MHT CET · Maths · Matrices
If \(A=\left[\begin{array}{ll}2 & -3 \\ 5 & -7\end{array}\right]\), then \(2 A-3 A^{-1}=\)
- A \(\left[\begin{array}{cc}25 & 15 \\ 25 & 20\end{array}\right]\)
- B \(\left[\begin{array}{cc}25 & 25 \\ -15 & -20\end{array}\right]\)
- C \(\left[\begin{array}{cc}25 & -15 \\ 25 & -20\end{array}\right]\)
- D \(\left[\begin{array}{cc}25 & -25 \\ -15 & -20\end{array}\right]\)
Answer & Solution
Correct Answer
(C) \(\left[\begin{array}{cc}25 & -15 \\ 25 & -20\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(A=\left[\begin{array}{ll}2 & -3 \\ 5 & -7\end{array}\right] \Rightarrow|A|=-14+15=1\)
\(\therefore A^{-1}=\left[\begin{array}{cc}-7 & 3 \\ -5 & 2\end{array}\right] \quad\) and \(\quad 2 A=\left[\begin{array}{cc}4 & -6 \\ 10 & -14\end{array}\right]\)
\(\therefore 2 A-3 A^{-1}=\left[\begin{array}{cc}4-(-21) & -6-9 \\ 10-(-15) & -14-6\end{array}\right]=\left[\begin{array}{cc}25 & -15 \\ 25 & -20\end{array}\right]\)
\(\therefore A^{-1}=\left[\begin{array}{cc}-7 & 3 \\ -5 & 2\end{array}\right] \quad\) and \(\quad 2 A=\left[\begin{array}{cc}4 & -6 \\ 10 & -14\end{array}\right]\)
\(\therefore 2 A-3 A^{-1}=\left[\begin{array}{cc}4-(-21) & -6-9 \\ 10-(-15) & -14-6\end{array}\right]=\left[\begin{array}{cc}25 & -15 \\ 25 & -20\end{array}\right]\)
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