Basis maths-linear algebra-Q13

+1 vote

The value of  \begin{vmatrix} 1+a& 1& 1& 1\\ 1&1+b &1 &1 \\ 1&1 &1+c &1 \\ 1&1 &1 &1+d \end{vmatrix}  is 

(A). abcd(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})

(B). abcd(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})

(C). 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}

(D). None of these

asked May 15 in Basic Maths by gbeditor (44,500 points)
reshown May 16 by gbeditor

1 Answer

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Best answer

We do these transformations :

Row 1 \leftarrow Row1-Row2 \\ Row 2 \leftarrow Row2-Row3 \\ Row 3 \leftarrow Row3-Row4

So, now we have :

\begin{vmatrix} a& -b& 0& 0\\ 0&b &-c &0 \\ 0&0 &c &-d \\ 1&1 &1 &1+d \end{vmatrix}

Expnading it, we get =

a[ bc(1+d) + cd +bd ] + b(cd)

abcd + abc + acd + abd + bcd

Which is same as Option A.

answered Jun 26 by deepak-gatebook (112,390 points)
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