ISI2015-DCG-3

Question

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

Answer 1

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

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

Splitting the 1st row

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

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

Splitting the 2nd Row of 1st determinant

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

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

Splitting the 3rd Row of 1st determinant

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

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

Splitting the 4th Row of 1st determinant

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

$=0+bcd$$+\begin{vmatrix} a & 0 & 0 & 0\\ 1+0 & 1+b & 1+0&1+0 \\ 1 &1 & 1+c &1 \\ 1 & 1& 1& 1+d \end{vmatrix}$

Splitting the 2nd Row of determinant

$=bcd$$+\begin{vmatrix} a & 0 & 0 & 0\\ 1 & 1 & 1&1 \\ 1+0 &1+0 & 1+c &1+0 \\ 1 & 1& 1& 1+d \end{vmatrix}$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 1 &1 & 1+c &1 \\ 1 & 1& 1& 1+d \end{vmatrix}$

Splitting the 3rd Row of 1st determinant

$=bcd$$+\begin{vmatrix} a & 0 & 0 & 0\\ 1 & 1 & 1&1 \\ 1 &1 & 1 &1 \\ 1 & 1& 1& 1+d \end{vmatrix}+\begin{vmatrix} a & 0 & 0 & 0\\ 1 & 1 & 1&1 \\ 0 &0 & c &0 \\ 1 & 1& 1& 1+d \end{vmatrix}$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 1 &1 & 1+c &1 \\ 1 & 1& 1& 1+d \end{vmatrix}$

$=bcd$$+0+\begin{vmatrix} a & 0 & 0 & 0\\ 1 & 1 & 1&1 \\ 0 &0 & c &0 \\ 1+0 & 1+0& 1+0& 1+d \end{vmatrix}$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 1 &1 & 1+c &1 \\ 1 & 1& 1& 1+d \end{vmatrix}$

Splitting the 4th Row of 1st determinant

$=bcd$$+\begin{vmatrix} a & 0 & 0 & 0\\ 1 & 1 & 1&1 \\ 0 &0 & c &0 \\ 1 & 1& 1& 1 \end{vmatrix}$$+\begin{vmatrix} a & 0 & 0 & 0\\ 1 & 1 & 1&1 \\ 0 &0 & c &0 \\ 0 & 0& 0& d \end{vmatrix}$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 1 &1 & 1+c &1 \\ 1 & 1& 1& 1+d \end{vmatrix}$

$=bcd+0+acd$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 1+0 &1+0 & 1+c &1+0 \\ 1 & 1& 1& 1+d \end{vmatrix}$

Splitting the 3rd Row of 1st determinant

$=bcd+acd$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 1 &1 & 1 &1 \\ 1+0 & 1+0& 1+0& 1+d \end{vmatrix}$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 0 &0 & c &0 \\ 1 & 1& 1& 1+d \end{vmatrix}$

Splitting the 4th Row of 1st determinant

$=bcd+acd$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 1 &1 & 1 &1 \\ 1 & 1& 1& 1 \end{vmatrix}$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 1 &1 & 1 &1 \\ 0 & 0& 0& d \end{vmatrix}$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 0 &0 & c &0 \\ 1 & 1& 1& 1+d \end{vmatrix}$

$=bcd+acd+0+abd$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 0 &0 & c &0 \\ 1+0 & 1+0& 1+0& 1+d \end{vmatrix}$

Splitting the 4th Row of  determinant

$=bcd+acd+abd$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 0 &0 & c &0 \\ 1 & 1& 1& 1 \end{vmatrix}$$+\begin{vmatrix} a & 0 & 0 & 0\\ 0 & b & 0&0 \\ 0 &0 & c &0 \\ 0 & 0& 0& d \end{vmatrix}$

$=bcd+acd+abd+abc+abcd$

$=abcd\left ( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} +\frac{1}{d}+1 \right )$

 

$\therefore$ Option $A.$ is the correct answer.

Answer 2

This problem can be solved using the transformation rules of the matrix.