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Question 111 of 121

Q.Solve the following equations by the method of reduction: x+3y+3z=12x+3y+3z=12; x+4y+4z=15x+4y+4z=15; x+3y+4z=13x+3y+4z=13

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2020Subjective· 4mImportance★★★★★
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Reduce the augmented matrix to (near) row-echelon form.

[133144134][xyz]=[121513]\begin{bmatrix}1 & 3 & 3\\ 1 & 4 & 4\\ 1 & 3 & 4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}12\\15\\13\end{bmatrix}

Augmented matrix: [133∣12144∣15134∣13]\begin{bmatrix}1 & 3 & 3 & | & 12\\ 1 & 4 & 4 & | & 15\\ 1 & 3 & 4 & | & 13\end{bmatrix}

R2→R2−R1R_2 \to R_2-R_1: [011∣3]\begin{bmatrix}0 & 1 & 1 & | & 3\end{bmatrix}

R3→R3−R1R_3 \to R_3-R_1: [001∣1]\begin{bmatrix}0 & 0 & 1 & | & 1\end{bmatrix}

Reduced system: …

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