====== Question 2, Exercise 2.3 ====== Solutions of Question 2 of Exercise 2.3 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 2(i)===== Find the inverse of the matrix by using elementary row operation. $$\begin{bmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}$$ ====Solution==== Let $$A=\begin{bmatrix} 4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}.$$ Then \begin{align}|A|&=\begin{vmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{vmatrix}\\ &=4(3)+2(6)+5(4+1) \\ & =49\neq 0. \end{align} This gives, $A$ is non-singular and $A^{-1}$ exists. Now \begin{align} & \left[\begin{matrix} 4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{matrix}\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right. \right] \\ \underset{\sim}{R} & \left[\begin{matrix} 1 & 4 & 14 \\ 0 & 5 & 6 \\ -1 & 2 & 3 \end{matrix}\left| \begin{matrix} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix} \right. \right]\text{ by }R_1+3R_3\text{ and }R_2+2R_3 \\ \underset{\sim}{R}&\left[\begin{matrix} 1 & 4 & 14 \\ 0 & 5 & 6 \\ 0 & 6 & 17 \end{matrix}\left| \begin{matrix} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 1 & 0 & 4 \end{matrix} \right. \right]\text{ by }R_3+R_1\\ \underset{\sim}{R}&\left[\begin{matrix} 1 & 4 & 14 \\ 0 & 1 & 11 \\ 0 & 6 & 17 \end{matrix}\left| \begin{matrix} 1 & 0 & 3 \\ 1 & -1 & 2 \\ 1 & 0 & 4 \end{matrix} \right. \right]\text{by}-R_2+R_3\\ \underset{\sim}{R}&\left[\begin{matrix} 1 & 0 & -30 \\ 0 & 1 & 11 \\ 0 & 0 & -49 \end{matrix}\left| \begin{matrix} -3 & 4 & -5 \\ 1 & -1 & 2 \\ -5 & 6 & -8 \end{matrix} \right. \right]\text{ by }R_3-6R_2 \text{ and } R_1-4R_2\\ \underset{\sim}{R}&\left[\begin{matrix} 1 & 0 & -30 \\ 0 & 1 & 11 \\ 0 & 0 & 1 \end{matrix}\left| \begin{matrix} -3 & 4 & -5 \\ 1 & -1 & 2 \\ \dfrac{5}{49} & -\dfrac{6}{49} & \dfrac{8}{49}\end{matrix} \right. \right]\text{by}-\dfrac{1}{49}R_3\\ \underset{\sim}{R}&\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\left| \begin{matrix} \dfrac{3}{49} & \dfrac{16}{49} & -\dfrac{5}{49} \\ -\dfrac{6}{49} & \dfrac{17}{49} & \dfrac{10}{49}\\ \dfrac{5}{49} & -\dfrac{6}{49} & \dfrac{8}{49} \end{matrix} \right. \right]\text{ by }R_1+30R_3\text{ and } R_2-11R_3\end{align} Thus we have \begin{align} A^{-1}&=\begin{bmatrix} \dfrac{3}{49} & \dfrac{16}{49} & -\dfrac{5}{49} \\ -\dfrac{6}{49} & \dfrac{17}{49} & \dfrac{10}{49} \\ \dfrac{5}{49} & -\dfrac{6}{49} & \dfrac{8}{49} \end{bmatrix}\\ \implies \quad A^{-1}&=\dfrac{1}{49}\begin{bmatrix} 3 & 16 & -5 \\ -6 & 17 & 10 \\ 5 & -6 & 8 \end{bmatrix} \end{align} =====Question 2(ii)===== Find the inverse of the matrix by using elementary row operation. $$\left[ \begin{matrix} 3 & -1 & 6 \\ 1 & 3 & 4 \\ -1 & 5 & 1 \\ \end{matrix} \right]$$ ====Solution==== Let $$A=\begin{bmatrix} 3 & -1 & 6 \\ 1 & 3 & 4 \\ -1 & 5 & 1 \end{bmatrix}.$$ Then \begin{align}|A|&=\left| \begin{matrix} 3 & -1 & 6 \\ 1 & 3 & 4 \\ -1 & 5 & 1 \end{matrix} \right|\\ &=3(3-20)+1(1+4)+6(5+3)\\ &=-51+4+48\\ &=1\ne 0.\end{align} This gives, $A$ is non-singular and $A^{-1}$ exists. Now \begin{align}&\left[ \begin{matrix} 3 & -1 & 6 \\ 1 & 3 & 4 \\ -1 & 5 & 1 \end{matrix}\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right. \right]\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 3 & 4 \\ 3 & -1 & 6 \\ -1 & 5 & 1 \\ \end{matrix}\left| \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right. \right]\text{ by }R_1 \leftrightarrow R_1\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 3 & 4 \\ 0 & -10 & -6 \\ 0 & 8 & 5 \end{matrix}\left| \begin{matrix} 0 & 1 & 0 \\ 1 & -3 & 0 \\ 0 & 1 & 1 \end{matrix} \right. \right]\text{ by }R_2-3R_1 \text{ and } R_3+R_1\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 3 & 4 \\ 0 & -2 & -1 \\ 0 & 8 & 5 \end{matrix}\left| \begin{matrix} 0 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & 1 \end{matrix} \right. \right]\text{ by }R_2+R_3\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & \dfrac{1}{2} \\ 0 & 8 & 5 \end{matrix}\left| \begin{matrix} 0 & 1 & 0 \\ -\dfrac{1}{2} & 1 & -\dfrac{1}{2} \\ 0 & 1 & 1 \end{matrix} \right. \right]\text{ by }-\dfrac{1}{2}R_2\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & \dfrac{1}{2} \\ 0 & 0 & 1 \end{matrix}\left| \begin{matrix} 0 & 1 & 0 \\ \dfrac{1}{2} & 1 & -\dfrac{1}{2} \\ 4 & -7 & 5 \end{matrix} \right. \right]\text{ by }R_3-8R_2\\ \underset{\sim}{R}&\left[ \begin{matrix}1 & 3 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\left| \begin{matrix} 0 & 1 & 0 \\ -\dfrac{5}{2} & \dfrac{9}{2} & -3 \\ 4 & -7 & 5 \end{matrix} \right. \right]\text{ by }R_2-\dfrac{1}{2}R_2\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\left| \begin{matrix} \dfrac{15}{2} & -\dfrac{25}{2} & 9 \\ -\dfrac{5}{2} & \dfrac{9}{2} & -3 \\ 4 & -7 & 5 \\ \end{matrix} \right. \right]\text{ by }R_1-3R_2\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\left| \begin{matrix} -\dfrac{17}{2} & \dfrac{31}{2} & -11 \\ -\dfrac{5}{2} & \dfrac{9}{2} & -3 \\ 4 & -7 & 5 \\ \end{matrix} \right. \right]\text{ by }R_1-4R_3.\end{align} Thus $$A^{-1}=\left[ \begin{matrix} -\dfrac{17}{2} & \dfrac{31}{2} & -11 \\ -\dfrac{5}{2} & \dfrac{9}{2} & -3 \\ 4 & -7 & 5 \end{matrix} \right]$$ =====Question 2(iii)===== Find the inverse of the matrix by using elementary row operation. $$\left[ \begin{matrix} 1 & 2 & -3 \\ 0 & -2 & 0 \\ -2 & -2 & 2 \\ \end{matrix} \right]$$ ====Solution==== Let $$A=\begin{bmatrix} 1 & 2 & -3 \\ 0 & -2 & 0 \\ -2 & -2 & 2 \end{bmatrix}.$$ Then \begin{align}|A|&=\left| \begin{matrix} 1 & 2 & -3 \\ 0 & -2 & 0 \\ -2 & -2 & 2 \end{matrix} \right|\\ &=1(-4)-0-3(-4)\\ &=8\neq 0.\end{align} This gives, $A$ is non-singular and $A^{-1}$ exists. Now \begin{align}&\left[\begin{matrix} 1 & 2 & -3 \\ 0 & -2 & 0 \\ -2 & -2 & 2 \\ \end{matrix}\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right.\right]\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 2 & -3 \\ 0 & -2 & 0 \\ 0 & 2 & -4 \end{matrix}\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix} \right. \right]\text{ by }R_3+2R_1 \\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & -2 & 0 \\ 0 & 2 & -4 \end{matrix}\left| \begin{matrix} -1 & 0 & -1 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix} \right. \right]\text{ by }R_1-R_3\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & -2 \end{matrix}\left| \begin{matrix} -1 & 0 & -1 \\ 0 & -\dfrac{1}{2} & 0 \\ 1 & 0 & \dfrac{1}{2} \end{matrix} \right. \right]\text{ by }-\dfrac{1}{2}R_2 \text{ and } \dfrac{1}{2}R_3\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{matrix}\left| \begin{matrix} -1 & 0 & -1 \\ 0 & -\dfrac{1}{2} & 0 \\ 1 & \dfrac{1}{2} & \dfrac{1}{2} \end{matrix} \right. \right]\text{ by }R_3-R_2\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\left| \begin{matrix} -1 & 0 & -1 \\ 0 & -\dfrac{1}{2} & 0 \\ -\dfrac{1}{2} & -\dfrac{1}{4} & -\dfrac{1}{4} \end{matrix} \right. \right]\text{ by }-\dfrac{1}{2}R_3\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\left| \begin{matrix} -\dfrac{1}{2} & \dfrac{1}{4} & -\dfrac{3}{4} \\ 0 & -\dfrac{1}{2} & 0 \\ -\dfrac{1}{2} & -\dfrac{1}{4} & -\dfrac{1}{4} \end{matrix} \right. \right]\text{ by }R_1-R_3\end{align} Thus $$A^{-1}=\begin{bmatrix} -\dfrac{1}{2} & \dfrac{1}{4} & -\dfrac{3}{4} \\ 0 & -\dfrac{1}{2} & 0 \\ -\dfrac{1}{2} & -\dfrac{1}{4} & -\dfrac{1}{4} \end{bmatrix}$$ =====Question 2(iv)===== Find the inverse of the matrix by using elementary row operation. $$\left[ \begin{matrix} 1 & 2 & -1 \\ 0 & -1 & 3 \\ 1 & 0 & 2 \\ \end{matrix} \right]$$ ====Solution==== Let $$A=\begin{bmatrix} 1 & 2 & -1 \\ 0 & -1 & 3 \\ 1 & 0 & 2 \end{bmatrix}.$$ Then \begin{align}|A|&=\left| \begin{matrix} 1 & 2 & -1 \\ 0 & -1 & 3 \\ 1 & 0 & 2 \end{matrix} \right|\\ &=-2+6-1\\ &=3\neq 0.\end{align} This gives, $A$ is non-singular and $A^{-1}$ exists. Now \begin{align}&\left[ \begin{matrix} 1 & 2 & -1 \\ 0 & -1 & 3 \\ 1 & 0 & 2 \end{matrix}\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right. \right]\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 2 & -1 \\ 0 & -1 & 3 \\ 0 & -2 & 3 \end{matrix}\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right.\right]\text{ by }R_3-R_1\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & -1 & 3 \\ 0 & -2 & 3 \end{matrix}\left| \begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right. \right]\text{by}R_1+R_3\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & -1 & 3 \\ 0 & 0 & -3 \end{matrix}\left| \begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & -2 & 1 \end{matrix} \right.\right]\text{ by }R_3-2R_2\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & -1 & 3 \\ 0 & 0 & 1 \end{matrix}\left| \begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ \dfrac{1}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \end{matrix} \right. \right]\text{ by }-\dfrac{1}{3}R_3\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix}\left| \begin{matrix} -\dfrac{2}{3} & -\dfrac{4}{3} & \dfrac{5}{3} \\ -1 & -1 & 1 \\ \dfrac{1}{3} & \dfrac{2}{3} & -\dfrac{1}{3}\end{matrix} \right. \right]\text{ by }R_1-2R_3 \text{ and } R_2-3R_3\\ \underset{\sim}{R}&\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\left| \begin{matrix} -\dfrac{2}{3} & -\dfrac{4}{3} & \dfrac{5}{3} \\ 1 & 1 & -1 \\ \dfrac{1}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \end{matrix} \right. \right]\text{ by }-R_2.\end{align} Thus $$A^{-1}=\begin{bmatrix} -\dfrac{2}{3} & -\dfrac{4}{3} & \dfrac{5}{3} \\ 1 & 1 & -1 \\ \dfrac{1}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \end{bmatrix}.$$ ====Go To==== [[ex2-3-p1 |< Question 1]] [[ex2-3-p3|Question 3 >]]