3, C31=3, C32=−2, C33=−1C_{11}=6,\ C_{12}=2,\ C_{13}=-2,\ C_{21}=-3,\ C_{22}=0,\ C_{23}=3,\ C_{31}=3,\ C_{32}=-2,\ C_{33}=-1.

adj(A)=[6−3320−2−23−1]\text{adj}(A)=\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix} (transpose of the cofactor matrix).

A−1=16[6−3320−2−23−1]A^{-1}=\dfrac{1}{6}\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix}.

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