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Q.Solve the system of linear equations using matrix method. x+y+z=6x+y+z=6, y+3z=11y+3z=11, x−2y+z=0x-2y+z=0. OR Obtain the inverse of the matrix using elementary operations: A=[20−1510013]A=\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}.

Jharkhand JacJAC Intermediate Board 2018Subjective· 6mImportance★★★★★
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Write the system as AX=BAX=B, confirm det⁡(A)≠0\det(A)\ne0, then solve by Cramer's rule (equivalent to the matrix/inverse method).

The system is:

x+y+z=6,y+3z=11,x−2y+z=0x+y+z=6,\qquad y+3z=11,\qquad x-2y+z=0

In matrix form AX=BAX=B:

A=[1110131−21],X=[xyz],B=[6110]A=\begin{bmatrix}1&1&1\\0&1&3\\1&-2&1\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}6\\11\\0\end{bmatrix}

Determinant of AA:

det⁡A=1(1⋅1−3⋅(−2))−1(0⋅1−3⋅1)+1(0⋅(−2)−1⋅1)=1(7)−1(−3)+1(−1)=7+3−1=9\det A = 1(1\cdot1-3\cdot(-2)) - 1(0\cdot1-3\cdot1) + 1(0\cdot(-2)-1\cdot1) = 1(7)-1(-3)+1(-1) = 7+3-1=9

Since det⁡A=9≠0\det A = 9\ne 0, a unique solution exists. Using Cramer's rule, replace each column of AA with BB in turn:

det⁡Ax=∣61111130−21∣=6(1+6)−1(11−0)+1(−22−0)=42−11−22=9  ⇒  x=99=1\det A_x = \begin{vmatrix}6&1&1\\11&1&3\\0&-2&1\end{vmatrix} = 6(1+6)-1(11-0)+1(-22-0) = 42-11-22=9 \;\Rightarrow\; x=\frac{9}{9}=1

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