The image you provided contains three matrix properties that can be proven using matrix algebra rules. I will go step by step to prove each one:

1. Proof of $\left( A^T \right)^{-1} = \left( A^{-1} \right)^T$

This property states that the inverse of the transpose of a matrix is equal to the transpose of the inverse of the matrix.

Proof:

• Let $A$ be an invertible matrix.

• The inverse of a matrix $A$ is defined as the matrix $A^{-1}$ such that: $A A^{-1} = A^{-1} A = I$ where $I$ is the identity matrix.

• Taking the transpose of both sides of $A A^{-1} = I$, we get: $\left( A A^{-1} \right)^T = I^T = I$

• By the property of transposes, $\left( A B \right)^T = B^T A^T$, so: $\left( A A^{-1} \right)^T = \left( A^{-1} \right)^T A^T = I$ This shows that $\left( A^{-1} \right)^T$ is the inverse of $A^T$.

Therefore, $\left( A^T \right)^{-1} = \left( A^{-1} \right)^T$

2. Proof of $\left( A^{-1} \right)^{-1} = A$

This property states that the inverse of the inverse of a matrix is the matrix itself.

Proof:

• By the definition of an inverse, if $A^{-1}$ is the inverse of $A$, then: $A A^{-1} = A^{-1} A = I$
• To find $\left( A^{-1} \right)^{-1}$, we need to find the matrix that, when multiplied by $A^{-1}$, gives the identity matrix: $A^{-1} \left( A^{-1} \right)^{-1} = I$ From this, we can conclude that $\left( A^{-1} \right)^{-1} = A$.

Therefore, $\left( A^{-1} \right)^{-1} = A$

3. Proof of $\left( ABC \right)^{-1} = C^{-1} B^{-1} A^{-1}$

This property states that the inverse of the product of matrices is the product of their inverses in reverse order.

Proof:

• Let $A$, $B$, and $C$ be invertible matrices.
• We need to show that: $\left( ABC \right) \left( C^{-1} B^{-1} A^{-1} \right) = I$
• Using the associative property of matrix multiplication, we calculate: $\left( ABC \right) \left( C^{-1} B^{-1} A^{-1} \right) = A \left( B \left( C C^{-1} \right) B^{-1} \right) A^{-1}$ Since $C C^{-1} = I$ and $B B^{-1} = I$, this simplifies to: $A \left( I I \right) A^{-1} = A A^{-1} = I$ Thus, $\left( ABC \right)^{-1} = C^{-1} B^{-1} A^{-1}$.

Therefore, $\left( ABC \right)^{-1} = C^{-1} B^{-1} A^{-1}$

Summary

We have successfully proven all three properties:

1. $\left( A^T \right)^{-1} = \left( A^{-1} \right)^T$
2. $\left( A^{-1} \right)^{-1} = A$
3. $\left( ABC \right)^{-1} = C^{-1} B^{-1} A^{-1}$

Let me know if you'd like further explanations or additional examples.

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Relative Questions:

1. What are the conditions for a matrix to be invertible?
2. Can you extend the property of matrix product inverses to more than three matrices?
3. How does the determinant of a matrix relate to its invertibility?
4. What is the difference between a matrix transpose and its inverse?
5. Can a non-square matrix have an inverse?

Tip:

For any matrix $A$, $A^{-1}$ exists if and only if $A$ is square and its determinant is non-zero.