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.

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 $$A A^{-1}=A^{-1} A=I$$

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 $$A A^{-1}=A^{-1} A=I$$

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.

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.

< Question 2