Question 7 and 8, Exercise 2.6
Solutions of Question 7 and 8 of Exercise 2.6 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 7
If $A=\left[\begin{array}{ccc}3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3\end{array}\right]$; find $A^{-1}$ and hence solve the system of equations.
$3 x+4 y+7 z=14 ; 2 x-y+3 z=4 ; \quad x+2 y-3 z=0$.
Solution.
Given \begin{align*} A &= \begin{bmatrix} 3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3 \end{bmatrix}\\ |A| &=3(-3) - 2(-26) + 19\\ &= -9 + 52 + 19\\ &= 62 \neq 0\end{align*} This system is consistent. Now to find $A^{-1}$, we calculate the cofactors of each element . \begin{align*} A_{11} &= (-1)^{1+1} \left| \begin{array}{cc} -1 & 2 \\ 3 & -3 \end{array} \right| = 3 - 6 = -3 \\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 4 & 2 \\ 7 & -3 \end{array} \right| = -(-12 - 14) = 26 \\ A_{13} &= (-1)^{1+3} \left| \begin{array}{cc} 4 & -1 \\ 7 & 3 \end{array} \right| = 12 + 7 = 19 \\ A_{21} &= (-1)^{2+1} \left| \begin{array}{cc} 2 & 1 \\ 3 & -3 \end{array} \right| = -(-6 - 3) = 9 \\ A_{22} &= (-1)^{2+2} \left| \begin{array}{cc} 3 & 1 \\ 7 & -3 \end{array} \right| = -9 - 7 = -16 \\ A_{23} &= (-1)^{2+3} \left| \begin{array}{cc} 3 & 2 \\ 7 & 3 \end{array} \right| = -(9 - 14) = 5 \\ A_{31} &= (-1)^{3+1} \left| \begin{array}{cc} 2 & 1 \\ -1 & 2 \end{array} \right| = 4 + 1 = 5 \\ A_{32} &= (-1)^{3+2} \left| \begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array} \right| = -(6 - 4) = -2 \\ A_{33} &= (-1)^{3+3} \left| \begin{array}{cc} 3 & 2 \\ 4 & -1 \end{array} \right| = -3 - 8 = -11 \\ A &= \begin{bmatrix} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{bmatrix}\\ \text{adj}(A) &= \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11 \end{bmatrix}\\ A^{-1} &= \frac{1}{62} \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11 \end{bmatrix}\\ A^{-1} &= \begin{bmatrix} \dfrac{-3}{62} & \dfrac{9}{62} & \dfrac{5}{62} \\ \dfrac{26}{62} & \dfrac{-16}{62} & \dfrac{-2}{62} \\ \dfrac{19}{62} & \dfrac{-5}{62} & \dfrac{-11}{62} \end{bmatrix}\\ A^{-1} &= \begin{bmatrix} \dfrac{-3}{62} & \dfrac{9}{62} & \dfrac{5}{62} \\ \dfrac{13}{32} & \dfrac{-8}{32} & \dfrac{-1}{32} \\ \dfrac{19}{62} & \dfrac{-5}{62} & \dfrac{-11}{62} \end{bmatrix} \end{align*} Therefore, the inverse of matrix $A $ is: $$A^{-1} = \begin{bmatrix} \dfrac{-3}{62} & \dfrac{9}{62} & \dfrac{5}{62} \\ \dfrac{13}{32} & \dfrac{-8}{32} & \dfrac{-1}{32} \\ \dfrac{19}{62} & \dfrac{-5}{62} & \dfrac{-11}{62} \end{bmatrix}$$ Now given the system of equations: \begin{align*} 3x + 4y + 7z &= 14 \\ 2x - y + 3z &= 4 \\ x + 2y - 3z &= 0 \end{align*} The associated augmented matrix for this system is: \begin{align*} A_b &= \begin{bmatrix} 3 & 4 & 7 & 14 \\ 2 & -1 & 3 & 4 \\ 1 & 2 & -3 & 0 \end{bmatrix}\\ &\sim \text{R} = \begin{bmatrix} 1 & 2 & -3 & 0\\ 2 & -1 & 3 & 4 \\ 3 & 4 & 7 & 14 \end{bmatrix} \quad R_1 \text{interchange} R_3\\ &\sim \text{R} = \begin{bmatrix} 1 & 2 & -3 & 0\\ 0 & -5 & 9 & 4 \\ 0 & -2 & 16 & 14 \end{bmatrix} \quad R_2-2R_1\quad \text{and} R_3-3R_1\\ &\sim \text{R} = \begin{bmatrix} 1 & 2 & -3 & 0\\ 0 & -5 & 9 & 4 \\ 0 & -1 & 8 & 7 \end{bmatrix} \quad \frac{1}{2}R_3\\ &\sim \text{R} = \begin{bmatrix} 1 & 0 & 13 & 14\\ 0 & 0 & -31 & -31 \\ 0 & -1 & 8 & 7 \end{bmatrix} \quad R_1+2R_3 \text{and}\quad R_2-5R_3\\ &\sim \text{R} = \begin{bmatrix} 1 & 0 & 13 & 14\\ 0 & -1 & 8 & 7\\ 0 & 0 & -31 & -31 \end{bmatrix} \quad R_2\text{interchange}\quad R_3\\ &\sim \text{R} = \begin{bmatrix} 1 & 0 & 13 & 14\\ 0 & -1 & 8 & 7\\ 0 & 0 & 1 & 1 \end{bmatrix} \quad \frac{-1}{31}R_2\text{interchange}\quad R_3\\ &\sim \text{R} = \begin{bmatrix} 1 & 0 & 0 & 11\\ 0 & -1 & 0 & -1\\ 0 & 0 & 1 & 1 \end{bmatrix} \quad R_2-8R_3\quad R_1-13R_3 \end{align*} From above equation we get, \begin{align*} x_1&=1\\ x_2&=1\\ x_3&=1 \end{align*} Now solutions of above equations are; $$ \begin{bmatrix} \dfrac{-3}{62} & \dfrac{9}{62} & \dfrac{5}{62} \\ \dfrac{13}{32} & \dfrac{-8}{32} & \dfrac{-1}{32} \\ \dfrac{19}{62} & \dfrac{-5}{62} & \dfrac{-11}{62} \end{bmatrix}\quad 1;1;1 $$
Question 8
Determine the value of $\lambda$ for which the following system has no solution, unique solution or infinitely many solutions.
$x+2 y-3 z=4 ; 3 x-y+5 z=2 ; 4 x+y+\left(\lambda^{2}-14\right) z=\lambda+2$
Solution.
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