Q68 / 125 · BITSAT 2011 · Mathematics
The matrix \(A^2+4 A-5 I\), where \(I\) is identity matrix and \(A=\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]\), equals :
- Correct answerA\(4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]\)
- B\(4\left[\begin{array}{cc}0 & -1 \\ 2 & 2\end{array}\right]\)
- C\(32\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]\)
- D\(32\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\)
Solution
\(\begin{aligned}
& A^2+4 A-5 I=A \times A+4 A-5 I \\
= & {\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right] \times\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]+4\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]-5\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] } \\
= & {\left[\begin{array}{cc}
9 & -4 \\
-8 & 17
\end{array}\right]+\left[\begin{array}{cc}
4 & 8 \\
16 & -12
\end{array}\right]-\left[\begin{array}{cc}
5 & 0 \\
0 & 5
\end{array}\right] } \\
= & {\left[\begin{array}{cc}
9+4-5 & -4+8-0 \\
-8+16-0 & 17-12-5
\end{array}\right]=\left[\begin{array}{cc}
8 & 4 \\
8 & 0
\end{array}\right] } \\
= & 4\left[\begin{array}{cc}
2 & 1 \\
2 & 0
\end{array}\right]
\end{aligned}\)
Hence, the correct answer is option: 'A'.