====== Question 8,9 & 10, Exercise 2.2 ======
Solutions of Questions 8,9 & 10 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
=====Question 8=====
Prove that $\left| \begin{matrix}1+x & y & z  \\x & 1+y & z  \\x & y & 1+z \end{matrix} \right|=1+x+y+z$
====Solution====
Let
$$L.H.S.=\left| \begin{matrix}
   1+x & y & z  \\
   x & 1+y & z  \\
   x & y & 1+z  \\
\end{matrix} \right|$$
Subtract third row from first row. We have,
$$=\left| \begin{matrix}
   1 & 0 & -1  \\
   x & 1+y & z  \\
   x & y & 1+z  \\
\end{matrix} \right|$$
$$=1(1+yz+y+z-yz)-0-1(xy-x-xy)$$
$$=1+y+z+x$$
$$=R.H.S.$$ 

=====Question 9=====
Prove that $\left| \begin{matrix}x & p & q  \\p & x & q  \\p & q & x  \end{matrix} \right|=( x-p )( x-q )( x+p+q )$
====Solution====
Let
$$L.H.S.=\left| \begin{matrix}
   x & p & q  \\
   p & x & q  \\
   p & q & x  \\
\end{matrix} \right|$$
Subtract second row from first row and subtract third row from second row.
$$=\left| \begin{matrix}
   x-p & p-x & 0  \\
   0 & x-q & q-x  \\
   p & q & x  \\
\end{matrix} \right|$$
$$=(x-p)(x^2-qx-q^2+qx)-(p-x)(-pq+px)+0$$
$$=(x-p)(x-q)(x+q)+p(x-p)(x-q)$$
$$=(x-p)(x-q)(x+q+p)$$
$$=(x-p)(x-q)(x+q+p)$$
$$=(x-p)(x-q)(x+q+p)$$
$$=R.H.S.$$ 

=====Question 10=====
Prove that $\left| \begin{matrix}1+a & 1 & 1  \\1 & 1+b & 1  \\1 & 1 & 1+c \end{matrix} \right|=abc( 1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} )$
====Solution====
Let
$$L.H.S.=\left| \begin{matrix}
   1+a & 1 & 1  \\
   1 & 1+b & 1  \\
   1 & 1 & 1+c  \\
\end{matrix} \right|$$
Subtract second row from first row and subtract third row from second row. We have,
$$=\left| \begin{matrix}
   a & -b & 0  \\
   0 & b & -c  \\
   1 & 1 & 1+c  \\
\end{matrix} \right|$$
$$=a(b+bc+c)+b(c)+0$$
$$=abc+bc+ac+ab$$
$$=abc(1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$$
$$=R.H.S.$$ 



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