====== Question 6, Exercise 2.6 ====== Solutions of Question 6 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 6(i)===== Solve the system of linear equation by matrix inversion method.FIXME\\ $5 x+3 y+z=6$\\ $2 x+y+3 z=19$\\ $x+2 y+4 z=25$\\ ** Solution. ** For this system of equations; we have \begin{align*} A &= \begin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 19 \\ 25 \end{bmatrix} \end{align*} And \begin{align*} |A|& = \left| \begin{array}{ccc} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array} \right|\\ &=5(4 - 6) - 3(8 - 3) + 1(4 - 1)\\ &= -10 - 15 + 3\\ &=-22 \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 & 3 \\ 2 & 4 \end{array} \right| = 4 - 6 = -2\\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 2 & 3 \\ 1 & 4 \end{array} \right| = -(8 - 3) = -5\\ A_{13} &= (-1)^{1+3} \left| \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array} \right| = 4 - 1 = 3\\ A_{21} &= (-1)^{2+1} \left| \begin{array}{cc} 3 & 1 \\ 2 & 4 \end{array} \right| = -(12 - 2) = -10\\ A_{22} &= (-1)^{2+2} \left| \begin{array}{cc} 5 & 1 \\ 1 & 4 \end{array} \right| = 20 - 1 = 19\\ A_{23} &= (-1)^{2+3} \left| \begin{array}{cc} 5 & 3 \\ 1 & 2 \end{array} \right| = -(10 - 3) = -7\\ A_{31} &= (-1)^{3+1} \left| \begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array} \right| = 9 - 1 = 8\\ A_{32} &= (-1)^{3+2} \left| \begin{array}{cc} 5 & 1 \\ 2 & 4 \end{array} \right| = -(20 - 2) = -18\\ A_{33} &= (-1)^{3+3} \left| \begin{array}{cc} 5 & 3 \\ 2 & 1 \end{array} \right| = 5 - 6 = -1\\ A&= \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix}\\ A &=\begin{bmatrix} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -18 & -1 \end{bmatrix}\\ \text{adj}(A)& = \begin{bmatrix} -2 & -10 & 8 \\ -5 & 19 & -18 \\ 3 & -7 & -1 \end{bmatrix}\\ A^{-1}& = \frac{1}{|A|} \text{adj}(A) \\ &= \frac{1}{-22} \begin{bmatrix} -2 & -10 & 8 \\ -5 & 19 & -18 \\ 3 & -7 & -1 \end{bmatrix}\\ A^{-1} &= \begin{bmatrix} \frac{2}{22} & \frac{10}{22} & \frac{-8}{22} \\ \frac{5}{22} & \frac{-19}{22} & \frac{18}{22} \\ \frac{-3}{22} & \frac{7}{22} & \frac{1}{22} \end{bmatrix}\\ \implies A^{-1} &= \begin{bmatrix} \frac{1}{11} & \frac{5}{11} & \frac{-4}{11} \\ \frac{5}{22} & \frac{-19}{22} & \frac{18}{22} \\ \frac{-3}{22} & \frac{7}{22} & \frac{1}{22} \end{bmatrix}\end{align*} Since \begin{align*} X &= A^{-1}B \\ \implies X &= A^{-1}B = \begin{bmatrix} \frac{1}{11} & \frac{5}{11} & \frac{-4}{11} \\ \frac{5}{22} & \frac{-19}{22} & \frac{18}{22} \\ \frac{-3}{22} & \frac{7}{22} & \frac{1}{22} \end{bmatrix} \begin{bmatrix} 6 \\ 19 \\ 25 \end{bmatrix}\\ \implies X &= \begin{bmatrix} \frac{6 + 95 - 100}{11} \\ \frac{30 - 361 + 450}{22} \\ \frac{-18 + 133 + 25}{22} \end{bmatrix}\\ \implies X &= \begin{bmatrix} \frac{1}{11} \\ \frac{119}{22} \\ \frac{140}{22} \end{bmatrix}\\ \implies X &= \begin{bmatrix} \frac{1}{11} \\ \frac{119}{11} \\ \frac{70}{11} \end{bmatrix} \end{align*} Therefore, the solution to the system of equations is: $$x = \frac{1}{11}, \quad y =\frac{119}{11}, \quad z = \frac{70}{11}$$ =====Question 6(ii)===== Solve the system of linear equation by matrix inversion method.\\ $x+2 y-3 z=5$\\ $2 x-3 y+2 z=1$\\ $-x+2 y-5 z=-3$ ** Solution. ** For this system of equations; we have \begin{align*} A& = \begin{bmatrix} 1 & 2 & -3 \\ 2 & -3 & 2 \\ -1 & 2 & -5 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ 1 \\ -3 \end{bmatrix}\\ |A| &= \left| \begin{array}{ccc} 1 & 2 & -3 \\ 2 & -3 & 2 \\ -1 & 2 & -5 \end{array} \right|\\ &= 1 (15 - 4) - 2 (-10 + 2) - 3 (4 - 3)\\ &= 11 + 16 - 3 = 24\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} -3 & 2 \\ 2 & -5 \end{array} \right| = 15 - 4 = 11\\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 2 & 2 \\ -1 & -5 \end{array} \right| = -(-10 + 2) = 8\\ A_{13} &= (-1)^{1+3} \left| \begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array} \right| = 4 - 3 = 1\\ A_{21} &= (-1)^{2+1} \left| \begin{array}{cc} 2 & -3 \\ 2 & -5 \end{array} \right| = -(-10 + 6) = 4\\ A_{22} &= (-1)^{2+2} \left| \begin{array}{cc} 1 & -3 \\ -1 & -5 \end{array} \right| = -5 - 3 = -8\\ A_{23} &= (-1)^{2+3} \left| \begin{array}{cc} 1 & 2 \\ -1 & 2 \end{array} \right| = -(2+ 2 )= -4\\ A_{31} &= (-1)^{3+1} \left| \begin{array}{cc} 2 & -3 \\ -3 & 2 \end{array} \right| = 4-9 = -5\\ A_{32} &= (-1)^{3+2} \left| \begin{array}{cc} 1 & -3 \\ 2 & 2 \end{array} \right| = -(2 + 6) = -8\\ A_{33} &= (-1)^{3+3} \left| \begin{array}{cc} 1 & 2 \\ 2 & -3 \end{array} \right| = -3 - 4 = -7\\ A&= \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix}\\ \text{Cofactor matrix} &= \begin{bmatrix} 11 & 8 & 1 \\ 4 & -8 & -4 \\ -5 & -8 & -7 \end{bmatrix}\\ \text{adj}(A) &= \begin{bmatrix} 11 & 4 & -5 \\ 8 & -8 & -8 \\ 1 & -4 & -7 \end{bmatrix}\\ A^{-1} &= \frac{1}{24} \begin{bmatrix} 11 & 4 & -5 \\ 8 & -8 & -8 \\ 1 & -4 & -7 \end{bmatrix}\\ \text{Now}\\ X &= A^{-1}B\\ &= \frac{1}{24} \begin{bmatrix} 11 & 4 & -5 \\ 8 & -8 & -8 \\ 1 & -4 & -7 \end{bmatrix} \begin{bmatrix} 5 \\ 1 \\ -3 \end{bmatrix}\\ &= \frac{1}{24} \begin{bmatrix} 55 + 4 + 15 \\ 40 - 8 + 24 \\ 5 - 4+ 21 \end{bmatrix}\\ &= \frac{1}{24} \begin{bmatrix} 74 \\ 56 \\ 22\end{bmatrix}\\ &= \begin{bmatrix} \frac{74}{24} \\ \frac{56}{24} \\ \frac{22}{24} \end{bmatrix}\\ &= \begin{bmatrix} \frac{37}{12} \\ \frac{7}{3} \\ \frac{11}{12} \end{bmatrix} \end{align*} Therefore, the solution to the system of equations is: $$ x = \frac{37}{12}, \quad y = \frac{7}{3}, \quad z = \frac{11}{12}$$ =====Question 6(iii)===== Solve the system of linear equation by matrix inversion method.\\ $-x+3 y-5 z=0$\\ $2 x+4 y-6 z=1$\\ $x-2 y+3 z=3$\\ ** Solution. ** Do yourself. =====Question 6(iv)===== Solve the system of linear equation by matrix inversion method.\\ $\dfrac{2}{x}+\dfrac{3}{y}+\dfrac{10}{z}=4$\\ $\dfrac{4}{x}-\dfrac{6}{y}+\dfrac{5}{z}=1$\\ $\dfrac{6}{x}+\dfrac{9}{y}-\dfrac{20}{z}=2$ ** Solution. ** Do yourself. ====Go to ==== [[math-11-nbf:sol:unit02:ex2-6-p5|< Question 5]] [[math-11-nbf:sol:unit02:ex2-6-p7|Question 7 & 8 >]]