# Question 2, Exercise 2.2

Solutions of Question 2 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.

Without evaluating state the reasons for the equalities. $\left|\begin{matrix} 1 & 2 & 0 \\3 & 1 & 0 \\-1 & 2 & 0 \end{matrix}\right|=0$.

Given $$\left| \begin{matrix} 1 & 2 & 0 \\ 3 & 1 & 0 \\ -1 & 2 & 0 \\ \end{matrix} \right|=0$$ Because elements of third column are zero.

Without evaluating state the reasons for the equalities. $\left| \begin{matrix}1 & 2 & 3 \\-8 & 4 & -12 \\2 & -1 & 3 \end{matrix} \right|=0$.

Given $$\left| \begin{matrix} 1 & 2 & 3 \\ -8 & 4 & -12 \\ 2 & -1 & 3 \\ \end{matrix} \right|=0$$ Taking $-4$ common from $R_2$, we have $$-4\left| \begin{matrix} 1 & 2 & 3 \\ 2 & -1 & 3 \\ 2 & -1 & 3 \\ \end{matrix} \right|=0$$ As elements of second and third rows are identical, so result is zero.

Without evaluating state the reasons for the equalities. $\left| \begin{matrix}1 & 3 & -2 \\3 & -1 & 1 \\2 & 1 & 4\end{matrix} \right|=\left| \begin{matrix}1 & 3 & 2 \\3 & -1 & 1 \\-2 & 1 & 4 \end{matrix} \right|$.

Given $$\left| \begin{matrix} 1 & 3 & -2 \\ 3 & -1 & 1 \\ 2 & 1 & 4 \\ \end{matrix} \right|=\left| \begin{matrix} 1 & 3 & 2 \\ 3 & -1 & 1 \\ -2 & 1 & 4 \\ \end{matrix} \right|$$ Right side is the transpose of left one.

Without evaluating state the reasons for the equalities. $\left| \begin{matrix}3 & 2 & 0 \\1 & 1 & -3 \\2 & 4 & -6 \end{matrix} \right|=-3\left| \begin{matrix}3 & 2 & 0 \\1 & 1 & 1 \\2 & 4 & 2 \end{matrix} \right|$.

Given $$\left| \begin{matrix}3 & 2 & 0 \\1 & 1 & -3 \\2 & 4 & -6\\ \end{matrix} \right|=-3\left| \begin{matrix}3 & 2 & 0 \\1 & 1 & 1 \\2 & 4 & 2\\ \end{matrix} \right|$$ Multiply $-3$ by third column of R.H.S.

Without evaluating state the reasons for the equalities. $\left| \begin{matrix}1 & 0 & -1 \\3 & 2 & 1 \\1 & -1 & 0 \end{matrix} \right|=-\left| \begin{matrix}1 & 0 & -1 \\1 & -1 & 0 \\3 & 2 & 1 \\\end{matrix} \right|$.

Given $$\left| \begin{matrix} 1 & 0 & -1 \\ 3 & 2 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right|=-\left| \begin{matrix} 1 & 0 & -1 \\ 1 & -1 & 0 \\ 3 & 2 & 1 \\ \end{matrix} \right|$$ Interchanging the second and third rows on L.H.S.

Without evaluating state the reasons for the equalities. $\left| \begin{matrix}2 & 0 & 1 \\3 & 1 & 2 \\1 & 2 & 2 \end{matrix} \right|=\left| \begin{matrix} 2 & 0 & 1 \\5 & 5 & 6 \\1 & 2 & 2 \end{matrix} \right|$.

Given $$\left| \begin{matrix} 2 & 0 & 1 \\ 3 & 1 & 2 \\ 1 & 2 & 2 \\ \end{matrix} \right|=\left| \begin{matrix} 2 & 0 & 1 \\ 5 & 5 & 6 \\ 1 & 2 & 2 \\ \end{matrix} \right|$$ Multiply the third row by $2$ and add it in 2nd row of L.H.S to get R.H.S.