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36. Factoring in accordance with property 5, then expanding in terms of the minors of column 2, we have   3   1  7 5 2 6   4   2 1 0 0 0 2 1 2 3   1  1    2    420 4  1  2   2 1 2 3  2    1  420; answer:  2
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37. No, because this is simply a special case of multiplying by 1 or 1 in property 7. 38. Factoring 2 from column 4, we have  3  4  2 1  6 1 3 4 7 2 0 4 5  3  4   2 0 0; by property 4: 1  6
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39. Note, rst, that no factoring can be done, and that no two rows or no two columns are identical. Therefore, using property 7, let s try to write the determinant in a more convenient form, so that most of the elements in one particular row or column are zeros. One way is as follows. First, to each element of column 4 add the corresponding element of column 2 multiplied by 2. Next, to each element of column 5 add the corresponding element of column 2 multiplied by 4. Taking these steps, the original determinant becomes   2   0    1   3   2 0 1 1 3 2 0 2 0 5 0 2 2 0 6 1 4 2 8   0 0   2   4 4   0     1 12   1   1 8   3     2 0 16 0 1 3 2 4 1 0 2 3 5 0 6 10  0  0   11   9    16
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In the new, but equivalent, determinant to the right, notice that all the elements in row 2 are zeros except the one element 1. This is good, because it is now easy to expand the determinant in terms of the elements of row 2; upon doing this, we are able, very easily, to reduce the given fth-order determinant to a single fourth-order determinant; thus   2   1  1   3   2 1 2 0 3 5 6 5 10  0  11    9   16 
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Let us now repeat the procedure to reduce the fourth-order to a third-order determinant; one way is as follows. First, multiply all the terms of column 3 by 3 ; this is permissible provided that we also multiply the determinant by 1/3 ;
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thus (basically making use of property 5)   2 1   1 2  1 3 3 0   2 3 15 18 15 30  0  11    9   16 
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Now take the following steps: rst, to each element of column 3 add the corresponding element of column 1 multiplied by 5 ; then, to each element of column 4 add the corresponding element of column 1 multiplied by 3 ; thus      2 1 15 10  2 1 25 0 6  6      1  1 2 18 5 11 3  2 13 8     1  1  3 3 3 0 15 15 9 9  3 0 0 0      2 3 30 10  2 3 20 10  16 6  In the determinant to the right notice that all the elements of row 3 are zeros except the one element 3. Hence it s now easy to expand the determinant in terms of the minors of the elements of row 3, and doing this reduces the fourth-order determinant to a single equivalent third-order determinant; thus,      1 25 3   1 25 6      2 13 8  2 2 13 4       3 20 5   3 20 10  The nal step is to expand the third-order determinant, on the right-hand side, into a sum of basic second-order determinants. Fundamentally, this is done by expanding the third-order determinant into a sum of three second-order determinants in the basic way. Or, using property 7, we can reduce the number of secondorder determinants to just one, as follows. First, to each element of column 2 add the corresponding element of column 1 multiplied by 25. Next, to each element of column 3 add the corresponding term of column 1 multiplied by 3. Doing this reduces the third-order determinant to a single equivalent second-order determinant; thus      1 0 0   63 10    2 2 63 10  2 1    55 4  2 252 550 596; answer:   3 55 4  * The foregoing solution shows how the use of property 7 can greatly reduce the amount of work needed to nd the value of a determinant. Thus, in one step we reduced the original 5th-order determinant to a 4th-order determinant. Then, in a second step we reduced the 4th-order determinant to a 3rd-order, and in a third step we reduced the 3rd-order to the basic 2nd-order form. Thus in three steps the given 5th-order determinant was reduced to one basic 2nd-order determinant. 40. Step 1 is already satis ed. Next (step 2) we have    5 2  20 6 26  D  3 4 
* At this point we could have factored a 1 from the third row, making the elements of the third row 3, 55, 4 and thus the multiplier of the determinant 2 instead of 2.
Step 3: Let us solve for x rst:   7 D0   25 Step 4:
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