document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1199095>Question 1199095</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Determinant matrix hard to type so photo linked below :)\r<BR>\n" );
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document.write( "Link: https://ibb.co/GP9XwKX<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 #832794 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</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=832794\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "The tutor @ikleyn offers a very efficient (perhaps the most efficient) method.\r<BR>\n" );
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document.write( "Here's an alternative pathway. <BR>\n" );
document.write( "It's much less efficient, but it might be handy to see another route. This longer route is not recommended for homework or exams. <BR>\n" );
document.write( "It can be treated as a thought exercise more or less.\r<BR>\n" );
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document.write( "One useful properties of adjugate matrices is the following<BR>\n" );
document.write( "Det( Adj(A) ) = ( Det(A) )^(n-1)<BR>\n" );
document.write( "where n is the number of rows and number of columns of the square matrix A.<BR>\n" );
document.write( "In this case we have n = 3.\r<BR>\n" );
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document.write( "Because of that formula above, we don't have to calculate the Adj(A) matrix itself.\r<BR>\n" );
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document.write( "We're given this inverse matrix<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%5E%28-1%29+=+%28matrix%283%2C3%2C1%2Ca%2C0%2C0%2C1%2F3%2C0%2C0%2Cb%2C1%29%29 ALIGN=MIDDLE  ALT=\"A%5E%28-1%29+=+%28matrix%283%2C3%2C1%2Ca%2C0%2C0%2C1%2F3%2C0%2C0%2Cb%2C1%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "I'll write it in this table format like so<BR>\n" );
document.write( "<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >a</TD><TD >0</TD></tr><tr><TD >0</TD><TD >1/3</TD><TD >0</TD></tr><tr><TD >0</TD><TD >b</TD><TD >1</TD></tr>\n" );
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document.write( "To find matrix A itself, aka the inverse of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%5E%28-1%29 ALIGN=MIDDLE  ALT=\"A%5E%28-1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, we need to append the 3x3 identity matrix to the right like so<BR>\n" );
document.write( "<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >a</TD><TD >0</TD><TD >;</TD><TD >1</TD><TD >0</TD><TD >0</TD></tr><tr><TD >0</TD><TD >1/3</TD><TD >0</TD><TD >;</TD><TD >0</TD><TD >1</TD><TD >0</TD></tr><tr><TD >0</TD><TD >b</TD><TD >1</TD><TD >;</TD><TD >0</TD><TD >0</TD><TD >1</TD></tr>\n" );
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document.write( "The column of semicolons is used to separate out the 3x3 blocks. \r<BR>\n" );
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document.write( "The goal is to turn the left-hand 3x3 block into the identity matrix. <BR>\n" );
document.write( "The updated right-hand block will be the matrix of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28A%5E%28-1%29%29%5E%28-1%29+=+A ALIGN=MIDDLE  ALT=\"%28A%5E%28-1%29%29%5E%28-1%29+=+A\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "We use row reduction operations to do this transformation.<BR>\n" );
document.write( "Notation like 3*R2 --> R2 means we triple each entry of row 2 (aka R2), and store the results in row 2.<BR>\n" );
document.write( "A slightly more complicated example is R3-b*R2 --> R3 which means \"temporarily multiply each item of R2 by the scalar 'b', then subtract those results from R3. Store the result in R3\".<BR>\n" );
document.write( "This operation will clear out the pivot below the '1' in R2.\r<BR>\n" );
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document.write( "Here's the set of row reduction steps<BR>\n" );
document.write( "<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >a</TD><TD >0</TD><TD >;</TD><TD >1</TD><TD >0</TD><TD >0</TD><TD ></TD></tr><tr><TD >0</TD><TD >1</TD><TD >0</TD><TD >;</TD><TD >0</TD><TD >3</TD><TD >0</TD><TD >3*R2 --> R2</TD></tr><tr><TD >0</TD><TD >b</TD><TD >1</TD><TD >;</TD><TD >0</TD><TD >0</TD><TD >1</TD><TD ></TD></tr>\n" );
document.write( "</TABLE>\r<BR>\n" );
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document.write( "<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >a</TD><TD >0</TD><TD >;</TD><TD >1</TD><TD >0</TD><TD >0</TD><TD ></TD></tr><tr><TD >0</TD><TD >1</TD><TD >0</TD><TD >;</TD><TD >0</TD><TD >3</TD><TD >0</TD><TD ></TD></tr><tr><TD >0</TD><TD >0</TD><TD >1</TD><TD >;</TD><TD >0</TD><TD >-3b</TD><TD >1</TD><TD >R3-b*R2 --> R3</TD></tr>\n" );
document.write( "</TABLE>\r<BR>\n" );
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document.write( "<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >0</TD><TD >0</TD><TD >;</TD><TD >1</TD><TD >-3a</TD><TD >0</TD><TD >R1-a*R2 --> R1</TD></tr><tr><TD >0</TD><TD >1</TD><TD >0</TD><TD >;</TD><TD >0</TD><TD >3</TD><TD >0</TD><TD ></TD></tr><tr><TD >0</TD><TD >0</TD><TD >1</TD><TD >;</TD><TD >0</TD><TD >-3b</TD><TD >1</TD><TD ></TD></tr>\n" );
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document.write( "The left-hand 3x3 block is now the 3x3 identity matrix, which means we conclude this process.<BR>\n" );
document.write( "The right-hand 3x3 block forms matrix A.\r<BR>\n" );
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document.write( "We find that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+%28matrix%283%2C3%2C1%2C-3a%2C0%2C0%2C3%2C0%2C0%2C-3b%2C1%29%29 ALIGN=MIDDLE  ALT=\"A+=+%28matrix%283%2C3%2C1%2C-3a%2C0%2C0%2C3%2C0%2C0%2C-3b%2C1%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > is the inverse of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%5E%28-1%29+=+%28matrix%283%2C3%2C1%2Ca%2C0%2C0%2C1%2F3%2C0%2C0%2Cb%2C1%29%29 ALIGN=MIDDLE  ALT=\"A%5E%28-1%29+=+%28matrix%283%2C3%2C1%2Ca%2C0%2C0%2C1%2F3%2C0%2C0%2Cb%2C1%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and vice versa.\r<BR>\n" );
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document.write( "To check this, you should find that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%2AA%5E%28-1%29+=+%28matrix%283%2C3%2C1%2C0%2C0%2C0%2C1%2C0%2C0%2C0%2C1%29%29 ALIGN=MIDDLE  ALT=\"A%2AA%5E%28-1%29+=+%28matrix%283%2C3%2C1%2C0%2C0%2C0%2C1%2C0%2C0%2C0%2C1%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%5E%28-1%29%2AA+=+%28matrix%283%2C3%2C1%2C0%2C0%2C0%2C1%2C0%2C0%2C0%2C1%29%29 ALIGN=MIDDLE  ALT=\"A%5E%28-1%29%2AA+=+%28matrix%283%2C3%2C1%2C0%2C0%2C0%2C1%2C0%2C0%2C0%2C1%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >. <BR>\n" );
document.write( "I'll leave these two calculations for the student to do.\r<BR>\n" );
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document.write( "After determining what matrix A is, we can now compute det(A)<BR>\n" );
document.write( "There are a few pathways, but perhaps the simplest (in my opinion) is to follow the steps similar to what is shown here<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Matrices-and-determiminant.faq.question.1198436.html\">https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Matrices-and-determiminant.faq.question.1198436.html</a><BR>\n" );
document.write( "I'll leave the steps for the student to do.\r<BR>\n" );
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document.write( "Using steps similar to those shown in the link above, if <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+%28matrix%283%2C3%2C1%2C-3a%2C0%2C0%2C3%2C0%2C0%2C-3b%2C1%29%29 ALIGN=MIDDLE  ALT=\"A+=+%28matrix%283%2C3%2C1%2C-3a%2C0%2C0%2C3%2C0%2C0%2C-3b%2C1%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "then<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=det%28A%29+=+3 ALIGN=MIDDLE  ALT=\"det%28A%29+=+3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Fortunately a lot of those 0's will make the determinant calculation fairly much easier.\r<BR>\n" );
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document.write( "Therefore,<BR>\n" );
document.write( "Det(Adj(A)) = ( Det(A) )^(n-1)<BR>\n" );
document.write( "Det(Adj(A)) = ( Det(A) )^(3-1)<BR>\n" );
document.write( "Det(Adj(A)) = ( Det(A) )^2<BR>\n" );
document.write( "Det(Adj(A)) = ( 3 )^2<BR>\n" );
document.write( "Det(Adj(A)) = 9\r<BR>\n" );
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document.write( "I don't recommend following this process for your homework or exams. <BR>\n" );
document.write( "This is simply another way to see why the final answer is 9. <BR>\n" );
document.write( "I recommend following the pathway @ikleyn had set out.<BR>\n" );
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