VI (10 points) provided $ n \, (n> 1 ) $ $ A $ th order square matrix satisfies: and the elements of each row are equal to $ C $, and $ | A | = d \ neq 0 $ test requirements. All $ a $ cofactor sum $ \ sum \ limits_ {i, j = 1} ^ nA_ {ij} $.
Proof a (matrix properties) provided $ \ alpha = (1,1, \ cdots, 1) '$, by the conditions found $ A \ alpha = c \ alpha $. Since $ A $ array is non-exclusive, so $ c \ neq 0 $ (or by the $ A \ alpha = 0 $ can Release $ \ alpha = 0 $, conflict), then $ A ^ {- 1} \ alpha = c ^ {- 1}. \ alpha $ $ note A ^ * = | A | A ^ {- 1} $, so $ A ^ * \ alpha = | A | A ^ {- 1} \ alpha = \ dfrac {d} {c} \ alpha $, i.e., $ A ^ * $ of elements per row sum is equal to $ \ dfrac {d} {c } $, so $ \ sum \ limits_ {i, j = 1} ^ nA_ {ij} = \ dfrac {nd} {c} $ .
Proof bis (determinant properties) The determinant $ | A | $ column other than the first column of all $ $ J $ J $ applied to the second column, by the nature of the condition and the available determinant: $$ | A | = \ begin {vmatrix } a_ {11} & \ cdots & c & \ cdots & a_ {1n} \\ a_ {21} & \ cdots & c & \ cdots & a_ {2n} \\ \ vdots & & \ vdots & & \ vdots \\ a_ {n1} & \ cdots & c & \ cdots & a_ {nn} \\ \ end {vmatrix}. $$ right determinant of expanding in the $ J $ columns, have $$ d = c (A_ {1j } + A_ {2j} + \ cdots + A_ {nj}), $$ then $ \ sum \ limits_ {i = 1} ^ nA_ {ij} = \ dfrac {d} {c} $, so $ \ sum \ limits_ {i, j = 1} ^ nA_ {ij} = \ dfrac {nd} {c} $.
Proof tris (determinant template) using the same white paper with a high generation token 1.32 embodiments, there is $ | A (t) | = | A | + t \ sum \ limits_ {i, j = 1} ^ nA_ {ij} . $ note $ a (- \ dfrac {c } {n}) $ , and the elements of each row are equal to zero, so that it is singular matrix, then $$ 0 = | a (- \ dfrac {c} {n}) | = | A | - \ frac {c} {n} \ sum \ limits_ {i, j = 1} ^ nA_ {ij} = d- \ frac {c} {n} \ sum \ limits_ {i, j = 1} ^ nA_ {ij}, $$ thereby obtain $ \ sum \ limits_ {i, j = 1} ^ nA_ {ij} = \ dfrac {nd} {c} $.
Proof tetrakis (reduced order equation) provided $ \ alpha = (1,1, \ cdots, 1) '$, by the reduced-order equation or higher substituting Example 1.8 white paper available $$ \ begin {vmatrix} A & \ alpha \\ \ alpha '& 1 \\ \ end {vmatrix} = | a | - \ sum \ limits_ {i, j = 1} ^ nA_ {ij} $$ other hand, the left front $ determinant. all columns added n $ $ n + 1 $ the first column, available $$ | A | - \ sum \ limits_ {i, j = 1} ^ nA_ {ij} = \ begin {vmatrix} A & \ alpha \ \ \ alpha '& 1 \\ \ end {vmatrix} = \ begin {vmatrix} A & (c + 1) \ alpha \\ \ alpha' & n + 1 \\ \ end {vmatrix} = (n + 1) | A | - (c + 1 ) \ sum \ limits_ {i, j = 1} ^ nA_ {ij}, $$ thereby obtain $ \ sum \ limits_ {i, j = 1} ^ nA_ {ij} = \ dfrac {nd} {c} $. $ \ Box $
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