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Exercise 2.3 · Q33

Q.If three numbers are added, their sum is 2. If 2 times the second number is subtracted from the sum of first and third number we get 8 and if three times the first number is added to the sum of second and third number we get 4. Find the numbers using matrices.

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Step 1: Let the three numbers be x,y,zx,y,z. "Their sum is 2": x+y+z=2x+y+z=2. "2 times the second number subtracted from the sum of first and third gives 8": (x+z)−2y=8(x+z)-2y=8, i.e. x−2y+z=8x-2y+z=8. "Three times the first number added to the sum of second and third gives 4": 3x+(y+z)=43x+(y+z)=4, i.e. 3x+y+z=43x+y+z=4.

Step 2: In matrix form, [1111−21311][xyz]=[284]\begin{bmatrix}1&1&1\\1&-2&1\\3&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\8\\4\end{bmatrix}, i.e. AX=BAX=B.

Step 3: ∣A∣|A|, expanding along row 1: 1(−2−1)−1(1−3)+1(1+6)=−3+2+7=6≠01(-2-1)-1(1-3)+1(1+6)=-3+2+7=6\neq0.

Step 4: Computing all nine cofactors and transposing gives adj A=[−3032−2072−3]\text{adj}\,A=\begin{bmatrix}-3&0&3\\2&-2&0\\7&2&-3\end{bmatrix}, so A−1=16[−3032−2072−3]A^{-1}=\dfrac{1}{6}\begin{bmatrix}-3&0&3\\2&-2&0\\7&2&-3\end{bmatrix}. …

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