document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1174343>Question 1174343</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Q−1: [5×2 marks] Answer each of the following as True or False justifying your answers:<BR>\n" );
document.write( "	If A and B are n×n matrices, then 〖(A+B)〗^2=A^2+2AB+B^2.<BR>\n" );
document.write( "	If A is an n×n invertible symmetric matrix, then (A^(-1) )^k is symmetric, for any positive integer k.<BR>\n" );
document.write( "	Let A be an n×n matrix. If A is invertible, then |A|≠0.<BR>\n" );
document.write( "	If A and B are n×n matrices, then |A+B^T |=|A^T+B|.<BR>\n" );
document.write( "	The vectors (a,0,0), (1,a,0) and (2,3,a) in R^3 are linearly independent for all a∈R.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #799775 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3816)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=799775\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "I'll do the first two problems to get you started.\r<BR>\n" );
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document.write( "Problem 1\r<BR>\n" );
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document.write( "(A+B)^2 = (A+B)*(A+B)<BR>\n" );
document.write( "(A+B)^2 = C*(A+B) ............ let C = A+B<BR>\n" );
document.write( "(A+B)^2 = C*A+C*B ............ Distribute<BR>\n" );
document.write( "(A+B)^2 = (A+B)*A+(A+B)*B ......... plug in C = A+B<BR>\n" );
document.write( "(A+B)^2 = A*A+B*A+A*B+B*B ...... distribute again; be careful about the order<BR>\n" );
document.write( "(A+B)^2 = A^2+B*A+A*B+B^2<BR>\n" );
document.write( "We stop here because B*A is <b>not</b> the same as A*B. So we cannot combine them to get 2AB or 2BA. Matrix multiplication is <b>not</b> commutative in general (there are some cases where it works, but overall it doesn't work).\r<BR>\n" );
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document.write( "Therefore, (A+B)^2 = A^2+2AB+B^2 is <font color=red>false</font> when A,B are two nxn matrices.<BR>\n" );
document.write( "It's only true when AB = BA.\r<BR>\n" );
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document.write( "Problem 2\r<BR>\n" );
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document.write( "The transpose operation of a matrix swaps the rows and columns. For instance, row 1 turns into column 1, and vice versa.\r<BR>\n" );
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document.write( "With matrix notation, the T indicates transpose.<BR>\n" );
document.write( "Writing A^T means \"transpose of matrix A\"\r<BR>\n" );
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document.write( "One useful property to take advantage of here is that<BR>\n" );
document.write( "(A*B)^T = B^T*A^T<BR>\n" );
document.write( "I won't prove that here, but these links are a good resource for that<BR>\n" );
document.write( "<a  rel=nofollow HREF=\"https://math.stackexchange.com/questions/3208939/transpose-of-product-of-matrices\">https://math.stackexchange.com/questions/3208939/transpose-of-product-of-matrices</a><BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://math.stackexchange.com/questions/1440305/how-to-prove-abt-bt-at\">https://math.stackexchange.com/questions/1440305/how-to-prove-abt-bt-at</a>\r<BR>\n" );
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document.write( "Matrix A is symmetric if and only if A = A^T<BR>\n" );
document.write( "Square both sides<BR>\n" );
document.write( "A = A^T<BR>\n" );
document.write( "A^2 = (A^T)^2<BR>\n" );
document.write( "A^2 = (A^T)*(A^T)<BR>\n" );
document.write( "A^2 = (A*A)^T .......... apply the transpose property mentioned<BR>\n" );
document.write( "A^2 = (A^2)^T<BR>\n" );
document.write( "This proves that if A is symmetric, then so is A^2<BR>\n" );
document.write( "This example can be extended to show that if A is symmetric, then so is A^3, and so on.<BR>\n" );
document.write( "By induction, if A is symmetric, then A^k is symmetric for any positive integer k.\r<BR>\n" );
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document.write( "If A is invertible, then A^(-1) multiplies with it to lead to the identity matrix I<BR>\n" );
document.write( "A*A^(-1) = A^(-1)*A = I\r<BR>\n" );
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document.write( "Let's focus on A*A^(-1) = I<BR>\n" );
document.write( "Apply the transpose operation to both sides<BR>\n" );
document.write( "A*A^(-1) = I<BR>\n" );
document.write( "[A*A^(-1)]^T = I^T<BR>\n" );
document.write( "[A^(-1)]^T*A^T = I<BR>\n" );
document.write( "we can see that for the matrix A^T, the matrix [A^(-1)]^T plays the role of the inverse.\r<BR>\n" );
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document.write( "Put another way, let B = [A^(-1)]^T<BR>\n" );
document.write( "We can see that<BR>\n" );
document.write( "B*A^T = I<BR>\n" );
document.write( "which leads to <BR>\n" );
document.write( "B*A^T = I<BR>\n" );
document.write( "(B*A^T)*(A^T)^(-1) = I*(A^T)^(-1)<BR>\n" );
document.write( "B * [A^T*(A^T)^(-1)] = (A^T)^(-1)<BR>\n" );
document.write( "B * I = (A^T)^(-1)<BR>\n" );
document.write( "B = (A^T)^(-1)<BR>\n" );
document.write( "Therefore,<BR>\n" );
document.write( "[A^(-1)]^T = (A^T)^(-1)<BR>\n" );
document.write( "A very similar situation occurs with A^T*B = I, but we'll be using left-multiplication to arrive at the same equation shown above. \r<BR>\n" );
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document.write( "What I've been building up to so far is to show that<ul><li>If matrix A is symmetric, then A^k is symmetric for any positive integer k</li><li>If matrix A is invertible, then [A^(-1)]^T = (A^T)^(-1)</li></ul>Such properties have probably been discussed and/or proven in your textbook at this point. If not, then likely they've been left as a homework exercise.\r<BR>\n" );
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document.write( "Let's get back to the problem at hand. We know matrix A is invertible. This means [A^(-1)]^T = (A^T)^(-1) by that second bullet point.\r<BR>\n" );
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document.write( "But we also know matrix A is symmetric. So A = A^T, which leads to<BR>\n" );
document.write( "[A^(-1)]^T = (A^T)^(-1)<BR>\n" );
document.write( "[A^(-1)]^T = A^(-1)<BR>\n" );
document.write( "A^(-1) = [A^(-1)]^T<BR>\n" );
document.write( "which shows that the inverse matrix A^(-1) is symmetric, but only if A is symmetric.<BR>\n" );
document.write( "If A is symmetric, then so is A^(-1)\r<BR>\n" );
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document.write( "But if A^(-1) is symmetric, we can use the first property mentioned in the bullet points above to conclude that ( A^(-1) )^k is also symmetric for any positive integer k. \r<BR>\n" );
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document.write( "Therefore, the statement that matrix A being symmetric and invertible leads to ( A^(-1) )^k also being symmetric, for any positive integer k, is a <font color=red>true statement</font>.<BR>\n" );
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