====== 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.
=====Questiopn 6(i)=====
Prov that $\left| \begin{matrix}a-b & b-c & c-a \\b-c & c-a & a-b \\c-a & a-b & b-c \end{matrix} \right|=0$
====Solution====
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}
=====Question 6(ii)=====
Prov that $\left| \begin{matrix}1 & a & a^3 \\1 & b & b^3 \\1 & c & c^3 \end{matrix} \right|=( a-b )( b-c )( c-a )( a+b+c )$
====Solution====
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,
$$=\left| \begin{matrix}
0 & a-b & a^3-b^3 \\
0 & b-c & b^3-c^3 \\
1 & c & c^3 \\
\end{matrix} \right|$$
$$=\left| \begin{matrix}
0 & ( a-b ) & ( a-b )( a^2+ab+b^2 ) \\
0 & ( b-c ) & ( b-c )( b^2+bc+c^2 ) \\
1 & c & c^3 \\
\end{matrix} \right|$$
$$=( a-b )( b-c )\left| \begin{matrix}
0 & 1 & ( a^2+ab+b^2 ) \\
0 & 1 & ( b^2+bc+c^2 ) \\
1 & c & c^3 \\
\end{matrix} \right|$$
$$=( a-b )( b-c )\left| \begin{matrix}
1 & ( a^2+ab+b^2 ) \\
1 & ( b^2+bc+c^2 ) \\
\end{matrix} \right|$$
$$=( a-b )( b-c )\left[ ( b^2+bc+c^2 )-( a^2+ab+b^2 ) \right]$$
$$=( a-b )( b-c )( b^2+bc+c^2-a^2-ab-b^2)$$
$$=( a-b )( b-c )( bc+c^2-a^2-ab )$$
$$=( a-b )( b-c )( b( c-a )+( c-a )( c+a ) )$$
$$=( a-b )( b-c )( c-a )( a+b+c )$$
$$=R.H.S. $$
=====Question 6(iii)=====
Prov that $\left| \begin{matrix}1 & a & a^2 \\1 & b & b^2 \\1 & c & c^2 \end{matrix} \right|=( a-b )( b-c )( c-a )$
====Solution====
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,
$$=\left| \begin{matrix}
0 & a-b & a^2-b^2 \\
0 & b-c & b^2-c^2 \\
1 & c & c^2 \\
\end{matrix} \right|$$
$$=\left| \begin{matrix}
0 & ( a-b ) & ( a-b)( a+b ) \\
0 & ( b-c ) & ( b-c )( b+c ) \\
1 & c & c^2 \\
\end{matrix} \right|$$
$$=( a-b )( b-c )\left| \begin{matrix}
0 & 1 & ( a+b ) \\
0 & 1 & ( b+c ) \\
1 & c & c^2 \\
\end{matrix} \right|$$
$$=( a-b )( b-c )\left| \begin{matrix}
1 & ( a+b ) \\
1 & ( b+c ) \\
\end{matrix} \right|$$
$$=( a-b )( b-c )( b+c-b-a )$$
$$=( a-b )( b-c )( c-a )$$
$$=R.H.S.$$
=====Question 6(iv)=====
Prove that $\left| \begin{matrix}-a^2 & ab & ac \\ab & -b^2 & bc \\ac & bc & -c^2 \end{matrix} \right|=4a^2b^2c^2$
====Solution====
\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}
=====Question 6(v)=====
Prov that $\left| \begin{matrix} bc & a^3 & \dfrac{1}{a} \\ca & b^3 & \dfrac{1}{b} \\ab & c^3 & \dfrac{1}{c}\end{matrix} \right|=0\quad a\ne 0,\,\,b\ne 0,\,\,c\ne 0$.
====Solution====
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.
$$=0$$
$$=R.H.S.$$
====Go To====
[[math-11-kpk:sol:unit02:ex2-2-p5 |< Question 5]]
[[math-11-kpk:sol:unit02:ex2-2-p7|Question 7 >]]