# Question 6, Exercise 2.2

Solutions of Question 6 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.

Prov that $\left| \begin{matrix}a-b & b-c & c-a \\b-c & c-a & a-b \\c-a & a-b & b-c \end{matrix} \right|=0$

Let \begin{align} L.H.S&=\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right| \\ &=\left| \begin{matrix} a-c & b-a & c-b \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right| \quad R_1+R_2 \\ &=-\left| \begin{matrix} c-a & a-b & b-c \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right|\quad -R_1 \\ &=0 \quad R_1 \cong R_3 \\ &=R.H.S. \end{align}

Prov that $\left| \begin{matrix}1 & a & a^3 \\1 & b & b^3 \\1 & c & c^3 \end{matrix} \right|=( a-b )( b-c )( c-a )( a+b+c )$

Let $$L.H.S.=\left| \begin{matrix} 1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3 \\ \end{matrix} \right|$$ Subtract first row from second row and second row from third row. We have, $$=\left| \begin{matrix} 0 & a-b & a^3-b^3 \\ 0 & b-c & b^3-c^3 \\ 1 & c & c^3 \\ \end{matrix} \right|$$ $$=\left| \begin{matrix} 0 & ( a-b ) & ( a-b )( a^2+ab+b^2 ) \\ 0 & ( b-c ) & ( b-c )( b^2+bc+c^2 ) \\ 1 & c & c^3 \\ \end{matrix} \right|$$ $$=( a-b )( b-c )\left| \begin{matrix} 0 & 1 & ( a^2+ab+b^2 ) \\ 0 & 1 & ( b^2+bc+c^2 ) \\ 1 & c & c^3 \\ \end{matrix} \right|$$ $$=( a-b )( b-c )\left| \begin{matrix} 1 & ( a^2+ab+b^2 ) \\ 1 & ( b^2+bc+c^2 ) \\ \end{matrix} \right|$$ $$=( a-b )( b-c )\left[ ( b^2+bc+c^2 )-( a^2+ab+b^2 ) \right]$$ $$=( a-b )( b-c )( b^2+bc+c^2-a^2-ab-b^2)$$ $$=( a-b )( b-c )( bc+c^2-a^2-ab )$$ $$=( a-b )( b-c )( b( c-a )+( c-a )( c+a ) )$$ $$=( a-b )( b-c )( c-a )( a+b+c )$$ $$=R.H.S.$$

Prov that $\left| \begin{matrix}1 & a & a^2 \\1 & b & b^2 \\1 & c & c^2 \end{matrix} \right|=( a-b )( b-c )( c-a )$

Let $$L.H.S.=\left| \begin{matrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{matrix} \right|$$ Subtract first row from second row and second row from third row, we have, $$=\left| \begin{matrix} 0 & a-b & a^2-b^2 \\ 0 & b-c & b^2-c^2 \\ 1 & c & c^2 \\ \end{matrix} \right|$$ $$=\left| \begin{matrix} 0 & ( a-b ) & ( a-b)( a+b ) \\ 0 & ( b-c ) & ( b-c )( b+c ) \\ 1 & c & c^2 \\ \end{matrix} \right|$$ $$=( a-b )( b-c )\left| \begin{matrix} 0 & 1 & ( a+b ) \\ 0 & 1 & ( b+c ) \\ 1 & c & c^2 \\ \end{matrix} \right|$$ $$=( a-b )( b-c )\left| \begin{matrix} 1 & ( a+b ) \\ 1 & ( b+c ) \\ \end{matrix} \right|$$ $$=( a-b )( b-c )( b+c-b-a )$$ $$=( a-b )( b-c )( c-a )$$ $$=R.H.S.$$

Prove that $\left| \begin{matrix}-a^2 & ab & ac \\ab & -b^2 & bc \\ac & bc & -c^2 \end{matrix} \right|=4a^2b^2c^2$

\begin{align}L.H.S.&=\left| \begin{matrix} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \\ \end{matrix} \right| \\ &=-a^2( b^2c^2-b^2c^2)-ab( -abc^2-abc^2 )\\ &\qquad +ac( ab^2c+ab^2c )\\ &=0+2a^2b^2c^2+2a^2b^2c^2\\ &=4a^2b^2c^2=R.H.S. \end{align}

Prov that $\left| \begin{matrix} bc & a^3 & \dfrac{1}{a} \\ca & b^3 & \dfrac{1}{b} \\ab & c^3 & \dfrac{1}{c}\end{matrix} \right|=0\quad a\ne 0,\,\,b\ne 0,\,\,c\ne 0$.

Let $$L.H.S.=\left| \begin{matrix} bc & a^3 & \dfrac{1}{a} \\ ca & b^3 & \dfrac{1}{b} \\ ab & c^3 & \dfrac{1}{c} \\ \end{matrix} \right|$$ Multiply first row by $a$, second row by $b$ and third row by $c.$ We have, $$=\dfrac{1}{abc}\left| \begin{matrix} abc & a^4 & a\dfrac{1}{a} \\ bca & b^4 & b\dfrac{1}{b} \\ cab & c^4 & c\dfrac{1}{c} \\ \end{matrix} \right|$$ $$=\dfrac{abc}{abc}\left| \begin{matrix} 1 & a^4 & 1 \\ 1 & b^4 & 1 \\ 1 & c^4 & 1 \\ \end{matrix} \right|$$ $$=\left| \begin{matrix} 1 & a^4 & 1 \\ 1 & b^4 & 1 \\ 1 & c^4 & 1 \\ \end{matrix} \right|$$ Two identical columns, so determinant is zero. $$=0$$ $$=R.H.S.$$