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Question 112 of 118

Q.Solve the system of linear equations 2x+5y=−22x+5y=-2, x+2y=−3x+2y=-3 by matrix inversion method.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2025Subjective· 3mImportance★★★★★
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Writes the system as AX=BAX=B, finds A−1A^{-1} using the 2×22\times2 inverse formula, and computes X=A−1BX=A^{-1}B.

  1. Write the system 2x+5y=−2, x+2y=−32x+5y=-2,\ x+2y=-3 in matrix form AX=BAX=B with A=[2512],X=[xy],B=[−2−3]A=\begin{bmatrix}2&5\\1&2\end{bmatrix},\quad X=\begin{bmatrix}x\\y\end{bmatrix},\quad B=\begin{bmatrix}-2\\-3\end{bmatrix}.
  2. Find ∣A∣=2(2)−5(1)=4−5=−1|A|=2(2)-5(1)=4-5=-1. Since ∣A∣≠0|A|\ne0, AA is invertible and the system has a unique solution.
  3. For A=[abcd]A=\begin{bmatrix}a&b\\c&d\end{bmatrix}, adj A=[d−b−ca]\text{adj}\,A=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}. Here adj A=[2−5−12]\text{adj}\,A=\begin{bmatrix}2&-5\\-1&2\end{bmatrix}.
  4. So A−1=1∣A∣adj A=1−1[2−5−12]=[−251−2]A^{-1}=\dfrac{1}{|A|}\text{adj}\,A=\dfrac{1}{-1}\begin{bmatrix}2&-5\\-1&2\end{bmatrix}=\begin{bmatrix}-2&5\\1&-2\end{bmatrix}.
  5. Solve X=A−1BX=A^{-1}B: …

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