====== Question 5, Exercise 2.2 ======
Solutions of Question 5 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 5(i)=====
Show that $\begin{vmatrix}a & b & c\\l & m & n\\x & y & z \end{vmatrix}=\begin{vmatrix}a & l & x\\b & m & y\\c & n & z \end{vmatrix}$
====Solution====
\begin{align}L.H.S.&=\begin{vmatrix}
a & b & c  \\
l & m & n  \\
x & y & z
\end{vmatrix}\\
&=\begin{vmatrix}
a & b & c  \\l & m & n  \\x & y & z
\end{vmatrix}^t \quad \because |A|=|A^t| \\
&=\begin{vmatrix}
a & l & x  \\
b & m & y  \\
c & n & z
\end{vmatrix} =R.H.S. \end{align} 


=====Question 5(ii)=====
Show that $\begin{vmatrix}a & b & c\\1-3a & 2-3b & 3-3c\\4 & 5 & 6 \end{vmatrix}=\begin{vmatrix}a & b & c\\1 & 2 & 3\\4 & 5 & 6 \end{vmatrix}.$
====Solution====
\begin{align}L.H.S.&=\begin{vmatrix}
a & b & c\\1-3a & 2-3b & 3-3c\\4 & 5 & 6
\end{vmatrix}\\
&=\begin{vmatrix}
a & b & c  \\
1-3a+3a & 2-3b+3b & 3-3c+3c  \\
4 & 5 & 6
\end{vmatrix} \text{ by } R_2-3R_1 \\
&=\begin{vmatrix}
a & b & c  \\1 & 2 & 3  \\4 & 5 & 6
\end{vmatrix}=R.H.S. \end{align} 

=====Question 5(iii)=====
Show that $\left| \begin{matrix}1 & 1 & 1  \\a & b & c  \\b+c & c+a & a+b \end{matrix} \right|=0$
====Solution====
\begin{align}L.H.S.&=\begin{vmatrix}
1 & 1 & 1  \\
a & b & c  \\
b+c & c+a & a+b
\end{vmatrix}\\
&=\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
a+b+c & b+c+a & c+a+b \end{vmatrix} \text{ by } R_3+R_2 \\
&=(a+b+c)\begin{vmatrix}
1 & 1 & 1\\ a & b & c\\ 1 & 1 & 1
\end{vmatrix} \text{taking common from }R_3 \\
&=(a+b+c)0 \text{ by } R_1\simeq R_3 \\
&=0=R.H.S.\end{align}



=====Question 5(iv)=====
Show that $\left| \begin{matrix}bc & ca & ab  \\a & b & c  \\a^2 & b^2 & c^2  \end{matrix} \right|=\left| \begin{matrix}1 & 1 & 1  \\
a^2 & b^2 & c^2  \\a^3 & b^3 & c^3 \end{matrix} \right|$
====Solution====
\begin{align}L.H.S.&=\begin{vmatrix}
 bc & ca & ab\\a & b & c \\a^2 & b^2 & c^2
\end{vmatrix}\\
&=\dfrac{1}{abc}\begin{vmatrix}
abc & bca & abc  \\
a^2 & b^2 & c^2  \\
a^3 & b^3 & c^3  \end{vmatrix} \text{ by } aR_1, bR_2, cR_3 \\
&=\dfrac{abc}{abc}\begin{vmatrix}
1 & 1 & 1  \\
a^2 & b^2 & c^2 \\
a^3 & b^3 & c^3
\end{vmatrix} \text{ by taking }abc\text{ common from }R_1 \\
&=\begin{vmatrix}
1 & 1 & 1\\
a^2 & b^2 & c^2\\
a^3 & b^3 & c^3 \end{vmatrix} =R.H.S. \end{align}



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