KCET · Maths · Matrices
If the determinant of the adjoint of a (real) matrix of order 3 is 25 , then the determinant of the inverse of the matrix is
- A \(0.2\)
- B \(\pm 5\)
- C \(\frac{1}{\sqrt[5]{625}}\)
- D \(\pm 0.2\)
Answer & Solution
Correct Answer
(D) \(\pm 0.2\)
Step-by-step Solution
Detailed explanation
Given, the determinant of the adjoint of a (real) matrix of order 3 is \(25 .\)
i.e., \(\quad|\operatorname{adj} A|=25...(i)\)
We know that,
\(|\operatorname{adj} A|=|A|^{n-1} \quad\) (here, \(n=3\) )
\(\Rightarrow \quad|A|^{3-1}=|A|^{2}=25 \quad\) [from Eq. (i)]
\(\Rightarrow \quad|A|=\pm 5\)
\(\therefore \quad\left|A^{-1}\right|=|A|^{-1}=\frac{1}{|A|}=\pm \frac{1}{5}=\pm 0.2\) (by property)
i.e., \(\quad|\operatorname{adj} A|=25...(i)\)
We know that,
\(|\operatorname{adj} A|=|A|^{n-1} \quad\) (here, \(n=3\) )
\(\Rightarrow \quad|A|^{3-1}=|A|^{2}=25 \quad\) [from Eq. (i)]
\(\Rightarrow \quad|A|=\pm 5\)
\(\therefore \quad\left|A^{-1}\right|=|A|^{-1}=\frac{1}{|A|}=\pm \frac{1}{5}=\pm 0.2\) (by property)
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