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the value of begin vmatri...
Question:
medium
The value of
\[ \begin{vmatrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{vmatrix} \textbf{is:} \]
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To calculate the determinant of a \( 3 \times 3 \) matrix, expand along any row or column and simplify using \( 2 \times 2 \) minors.
CBSE Class XII - 2024
CBSE Class XII
Updated On:
Feb 25, 2026
\( 0 \)
\( 2 \)
\( 7 \)
\( -2 \)
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The Correct Option is
A
Solution and Explanation
Step 1:
Formulate the determinant.
The determinant to be calculated is: \[ \begin{vmatrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{vmatrix}. \]
Step 2:
Expand the determinant along the first row.
The expansion along the first row is: \[ \begin{vmatrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{vmatrix} = 8 \begin{vmatrix} 3 & 5 \\ 4 & 3 \end{vmatrix} - 2 \begin{vmatrix} 12 & 5 \\ 16 & 3 \end{vmatrix} + 7 \begin{vmatrix} 12 & 3 \\ 16 & 4 \end{vmatrix}. \]
Step 3:
Compute the \( 2 \times 2 \) minors.
1. For \( \begin{vmatrix} 3 & 5 \\ 4 & 3 \end{vmatrix} \): \[ \begin{vmatrix} 3 & 5 \\ 4 & 3 \end{vmatrix} = (3)(3) - (5)(4) = 9 - 20 = -11. \] 2. For \( \begin{vmatrix} 12 & 5 \\ 16 & 3 \end{vmatrix} \): \[ \begin{vmatrix} 12 & 5 \\ 16 & 3 \end{vmatrix} = (12)(3) - (5)(16) = 36 - 80 = -44. \] 3. For \( \begin{vmatrix} 12 & 3 \\ 16 & 4 \end{vmatrix} \): \[ \begin{vmatrix} 12 & 3 \\ 16 & 4 \end{vmatrix} = (12)(4) - (3)(16) = 48 - 48 = 0. \]
Step 4:
Substitute the minor values back into the determinant expression.
\[ \begin{vmatrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{vmatrix} = 8(-11) - 2(-44) + 7(0) = -88 + 88 + 0 = 0. \]
Conclusion:
The determinant evaluates to \( 0 \).
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