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The value of ∣265240219240225198219198181∣\begin{vmatrix}265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181\end{vmatrix} is

Solution

✅ Correct Option: 3

The given determinant is:

∣265240219240225198219198181∣\begin{vmatrix}265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181\end{vmatrix}

A determinant equals zero when the rows (or columns) are linearly dependent, meaning one row can be expressed as a combination of other rows.


Applying R₁ → R₁ - R₂:

∣265−240240−225219−198240225198219198181∣\begin{vmatrix}265-240 & 240-225 & 219-198 \\ 240 & 225 & 198 \\ 219 & 198 & 181\end{vmatrix}

=∣251521240225198219198181∣= \begin{vmatrix}25 & 15 & 21 \\ 240 & 225 & 198 \\ 219 & 198 & 181\end{vmatrix}


Applying R₂ → R₂ - R₃:

∣251521240−219225−198198−181219198181∣\begin{vmatrix}25 & 15 & 21 \\ 240-219 & 225-198 & 198-181 \\ 219 & 198 & 181\end{vmatrix}

=∣251521212717219198181∣= \begin{vmatrix}25 & 15 & 21 \\ 21 & 27 & 17 \\ 219 & 198 & 181\end{vmatrix}


Applying R₁ → R₁ - R₂:

∣25−2115−2721−17212717219198181∣\begin{vmatrix}25-21 & 15-27 & 21-17 \\ 21 & 27 & 17 \\ 219 & 198 & 181\end{vmatrix}

=∣4−124212717219198181∣= \begin{vmatrix}4 & -12 & 4 \\ 21 & 27 & 17 \\ 219 & 198 & 181\end{vmatrix}

The first row can be written as 4(1,−3,1)4(1, -3, 1). Continuing this process reveals that the rows are linearly dependent.


Therefore, the value of the determinant is 00.

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