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Question 45 of 47

Q.Solve by matrix inversion method.
2x+3y−5=02x+3y-5=0; x−2y+1=0x-2y+1=0.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2025Subjective· 3mImportance★★★★★
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A=[231−2]A=\begin{bmatrix}2&3\\1&-2\end{bmatrix}, ∣A∣=−7|A|=-7; X=A−1BX=A^{-1}B gives x=1, y=1x=1,\ y=1.

Given system: 2x+3y−5=0, x−2y+1=02x+3y-5=0,\ x-2y+1=0, i.e. 2x+3y=5, x−2y=−12x+3y=5,\ x-2y=-1.

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

A=[231−2],X=[xy],B=[5−1].A=\begin{bmatrix}2&3\\1&-2\end{bmatrix},\quad X=\begin{bmatrix}x\\y\end{bmatrix},\quad B=\begin{bmatrix}5\\-1\end{bmatrix}.

Step 2 — find ∣A∣|A| and A−1A^{-1}.

∣A∣=(2)(−2)−(3)(1)=−4−3=−7eq0.|A|=(2)(-2)-(3)(1)=-4-3=-7 eq0.

A−1=1∣A∣ adj(A)=1−7[−2−3−12].A^{-1}=\frac{1}{|A|}\,\text{adj}(A)=\frac{1}{-7}\begin{bmatrix}-2&-3\\-1&2\end{bmatrix}.

Step 3 — compute X=A−1BX=A^{-1}B. …

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