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Matrices (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Matrices (p-1)2 Marks
Question
If $A=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]$, then find $A^3$.
✓

Answer

$
\begin{aligned}
& A^2=A \cdot A=\left(\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right)\left(\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right) \\
& =\left[\begin{array}{rr}
1-2 & 2+6 \\
-1-3 & -2+9
\end{array}\right]=\left[\begin{array}{ll}
-1 & 8 \\
-4 & 7
\end{array}\right] \\
& \therefore A^3=A^2 \cdot A=\left[\begin{array}{ll}
-1 & 8 \\
-4 & 7
\end{array}\right]\left[\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right] \\
& =\left(\begin{array}{ll}
-1-8 & -2+24 \\
-4-7 & -8+21
\end{array}\right) \\
& \therefore A^3=\left(\begin{array}{rr}
-9 & 22 \\
-11 & 13
\end{array}\right)\\
&
\end{aligned}
$

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