#### Explain solution RD Sharma class 12 chapter 5 Determinants exercise multiple choise question 32 maths

Correct option (d)

Hint:

We know that,

If each element of a row or column of determinant is expressed as a sum of two or more terms, then the determinant can be expressed as the sum of two or more determinant.

Given:

Given that

$\begin{vmatrix} a &p &x \\ b &q &y \\ c &r &z \end{vmatrix}=16$

We have to find the value of

$\left|\begin{array}{lll} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{array}\right|$

Solution:

By using properties we can write,

$\left|\begin{array}{lll} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{array}\right|=\left|\begin{array}{lll} p & x & a \\ q & y & b \\ r & z & c \end{array}\right|+\left|\begin{array}{ccc} x & a & p \\ y & b & q \\ z & c & r \end{array}\right|$

Applying C1↔C3  in first determinant and applying C2↔C1  in second determinant.

$=(-)\left|\begin{array}{lll} a & x & p \\ b & y & q \\ c & z & r \end{array}\right|+(-)\left|\begin{array}{lll} a & x & p \\ b & y & q \\ c & z & r \end{array}\right|$

Again applying C2↔C3  in first determinant and C2↔C3  in second determinant.

$=\left | a\: p\: x\: b\: q\: y\: c\: r\: z \right |+\left | a\: p\: x\: b\: q\: y\: c\: r\: z \right |$

$=16+16 = 32$

$\left|\begin{array}{lll} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{array}\right|=32$
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