document.write( "Question 867083: $5,000 is distributed among three investment types: at 8%, 3%, and 1%. the total of amounts invested at 8% and at 1% is equal the amount invested at 3%. the total interest after one year from all three accounts is $144. How much was invested at each account? Solve using matrices. \n" ); document.write( "
Algebra.Com's Answer #522740 by richwmiller(17219)\"\" \"About 
You can put this solution on YOUR website!
a+b+c=5000
\n" ); document.write( ".01a+.03b+.08c=144
\n" ); document.write( ".01a-.03b+.08c=0\r
\n" ); document.write( "\n" ); document.write( "Add (-1/100 * row1) to row2
\n" ); document.write( "1 1 1 5000
\n" ); document.write( "0 1/50 7/100 94
\n" ); document.write( "1/100 -3/100 8/100 0\r
\n" ); document.write( "\n" ); document.write( "Add (-1/100 * row1) to row3
\n" ); document.write( "1 1 1 5000
\n" ); document.write( "0 1/50 7/100 94
\n" ); document.write( "0 -1/25 7/100 -50\r
\n" ); document.write( "\n" ); document.write( "Divide row2 by 1/50
\n" ); document.write( "1 1 1 5000
\n" ); document.write( "0 1 7/2 4700
\n" ); document.write( "0 -1/25 7/100 -50\r
\n" ); document.write( "\n" ); document.write( "Add (1/25 * row2) to row3
\n" ); document.write( "1 1 1 5000
\n" ); document.write( "0 1 7/2 4700
\n" ); document.write( "0 0 21/100 138\r
\n" ); document.write( "\n" ); document.write( "Divide row3 by 21/100
\n" ); document.write( "1 1 1 5000
\n" ); document.write( "0 1 7/2 4700
\n" ); document.write( "0 0 1 4600/7\r
\n" ); document.write( "\n" ); document.write( "Add (-7/2 * row3) to row2
\n" ); document.write( "1 1 1 5000
\n" ); document.write( "0 1 0 2400
\n" ); document.write( "0 0 1 4600/7\r
\n" ); document.write( "\n" ); document.write( "Add (-1 * row3) to row1
\n" ); document.write( "1 1 0 30400/7
\n" ); document.write( "0 1 0 2400
\n" ); document.write( "0 0 1 4600/7\r
\n" ); document.write( "\n" ); document.write( "Add (-1 * row2) to row1
\n" ); document.write( "1 0 0 13600/7
\n" ); document.write( "0 1 0 2400
\n" ); document.write( "0 0 1 4600/7\r
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Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables

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\n" ); document.write( " First let \"A=%28matrix%283%2C3%2C1%2C1%2C1%2C0.01%2C0.03%2C0.08%2C0.01%2C-0.03%2C0.08%29%29\". This is the matrix formed by the coefficients of the given system of equations.
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\n" ); document.write( " Take note that the right hand values of the system are \"5000\", \"144\", and \"0\" and they are highlighted here:
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\n" ); document.write( " These values are important as they will be used to replace the columns of the matrix A.
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\n" ); document.write( " Now let's calculate the the determinant of the matrix A to get \"abs%28A%29=0.0042\". To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.
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\n" ); document.write( " Notation note: \"abs%28A%29\" denotes the determinant of the matrix A.
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\n" ); document.write( " Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix \"A%5Bx%5D\" (since we're replacing the 'x' column so to speak).
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\n" ); document.write( " Now compute the determinant of \"A%5Bx%5D\" to get \"abs%28A%5Bx%5D%29=8.16\". Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.
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\n" ); document.write( " To find the first solution, simply divide the determinant of \"A%5Bx%5D\" by the determinant of \"A\" to get: \"x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%288.16%29%2F%280.0042%29=1942.85714285714\"
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\n" ); document.write( " So the first solution is \"x=1942.85714285714\"
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\n" ); document.write( " We'll follow the same basic idea to find the other two solutions. Let's reset by letting \"A=%28matrix%283%2C3%2C1%2C1%2C1%2C0.01%2C0.03%2C0.08%2C0.01%2C-0.03%2C0.08%29%29\" again (this is the coefficient matrix).
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\n" ); document.write( " Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix \"A%5By%5D\" (since we're replacing the 'y' column in a way).
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\n" ); document.write( " Now compute the determinant of \"A%5By%5D\" to get \"abs%28A%5By%5D%29=10.08\".
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\n" ); document.write( " To find the second solution, divide the determinant of \"A%5By%5D\" by the determinant of \"A\" to get: \"y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%2810.08%29%2F%280.0042%29=2400\"
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\n" ); document.write( " So the second solution is \"y=2400\"
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\n" ); document.write( " Let's reset again by letting \"A=%28matrix%283%2C3%2C1%2C1%2C1%2C0.01%2C0.03%2C0.08%2C0.01%2C-0.03%2C0.08%29%29\" which is the coefficient matrix.
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\n" ); document.write( " Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix \"A%5Bz%5D\"
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\n" ); document.write( " Now compute the determinant of \"A%5Bz%5D\" to get \"abs%28A%5Bz%5D%29=2.76\".
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\n" ); document.write( " To find the third solution, divide the determinant of \"A%5Bz%5D\" by the determinant of \"A\" to get: \"z=%28abs%28A%5Bz%5D%29%29%2F%28abs%28A%29%29=%282.76%29%2F%280.0042%29=657.142857142857\"
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\n" ); document.write( " So the third solution is \"z=657.142857142857\"
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\n" ); document.write( " Final Answer:
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\n" ); document.write( " So the three solutions are \"x=1942.85714285714\", \"y=2400\", and \"z=657.142857142857\" giving the ordered triple (1942.85714285714, 2400, 657.142857142857)
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\n" ); document.write( " Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.
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