%5Et-%3D-(a%5Et)%5E-1-matrix-proof.jpeg "(a^-1)^t = (a^t)^-1 Matrix Proof")
Proof: $(A^{-1})^t = (A^t)^{-1}$ for Matrices
In this article, we will prove that $(A^{-1})^t = (A^t)^{-1}$ for matrices, where $A$ is a square matrix and $t$ is a positive integer.
Preliminaries
Before we begin the proof, let's recall some basic properties of matrix multiplication and inversion.
- For any square matrix $A$, the inverse of $A$ is denoted by $A^{-1}$ and satisfies $AA^{-1} = A^{-1}A = I$, where $I$ is the identity matrix.
- For any two square matrices $A$ and $B$, the product $AB$ is also a square matrix, and $(AB)^{-1} = B^{-1}A^{-1}$.
Proof
We will prove the statement by induction on $t$.
Base Case: $t = 1$
For $t = 1$, we have
$(A^{-1})^1 = A^{-1}$and$(A^1)^{-1} = A^{-1}.$Therefore, the statement is true for $t = 1$.
Inductive Step: $t \ge 2$
Assume that the statement is true for some positive integer $k$, i.e.,
$(A^{-1})^k = (A^k)^{-1}.$We need to show that the statement is true for $k + 1$, i.e.,
$(A^{-1})^{k+1} = (A^{k+1})^{-1}.$Using the associative property of matrix multiplication, we can write
$(A^{-1})^{k+1} = A^{-1}(A^{-1})^k.$Substituting the inductive hypothesis, we get
$(A^{-1})^{k+1} = A^{-1}(A^k)^{-1}.$Using the property of matrix inversion, we can rewrite the right-hand side as
$(A^{-1})^{k+1} = (A^kA)^{-1} = (A^{k+1})^{-1}.$Therefore, the statement is true for $k + 1$, and hence, by mathematical induction, it is true for all positive integers $t$.
Conclusion
In conclusion, we have shown that $(A^{-1})^t = (A^t)^{-1}$ for all positive integers $t$ and square matrices $A$. This result is useful in various applications of linear algebra and matrix theory.
