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.

Prove that $\left| \begin{matrix}1+x & y & z \\x & 1+y & z \\x & y & 1+z \end{matrix} \right|=1+x+y+z$

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, $$=\left| \begin{matrix} 1 & 0 & -1 \\ x & 1+y & z \\ x & y & 1+z \\ \end{matrix} \right|$$ $$=1(1+yz+y+z-yz)-0-1(xy-x-xy)$$ $$=1+y+z+x$$ $$=R.H.S.$$

Prove that $\left| \begin{matrix}x & p & q \\p & x & q \\p & q & x \end{matrix} \right|=( x-p )( x-q )( x+p+q )$

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. $$=\left| \begin{matrix} x-p & p-x & 0 \\ 0 & x-q & q-x \\ p & q & x \\ \end{matrix} \right|$$ $$=(x-p)(x^2-qx-q^2+qx)-(p-x)(-pq+px)+0$$ $$=(x-p)(x-q)(x+q)+p(x-p)(x-q)$$ $$=(x-p)(x-q)(x+q+p)$$ $$=(x-p)(x-q)(x+q+p)$$ $$=(x-p)(x-q)(x+q+p)$$ $$=R.H.S.$$

Prove that $\left| \begin{matrix}1+a & 1 & 1 \\1 & 1+b & 1 \\1 & 1 & 1+c \end{matrix} \right|=abc( 1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} )$

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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