document.write( "Question 1195865: Answer each of the following as True or False justifying your answers:
\n" ); document.write( " If S and K are n×n matrices such that S is symmetric and K is skew symmetric, then |S+K|=|S-K|. If |■(a&a&a-1&c@a&a&a-1&b@a&a&a+1&b@a&a+1&b&c)|=0, then a=b=c=0. If A,B and C are non-singular n×n matrices such that AB=C, BC=A and CA=B, then |ABC|=1. If A and B are 3×3 matrices, then AB-AB^T is a non-singular matrix. The vector U=(2,-1,-1) in R^3 is a linear combination of the vectors V_1=(1,2,7),V_2=(2,5,17) and V_3=(-1,2,5). \n" ); document.write( "

Algebra.Com's Answer #828452 by ikleyn(52775) $\"\"$ $\"About$

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\n" ); document.write( "Answer each of the following as True or False justifying your answers:\r
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\n" ); document.write( "\n" ); document.write( "(a) If S and K are n×n matrices such that S is symmetric and K is skew symmetric,
\n" ); document.write( "then |S+K|=|S-K|. \r
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\n" ); document.write( "\n" ); document.write( "(b) If |■(a&a&a-1&c@a&a&a-1&b@a&a&a+1&b@a&a+1&b&c)|=0, then a=b=c=0. \r
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\n" ); document.write( "\n" ); document.write( "(c) If A,B and C are non-singular n×n matrices such that AB=C, BC=A and CA=B, then |ABC|=1. \r
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\n" ); document.write( "\n" ); document.write( "(d) If A and B are 3×3 matrices, then AB-AB^T is a non-singular matrix. \r
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\n" ); document.write( "\n" ); document.write( "(e) The vector U=(2,-1,-1) in R^3 is a linear combination of the vectors
\n" ); document.write( "V_1=(1,2,7), V_2=(2,5,17) and V_3=(-1,2,5).
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document.write( "(c)  We are given for non-singular nxn-matrices A, B and C\r\n" );
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document.write( "         AB = C,  BC = A, CA = B.\r\n" );
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document.write( "     Multiply all three equations (both sides).  You will get then\r\n" );
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document.write( "         A*B*B*C*C*A = A*B*C.\r\n" );
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document.write( "     It implies for determinants\r\n" );
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document.write( "         |A|*|B|*|B|*|C|*|C|*|A| = |A|*|B|*|C|,  or\r\n" );
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document.write( "         |A*B*C|^2 = |A*B*C|.\r\n" );
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document.write( "     Since |ABC| =/= 0 (due to non-singularity), we can reduce/cancel common factor |A*B*C| in both sides.\r\n" );
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document.write( "     We then get  |A*B*C| = 1, QED.\r\n" );
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document.write( "     ANSWER.  The statement is TRUE.\r\n" );
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document.write( "(d)  The statement is FALSE.\r\n" );
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document.write( "     To prove that it is false, consider A= 0, B= 0  (zero matrices).\r\n" );
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document.write( "(e)  We want to check if there are real numbers \"a\", \"b\" and \"c\" such that\r\n" );
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document.write( "        U = a*V_1 + b*V_2 + c*V_3.\r\n" );
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document.write( "     It is the same as to check, if this system of three linear equations has a solution\r\n" );
document.write( "     for real numbers \"a\", \"b\" and \"c\"\r\n" );
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document.write( "        1*a +  2*b - 1*c =  2,   (E1)\r\n" );
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document.write( "        2*a +  5*b + 2*c = -1,   (E2)\r\n" );
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document.write( "        7*a + 17*b + 5*c = -1.   (E3)\r\n" );
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document.write( "     Notice that for left sides of equations (E1), (E2), (E3) we have their linear combination 1*E1 + 3*E2 = E3.\r\n" );
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document.write( "     For right side terms,  1*E1 + 3*E2 = 1*2 + 3*(-1) = 2 - 3 = -1  is the same as E3.\r\n" );
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document.write( "     So, we can expect that the system has a solution.\r\n" );
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document.write( "     Now, I am laizy to check it directly, and we live in XXI century - so I use an online solver \r\n" );
document.write( "          https://www.emathhelp.net/calculators/linear-algebra/gauss-jordan-elimination-calculator/\r\n" );
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document.write( "     It confirms that the system (E1), (E2), (E3) has the solution a= 12, b= -5, c= 0.\r\n" );
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document.write( "     ANSWER.  Vector U=(2,-1,-1) in R^3 is a linear combination of the vectors \r\n" );
document.write( "              V_1=(1,2,7), V_2=(2,5,17) and V_3=(-1,2,5):  U = 12*U_1 - 5*V_2.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Parts (c), (d) and (e) are solved, answered and explained.\r
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