KCET · Maths · Matrices
If \(P=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]\) is the adjoint of a \(3 \times 3\)
matrix \(A\) and \(|A|=4\), then \(\alpha\) is equal to
- A \(4\)
- B \(5\)
- C \(11\)
- D \(0\)
Answer & Solution
Correct Answer
(C) \(11\)
Step-by-step Solution
Detailed explanation
\(P=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]\)
For \(3 \times 3\), matrix,
\(|\operatorname{adj} A|=|A|^2\)
\(|\operatorname{adj} A|=(4)^2=16\)
Now, \(\quad P=\left|\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right|=16\)
\(\Rightarrow \quad 1(12-12)-\alpha(4-6)+3(4-6)=16\)
\(\Rightarrow \quad 0+2 \alpha-6=16\)
\(\Rightarrow \quad 2 \alpha=22\)
Hence, \(\alpha=11\)
For \(3 \times 3\), matrix,
\(|\operatorname{adj} A|=|A|^2\)
\(|\operatorname{adj} A|=(4)^2=16\)
Now, \(\quad P=\left|\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right|=16\)
\(\Rightarrow \quad 1(12-12)-\alpha(4-6)+3(4-6)=16\)
\(\Rightarrow \quad 0+2 \alpha-6=16\)
\(\Rightarrow \quad 2 \alpha=22\)
Hence, \(\alpha=11\)
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