Question:
Let \(A = \left[ \begin{array}{ccc}3 & 1 & 2 \\ 1 & 2 & 0 \\ 1 & 1 & 4\end{array} \right]\) and \(pC_{11}+4C_{21}-5C_{32} = -2\), where \(C_{ij}\) denotes the cofactor of an element \(a_{ij}\) of matrix \(A\), then the value of \(p\) is :
Let \(A = \left[ \begin{array}{ccc}3 & 1 & 2 \\ 1 & 2 & 0 \\ 1 & 1 & 4\end{array} \right]\) and \(pC_{11}+4C_{21}-5C_{32} = -2\), where \(C_{ij}\) denotes the cofactor of an element \(a_{ij}\) of matrix \(A\), then the value of \(p\) is :
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Cofactor C_ij is (-1)^(i+j) times the minor M_ij.
Updated On: Oct 1, 2026
- \(-2\)
- \(2\)
- \(4\)
- \(3\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
The cofactor of \(a_{ij}\) is \(C_{ij}=(-1)^{i+j}M_{ij}\), where \(M_{ij}\) is the determinant left after deleting row \(i\) and column \(j\).
Step 2: Find the three cofactors
\(C_{11}=+\begin{vmatrix}2&0\\1&4\end{vmatrix}=8\).
\(C_{21}=-\begin{vmatrix}1&2\\1&4\end{vmatrix}=-(4-2)=-2\).
\(C_{32}=-\begin{vmatrix}3&2\\1&0\end{vmatrix}=-(0-2)=2\).
Step 3: Substitute
\[ p(8)+4(-2)-5(2)=-2 \]
\[ 8p-8-10=-2\Rightarrow8p=16\Rightarrow p=2 \]
Step 4: Conclusion
The value is \(p=2\), option (B).
Final Answer:
The cofactors are 8, -2 and 2, which give p = 2, option (B). \[ \boxed{2} \]
The cofactor of \(a_{ij}\) is \(C_{ij}=(-1)^{i+j}M_{ij}\), where \(M_{ij}\) is the determinant left after deleting row \(i\) and column \(j\).
Step 2: Find the three cofactors
\(C_{11}=+\begin{vmatrix}2&0\\1&4\end{vmatrix}=8\).
\(C_{21}=-\begin{vmatrix}1&2\\1&4\end{vmatrix}=-(4-2)=-2\).
\(C_{32}=-\begin{vmatrix}3&2\\1&0\end{vmatrix}=-(0-2)=2\).
Step 3: Substitute
\[ p(8)+4(-2)-5(2)=-2 \]
\[ 8p-8-10=-2\Rightarrow8p=16\Rightarrow p=2 \]
Step 4: Conclusion
The value is \(p=2\), option (B).
Final Answer:
The cofactors are 8, -2 and 2, which give p = 2, option (B). \[ \boxed{2} \]
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