SOLUTION: Suppose A and B are invertible square matrices of the same size. Show that (A^−1 + B^−1)^−1 = A(A + B)^−1 B

Question 1164528: Suppose A and B are invertible square matrices of the same size. Show that
(A^−1 + B^−1)^−1 = A(A + B)^−1 B
Answer by CPhill(2261) (Show Source): You can put this solution on YOUR website!
To show that $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1} B$, we can use the property that two matrices are inverses if their product equals the identity matrix $I$.
Let $X = (A^{-1} + B^{-1})$ and $Y = A(A + B)^{-1} B$. We want to prove that $Y = X^{-1}$, or equivalently, that $Y \cdot X = I$.
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### **Proof**
**Step 1: Set up the product $Y \cdot X$**
Substitute the expressions for $Y$ and $X$:
$$Y \cdot X = \left[ A(A + B)^{-1} B \right] \cdot \left[ A^{-1} + B^{-1} \right]$$
**Step 2: Distribute the factor $B$ into the parentheses**
We distribute $B$ into the terms inside the second bracket:
$$Y \cdot X = A(A + B)^{-1} (B \cdot A^{-1} + B \cdot B^{-1})$$
Since $B \cdot B^{-1} = I$, the expression simplifies to:
$$Y \cdot X = A(A + B)^{-1} (B A^{-1} + I)$$
**Step 3: Distribute the factor $A$ into the expression**
To make the terms inside the rightmost parentheses match $(A + B)$, we need to manipulate the expression. Let's factor out an $A^{-1}$ from the right side of $(B A^{-1} + I)$:
$$(B A^{-1} + I) = (B + A)A^{-1}$$
Substitute this back into the equation:
$$Y \cdot X = A(A + B)^{-1} (B + A)A^{-1}$$
**Step 4: Simplify using the property of Inverses**
Since $(A + B)^{-1}$ and $(B + A)$ are inverses of each other (and matrix addition is commutative, so $A+B = B+A$):
$$(A + B)^{-1} (A + B) = I$$
Now the equation looks like this:
$$Y \cdot X = A \cdot I \cdot A^{-1}$$
$$Y \cdot X = A \cdot A^{-1}$$
**Step 5: Final Identity**
$$Y \cdot X = I$$
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### **Conclusion**
Since multiplying $(A^{-1} + B^{-1})$ by $A(A + B)^{-1} B$ results in the identity matrix $I$, the two expressions are indeed inverses of each other. Therefore:
$$(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1} B$$
$\blacksquare$
> **Insight:** This identity is the matrix version of the "product over sum" rule used for parallel resistors in physics: $\frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} = \frac{R_1 R_2}{R_1 + R_2}$. Just remember that with matrices, the order of multiplication is crucial!
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