====== Question 3, Exercise 2.5 ====== Solutions of Question 3 of Exercise 2.5 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. =====Question 3(i)===== With the help of row operations, find the inverse of the matrix $\left[\begin{array}{ccc}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4\end{array}\right]$ if it exists. Also verify your answer by showing that $A A^{-1}=A^{-1} A=I$.\\ ** Solution. ** Let \begin{align*} A&=\left[ \begin{array}{ccc} 0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4 \end{array} \right]\\ |A|&=0+1(-4)-1(1-3)\\ &=-4+3\\ &=-1\neq 0\end{align*} So $A$ is non singular. Now consider \begin{align*} &\quad\left[ \begin{array}{ccc|ccc} 0 & -1 & -1 & 1 & 0 & 0 \\ -1 & 3 & 0 & 0 & 1 & 0 \\ 1 & -1 & 4 & 0 & 0 & 1 \end{array} \right]\\ \sim &{\text{R}} \left[ \begin{array}{ccc|ccc} 1 & -1 & 4 & 0 & 0 & 1 \\ -1 & 3 & 0 & 0 & 1 & 0 \\ 0 & -1 & -1 & 1 & 0 & 0 \end{array} \right] \quad \text{by swapping } R1 \text{ and } R3\\ \sim&{\text{R}} \left[ \begin{array}{ccc|ccc} 1 & -1 & 4 & 0 & 0 & 1 \\ 0 & 2 & 4 & 0 & 1 & 1 \\ 0 & -1 & -1 & 1 & 0 & 0 \end{array} \right] \quad \text{by } R2 + R1 \\ \sim&{\text{R}} \left[ \begin{array}{ccc|ccc} 1 & 0 & 5 & -1 & 0 & 1 \\ 0 & 1 & 3 & 1 & 1 & 1 \\ 0 & 0 & 2 & 2 & 1 & 1 \end{array} \right] \quad \text{by }R1 - R3\quad 2R3 + R2 \\ \sim&{\text{R}} \left[ \begin{array}{ccc|ccc} 1 & 0 & 5 & -1 & 0 & 1 \\ 0 & 1 & 3 & 1 & 1 & 1 \\ 0 & 0 & 2 & 2 & 1 & 1 \end{array} \right] \quad \text{by } R2-R_3 \quad \frac{1}{2} R3\\ \sim&{\text{R}} \left[ \begin{array}{ccc|ccc} 1 & 0 & 5 & -1 & 0 & 1 \\ 0 & 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 & \frac{1}{2} & \frac{1}{2} \end{array} \right] \quad \text{by } R1 + R2 \text{ and } \frac{1}{2} R3\\ \sim&{\text{R}} \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & -6 & -\frac{5}{2} & -\frac{3}{2} \\ 0 & 1 & 0 & -2 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & 1 & 1 & \frac{1}{2} & \frac{1}{2} \end{array} \right] \quad \text{by } R2 -R3\\ A^{-1} &= \left[ \begin{array}{ccc} -6 & -\frac{5}{2} & -\frac{3}{2} \\ -2 & -\frac{1}{2} & -\frac{1}{2} \\ 1 & \frac{1}{2} & \frac{1}{2} \end{array} \right]\\ \end{align*} To verify, we need to show that \( A A^{-1} = A^{-1} A = I \): \begin{align*} AA^{-1}& = \left[ \begin{array}{ccc} 0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4 \end{array} \right] \left[ \begin{array}{ccc} -6 & -\frac{5}{2} & -\frac{3}{2} \\ -2 & -\frac{1}{2} & -\frac{1}{2} \\ 1 & \frac{1}{2} & \frac{1}{2} \end{array} \right]\\ &= \left[ \begin{array}{ccc} 0 + 2 - 1 & 0 + \frac{1}{2} - \frac{1}{2} & 0 + \frac{1}{2} - \frac{1}{2} \\ 6 - 6 + 0 & \frac{5}{2} - \frac{3}{2} + 0 & \frac{3}{2} - \frac{3}{2} + 0 \\ -6 + 2 + 4 & -\frac{5}{2} + \frac{1}{2} + 2 & -\frac{3}{2} + \frac{1}{2} + 2 \end{array} \right]\\ &= \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = I\\ A^{-1} A &= \left[ \begin{array}{ccc} -6 & -\frac{5}{2} & -\frac{3}{2} \\ -2 & -\frac{1}{2} & -\frac{1}{2} \\ 1 & \frac{1}{2} & \frac{1}{2} \end{array} \right] \left[ \begin{array}{ccc} 0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4 \end{array} \right]\\ &= \left[ \begin{array}{ccc} 0 + \frac{5}{2} - \frac{3}{2} & 6 - \frac{15}{2} + \frac{3}{2} & 6 + 0 - 6 \\ 0 + \frac{1}{2} - \frac{1}{2} & 2 - \frac{3}{2} + \frac{1}{2} & 2 + 0 - 2 \\ 0 - \frac{1}{2} + \frac{1}{2} & -1 + \frac{3}{2} - \frac{1}{2} & -1 + 0 + 2 \end{array} \right]\\ &= \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = I \end{align*} Hence $$A A^{-1}=A^{-1} A=I$$ =====Question 3(ii)===== With the help of row operations, find the inverse of the matrix $\left[\begin{array}{ccc}1 & 2 & 5 \\ -3 & 0 & 1 \\ 4 & 2 & 5\end{array}\right]$ if it exists. Also verify your answer by showing that $A A^{-1}=A^{-1} A=I$.\\ ** Solution. ** \begin{align*} A &= \left[ \begin{array}{ccc} 1 & 2 & 5 \\ -3 & 0 & 1 \\ 4 & 2 & 5 \end{array} \right] \\ |A|&=1(-2)-2(-15-4)+5(-6)\\ &=-2+19-3-19\neq 0 \end{align*} Now we will find $A^{-1}$ \begin{align*} &\quad\left[ \begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ -3 & 0 & 1 & 0 & 1 & 0 \\ 4 & 2 & 5 & 0 & 0 & 1 \end{array} \right]\\ =&\left[ \begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 6 & 16 & 3 & 1 & 0 \\ 4 & 2 & 5 & 0 & 0 & 1 \end{array} \right]\quad R_2+ 3R_1\\ =&\left[ \begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 6 & 16 & 3 & 1 & 0 \\ 0 & -6 & -15 & -4 & 0 & 1 \end{array} \right]\quad R_3-4R_1\\ =&\left[ \begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 6 & 16 & 3 & 1 & 0 \\ 0 & 0 & 1 & -1 & 1 & 1 \end{array} \right]\quad R_3+ R_2\\ =&\left[ \begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 1 & \frac{8}{3} & \frac{1}{2} & \frac{1}{6} & 0 \\ 0 & 0 & 1 & -1 & 1 & 1 \end{array} \right]\quad \frac{1}{6}R_2\\ =&\left[ \begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ 0 & 0 & 1 & -1 & 1 & 1 \end{array} \right]\quad R_2-\frac{8}{3} R_3\\ =&\left[ \begin{array}{ccc|ccc} 1 & 2 & 0 & 6 & -5 & -5 \\ 0 & 1 & 0 & \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ 0 & 0 & 1 & -1 & 1 & 1 \end{array} \right]\quad R_1-5R_3\\ =&\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 &- \frac{1}{3} & 0 &\frac{1}{3} \\ 0 & 1 & 0 & \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ 0 & 0 & 1 & -1 & 1 & 1 \end{array} \right]\quad R_1-2R_2 \end{align*} Thus, the inverse of $A $ is: \begin{align*} A^{-1}& = \left[ \begin{array}{ccc} - \frac{1}{3} & 0 &\frac{1}{3} \\ \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ -1 & 1 & 1 \end{array} \right] \end{align*} Now we show that $A A^{-1}=A^{-1} A=I$ \begin{align*} AA^{-1}& = \left[ \begin{array}{ccc} 1 & 2 & 5 \\ -3 & 0 & 1 \\ 4 & 2 & 5 \end{array} \right] \left[ \begin{array}{ccc} - \frac{1}{3} & 0 &\frac{1}{3} \\ \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ -1 & 1 & 1 \end{array} \right]\\ &= \left[ \begin{array}{ccc} -\frac{1}{3} + \frac{19}{3} - \frac{15}{3} & 0 - 5 + 5 & \frac{1 - 16 + 15}{3} \\ 1 + 0 - 1& 0 + 0 + 1 & -1 + 0 + 1 \\ -\frac{4}{3} + \frac{19}{3} - \frac{15}{3} & 0 - 5 + 5 & \frac{4 - 16 + 15}{3} \end{array} \right] \\ &= \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]= I\\ A^{-1}A& = \left[ \begin{array}{ccc} - \frac{1}{3} & 0 &\frac{1}{3} \\ \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ -1 & 1 & 1 \end{array} \right] \left[ \begin{array}{ccc} 1 & 2 & 5 \\ -3 & 0 & 1 \\ 4 & 2 & 5 \end{array} \right]\\ &= \left[ \begin{array}{ccc} -\frac{1}{3} + 0 + \frac{4}{3} & -\frac{2}{3} + 0 + \frac{2}{3} & -\frac{5}{3} + 0 + \frac{5}{3} \\ \frac{19}{6} + \frac{45}{6} - \frac{64}{6}& \frac{38}{6} - \frac{32}{6} & \frac{95}{6} - \frac{15}{6} - \frac{80}{6} \\ -1 - 3 + 4 & - 2 + 0 + 2 & -5 + 1 + 5 \end{array} \right] \\ &= \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = I \end{align*} Hence $$A A^{-1}=A^{-1} A=I$$ =====Question 3(iii)===== With the help of row operations, find the inverse of the matrix $\left[\begin{array}{ccc}-5 & 2 & 3 \\ -1 & -2 & 3 \\ 1 & -2 & 3\end{array}\right]$ if it exists. Also verify your answer by showing that $A A^{-1}=A^{-1} A=I$. ** Solution. ** Do yourself. =====Question 3(iv)===== With the help of row operations, find the inverse of the matrix $\left[\begin{array}{ccc}0 & 1 & 3 \\ 3 & 2 & 4 \\ 6 & -1 & 2\end{array}\right]$ if it exists. Also verify your answer by showing that $A A^{-1}=A^{-1} A=I$. ** Solution. ** Do yourself. ====Go to ==== [[math-11-nbf:sol:unit02:ex2-5-p2|< Question 2]]