For \(M=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]\) and for any \(n \in \mathbf{N}\) the…
For \(M=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]\) and for any \(n \in \mathbf{N}\) the matrix \(M^{n+1}-M^n=\)
- \(\left[\begin{array}{cc}2 & 4 \\ 1 & -2\end{array}\right]\)
- \(\left[\begin{array}{ll}2 & -4 \\ 1 & -2\end{array}\right]\)
- \(\left[\begin{array}{cc}2 & -4 \\ 1 & 2\end{array}\right]\)
- \(\left[\begin{array}{ll}2 & 4 \\ 1 & 2\end{array}\right]\)
Solution
\(\begin{aligned}
M & =\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right] \\
M^2 & =\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right] \\
& =\left[\begin{array}{ll}
9-4 & -12+4 \\
3-1 & -4+1
\end{array}\right]=\left[\begin{array}{ll}
5 & -8 \\
2 & -3
\end{array}\right] \\
M^2-M & =\left[\begin{array}{ll}
5 & -8 \\
2 & -3
\end{array}\right]-\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]=\left[\begin{array}{ll}
2 & -4 \\
1 & -2
\end{array}\right]
\end{aligned}\)
So; \(M^{n+1}-M=M^2-M=\left[\begin{array}{ll}2 & -4 \\ 1 & -2\end{array}\right]\)
Asked in: AP EAMCET 2020 (17 Sep Shift 2)
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