If A = [[0.8, 0.6], [-0.6, 0.8]], find A^3
Understand the Problem
The question requires us to find A cubed (A^3) given matrix A, we calculate A^3 by multiplying matrix A by itself three times (AAA).
Answer
$\begin{bmatrix} -0.352 & 0.936 \\ -0.936 & -0.352 \end{bmatrix}$
Answer for screen readers
$A^3 = \begin{bmatrix} -0.352 & 0.936 \ -0.936 & -0.352 \end{bmatrix}$
Steps to Solve
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Calculate $A^2$ To find $A^3$, we first need to calculate $A^2$ by multiplying matrix $A$ by itself: $$ A^2 = A \cdot A = \begin{bmatrix} 0.8 & 0.6 \ -0.6 & 0.8 \end{bmatrix} \cdot \begin{bmatrix} 0.8 & 0.6 \ -0.6 & 0.8 \end{bmatrix} $$ $$ A^2 = \begin{bmatrix} (0.8 \cdot 0.8 + 0.6 \cdot -0.6) & (0.8 \cdot 0.6 + 0.6 \cdot 0.8) \ (-0.6 \cdot 0.8 + 0.8 \cdot -0.6) & (-0.6 \cdot 0.6 + 0.8 \cdot 0.8) \end{bmatrix} $$ $$ A^2 = \begin{bmatrix} (0.64 - 0.36) & (0.48 + 0.48) \ (-0.48 - 0.48) & (-0.36 + 0.64) \end{bmatrix} $$ $$ A^2 = \begin{bmatrix} 0.28 & 0.96 \ -0.96 & 0.28 \end{bmatrix} $$
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Calculate $A^3$ Now, we can find $A^3$ by multiplying $A^2$ by $A$: $$ A^3 = A^2 \cdot A = \begin{bmatrix} 0.28 & 0.96 \ -0.96 & 0.28 \end{bmatrix} \cdot \begin{bmatrix} 0.8 & 0.6 \ -0.6 & 0.8 \end{bmatrix} $$ $$ A^3 = \begin{bmatrix} (0.28 \cdot 0.8 + 0.96 \cdot -0.6) & (0.28 \cdot 0.6 + 0.96 \cdot 0.8) \ (-0.96 \cdot 0.8 + 0.28 \cdot -0.6) & (-0.96 \cdot 0.6 + 0.28 \cdot 0.8) \end{bmatrix} $$ $$ A^3 = \begin{bmatrix} (0.224 - 0.576) & (0.168 + 0.768) \ (-0.768 - 0.168) & (-0.576 + 0.224) \end{bmatrix} $$ $$ A^3 = \begin{bmatrix} -0.352 & 0.936 \ -0.936 & -0.352 \end{bmatrix} $$
$A^3 = \begin{bmatrix} -0.352 & 0.936 \ -0.936 & -0.352 \end{bmatrix}$
More Information
Matrix exponentiation is analogous to scalar exponentiation, but requires matrix multiplication. It is used extensively in linear algebra, physics and engineering.
Tips
A common mistake is performing scalar exponentiation on the individual elements of the matrix, rather than performing matrix multiplication. Matrix multiplication must be performed to get the correct answer. Calculation errors are also very common.
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