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Q.Solve the following system of Linear equations by matrix method :
x+y+z=6x + y + z = 6
y+3z=11y + 3z = 11
x−2y+z=0x - 2y + z = 0.

Karnataka PUCKarnataka II PUC Board 2024Subjective· 5mImportance★★★★★
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Write the system as AX=BAX=B and solve X=A−1BX=A^{-1}B; the solution is x=1, y=2, z=3x=1,\ y=2,\ z=3.

Step 1 — Express in matrix form AX=BAX=B.

The equations x+y+z=6, y+3z=11, x−2y+z=0x+y+z=6,\ y+3z=11,\ x-2y+z=0 give

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}.

Step 2 — Evaluate ∣A∣|A|.

∣A∣=1(1⋅1−3⋅(−2))−1(0⋅1−3⋅1)+1(0⋅(−2)−1⋅1)|A|=1\big(1\cdot1-3\cdot(-2)\big)-1\big(0\cdot1-3\cdot1\big)+1\big(0\cdot(-2)-1\cdot1\big)

=1(1+6)−1(0−3)+1(0−1)=7+3−1=9.=1(1+6)-1(0-3)+1(0-1)=7+3-1=9.

Since ∣A∣=9≠0|A|=9\neq0, A−1A^{-1} exists and the system has a unique solution.

Step 3 — Cofactors and adjoint.

The cofactors are

A11=7, A12=3, A13=−1,A_{11}=7,\ A_{12}=3,\ A_{13}=-1,

A21=−3, A22=0, A23=3,A_{21}=-3,\ A_{22}=0,\ A_{23}=3,

A31=2, A32=−3, A33=1.A_{31}=2,\ A_{32}=-3,\ A_{33}=1.

The adjoint is the transpose of the cofactor matrix:

adj(A)=[7−3230−3−131].\text{adj}(A)=\begin{bmatrix}7&-3&2\\3&0&-3\\-1&3&1\end{bmatrix}.

Step 4 — Form A−1A^{-1} and solve X=A−1BX=A^{-1}B.

A−1=1∣A∣ adj(A)=19[7−3230−3−131].A^{-1}=\frac{1}{|A|}\,\text{adj}(A)=\frac19\begin{bmatrix}7&-3&2\\3&0&-3\\-1&3&1\end{bmatrix}. …

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