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75 3; answer; value of y: 25  1  0 D 3  0 1 4 4  4  2  55  1
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Next for the value of z,
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thus, z D 0 =D
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55 2:2; answer; value of z: 25 Thus the correct answers are x 4:8, y 3:0, and z 2:2, which you can verify by replacing x, y, and z with these values in the original three equations.
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43. Step 1 is satis ed. Next, for step 2, we have    1 1 1 1    3 2 4 4   D    2 5 7 0    0 3 2 3  The above 4th-order determinant can be reduced to a single 3rd-order by applying property 7; one procedure is as follows. First, to each element of row 2 add the corresponding element of row 1 multiplied by 4. Then, to each element of row 4 add the corresponding element of row 1 multiplied by 3, thus getting      1 1 1 1  1 6 0       1 6 0 0      5 7 D    2    2 5 7 0  3   6 5   3 6 5 0 Now let us reduce the 3rd-order determinant to a single 2nd-order; one way is as follows. To each element of column 2 add the corresponding element of column 1 multiplied by 6, thus giving us    1 0 0    17 7       D  2 17 7    12 5  169    3 12 5  Thus, D 169 for this problem. We now go to step 3 to nd the values of w, x, y, and z, in that order, as follows. For w:    4 1 1 1    0 2 4 4   D0   338  12 5 7 0    5 3 2 3  thus, w D 0 =D For x: 338 2. 169   1   3  0 D   2   0 507 3 169 4 0 12 5  1  4   507 0  3 
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1 4 7 2
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thus, x D 0 =D
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For y:   1   3  0 D   2   0 thus, y D 0 =D For z:   1   3  0 D   2   0 thus, z D 0 =D 676 4. 169 1 2 5 3 1 4 7 2  4   0   676 12   5 169 1. 169 1 2 5 3 4 0 12 5  1  4   169 0  3 
Thus the correct answers are w 2, x 3, y 1, and z 4. 44. First you should verify that   3 2   D  1 1   2 1  5    1  0  4 
which assures us that the given system does have non-trivial solutions. Then, from inspection of the equations a 3; d 1; g 2; b 2; e 1; h 1; c 5 f 1 i 4
which, upon substitution into eq. (55), gives the answers x 3, y 2, and z 1. Note: An important point concerning systems of homogeneous linear equations can be shown as follows. Let k be any constant, and let us multiply through each of the three equations of eq. (53) by k, thus getting a kx b ky c kz 0 d kx e ky f kz 0 g kx h ky i kz 0 Note that the last three equations have exactly the same form as eq. (53); thus, if x, y, and z represent values that satisfy eq. (53), the values of kx, ky, and kz also satisfy eq. (53). It s apparent that the foregoing is true for any system of n homogeneous linear equations; thus, if x1 , x2 ; . . . ; xn are found to be a solution set of such a system, then kx1 , kx2 ; . . . ; kxn is also a solution set to the system, where k is any constant.
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