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.

Solve the system of linear equation by matrix inversion method.
$5 x+3 y+z=6$
$2 x+y+3 z=19$
$x+2 y+4 z=25$

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: $$x = \frac{1}{11}, \quad y =\frac{119}{11}, \quad z = \frac{70}{11}$$

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: $$ x = \frac{37}{12}, \quad y = \frac{7}{3}, \quad z = \frac{11}{12}$$

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.

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.

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