document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1164528>Question 1164528</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose A and B are invertible square matrices of the same size. Show that<BR>\n" );
document.write( "(A^−1 + B^−1)^−1 = A(A + B)^−1 B </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #854345 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2261)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=854345\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 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 $I$.\r<BR>\n" );
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document.write( "Let $X = (A^{-1} + B^{-1})$ and $Y = A(A + B)^{-1} B$. We want to prove that $Y = X^{-1}$, or equivalently, that $Y \cdot X = I$.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **Proof**\r<BR>\n" );
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document.write( "**Step 1: Set up the product $Y \cdot X$**<BR>\n" );
document.write( "Substitute the expressions for $Y$ and $X$:<BR>\n" );
document.write( "$$Y \cdot X = \left[ A(A + B)^{-1} B \right] \cdot \left[ A^{-1} + B^{-1} \right]$$\r<BR>\n" );
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document.write( "**Step 2: Distribute the factor $B$ into the parentheses**<BR>\n" );
document.write( "We distribute $B$ into the terms inside the second bracket:<BR>\n" );
document.write( "$$Y \cdot X = A(A + B)^{-1} (B \cdot A^{-1} + B \cdot B^{-1})$$<BR>\n" );
document.write( "Since $B \cdot B^{-1} = I$, the expression simplifies to:<BR>\n" );
document.write( "$$Y \cdot X = A(A + B)^{-1} (B A^{-1} + I)$$\r<BR>\n" );
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document.write( "**Step 3: Distribute the factor $A$ into the expression**<BR>\n" );
document.write( "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 $(B A^{-1} + I)$:<BR>\n" );
document.write( "$$(B A^{-1} + I) = (B + A)A^{-1}$$<BR>\n" );
document.write( "Substitute this back into the equation:<BR>\n" );
document.write( "$$Y \cdot X = A(A + B)^{-1} (B + A)A^{-1}$$\r<BR>\n" );
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document.write( "**Step 4: Simplify using the property of Inverses**<BR>\n" );
document.write( "Since $(A + B)^{-1}$ and $(B + A)$ are inverses of each other (and matrix addition is commutative, so $A+B = B+A$):<BR>\n" );
document.write( "$$(A + B)^{-1} (A + B) = I$$<BR>\n" );
document.write( "Now the equation looks like this:<BR>\n" );
document.write( "$$Y \cdot X = A \cdot I \cdot A^{-1}$$<BR>\n" );
document.write( "$$Y \cdot X = A \cdot A^{-1}$$\r<BR>\n" );
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document.write( "**Step 5: Final Identity**<BR>\n" );
document.write( "$$Y \cdot X = I$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **Conclusion**<BR>\n" );
document.write( "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:<BR>\n" );
document.write( "$$(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1} B$$<BR>\n" );
document.write( "$\blacksquare$\r<BR>\n" );
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document.write( "> **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! </SPAN>\n" );
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