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 |abclmnxyz|=|alxbmycnz|
Solution
L.H.S.=|abclmnxyz|=|abclmnxyz|t∵
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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