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Question 44 of 47

Q.Solve : ∣7411−35x−x31∣=0\begin{vmatrix} 7 & 4 & 11 \\ -3 & 5 & x \\ -x & 3 & 1 \end{vmatrix}=0

Puducherry TnboardTamil Nadu HSC First Year (DGE) Commerce Board 2025Subjective· 2mImportance★★★★★
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Expand the determinant along the first row, simplify to 2x2−17x+26=02x^2-17x+26=0, and solve: x=2x=2 or x=132x=\tfrac{13}{2}.

Given: ∣7411−35x−x31∣=0.\begin{vmatrix} 7 & 4 & 11 \\ -3 & 5 & x \\ -x & 3 & 1 \end{vmatrix}=0.

Step 1 — expand along the first row.

7∣5x31∣−4∣−3x−x1∣+11∣−35−x3∣=0.7\begin{vmatrix}5 & x\\3 & 1\end{vmatrix}-4\begin{vmatrix}-3 & x\\-x & 1\end{vmatrix}+11\begin{vmatrix}-3 & 5\\-x & 3\end{vmatrix}=0.

Step 2 — evaluate each 2×22\times2 minor.

7(5−3x)−4(−3+x2)+11(−9+5x)=0.7(5-3x)-4(-3+x^2)+11(-9+5x)=0.

Step 3 — simplify.

35−21x+12−4x2−99+55x=0  ⇒  −4x2+34x−52=0.35-21x+12-4x^2-99+55x=0\;\Rightarrow\;-4x^2+34x-52=0.

Divide by −2-2: …

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