Exercise 1.7 Solutions – Applications of Matrices and Determinants (12th Maths)

Q1 (i) Solve the following system of homogeneous equations

$\textbf{1(i). 3x+2y+7z=0,\; 4x-3y-2z=0,\; 5x+9y+23z=0}$

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$\text{Matrix form}$

$\begin{pmatrix}3 & 2 & 7\\4 & -3 & -2\\5 & 9 & 23\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$

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$\text{Augmented matrix } (A|B)$

$\left[\begin{array}{ccc|c}3 & 2 & 7 & 0\\4 & -3 & -2 & 0\\5 & 9 & 23 & 0\end{array}\right]$

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$R_2 \rightarrow 3R_2 - 4R_1$

$R_3 \rightarrow 3R_3 - 5R_1$

$\left[\begin{array}{ccc|c}3 & 2 & 7 & 0\\0 & -17 & -34 & 0\\0 & 17 & 34 & 0\end{array}\right]$

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$R_3 \rightarrow R_3 + R_2$

$\left[\begin{array}{ccc|c}3 & 2 & 7 & 0\\0 & -17 & -34 & 0\\0 & 0 & 0 & 0\end{array}\right]$

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$\rho(A)=2,\quad \rho(A|B)=2<3$

$\text{The system is consistent and has infinitely many solutions}$

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$\text{Equivalent equations}$

$3x+2y+7z=0 \;\rightarrow\; (1)$

$-17y-34z=0 \Rightarrow 17y+34z=0$

$y+2z=0 \;\rightarrow\; (2)$

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$\text{Let } z=t$

$\text{Substitute in (2): } y+2t=0 \Rightarrow y=-2t$

$\text{Substitute } y \text{ and } z \text{ in (1)}$

$3x+2(-2t)+7t=0$

$3x-4t+7t=0$

$3x+3t=0 \Rightarrow x=-t$

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$\therefore\ x=-t,\; y=-2t,\; z=t,\quad t\in\mathbb{R}$

Q1 (ii) Solve the following system of homogeneous equations

$\textbf{1(ii). 2x+3y-z=0,\; x-y-2z=0,\; 3x+y+3z=0}$

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$\text{Matrix form}$

$\begin{pmatrix}2 & 3 & -1\\1 & -1 & -2\\3 & 1 & 3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$

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$\text{Augmented matrix } (A|B)$

$\left[\begin{array}{ccc|c}2 & 3 & -1 & 0\\1 & -1 & -2 & 0\\3 & 1 & 3 & 0\end{array}\right]$

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$R_1 \leftrightarrow R_2$

$\left[\begin{array}{ccc|c}1 & -1 & -2 & 0\\2 & 3 & -1 & 0\\3 & 1 & 3 & 0\end{array}\right]$

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$R_2 \rightarrow R_2 - 2R_1$

$R_3 \rightarrow R_3 - 3R_1$

$\left[\begin{array}{ccc|c}1 & -1 & -2 & 0\\0 & 5 & 3 & 0\\0 & 4 & 9 & 0\end{array}\right]$

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$R_3 \rightarrow 5R_3 - 4R_2$

$\left[\begin{array}{ccc|c}1 & -1 & -2 & 0\\0 & 5 & 3 & 0\\0 & 0 & 33 & 0\end{array}\right]$

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$\text{The above matrix is in echelon form}$

$\rho(A)=\rho(A|B)=3=n$

$\text{The system is consistent with a unique solution (trivial solution)}$

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$\text{Equivalent equations}$

$x-y-2z=0 \;\rightarrow\; (1)$

$5y+3z=0 \;\rightarrow\; (2)$

$33z=0 \;\rightarrow\; (3)$

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$z=0$

$\text{Substitute in (2): } 5y+3(0)=0 \Rightarrow y=0$

$\text{Substitute } y,z \text{ in (1): } x-0-0=0 \Rightarrow x=0$

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$\therefore\ (x,y,z)=(0,0,0)$

Q2

$\textbf{2. Determine the values of } \lambda \text{ for which the system}$

$x+y+3z=0,\; 4x+3y+\lambda z=0,\; 2x+y+2z=0$

$\text{has (i) a unique solution \quad (ii) a non-trivial solution}$

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$\text{Matrix form}$

$\begin{pmatrix}1 & 1 & 3\\4 & 3 & \lambda\\2 & 1 & 2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$

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$\text{Augmented matrix } (A|B)$

$\left[\begin{array}{ccc|c}1 & 1 & 3 & 0\\4 & 3 & \lambda & 0\\2 & 1 & 2 & 0\end{array}\right]$

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$R_2 \leftrightarrow R_3$

$\left[\begin{array}{ccc|c}1 & 1 & 3 & 0\\2 & 1 & 2 & 0\\4 & 3 & \lambda & 0\end{array}\right]$

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$R_2 \rightarrow R_2 - 2R_1$

$R_3 \rightarrow R_3 - 4R_1$

$\left[\begin{array}{ccc|c}1 & 1 & 3 & 0\\0 & -1 & -4 & 0\\0 & -1 & \lambda -12 & 0\end{array}\right]$

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$R_3 \rightarrow R_3 - R_2$

$\left[\begin{array}{ccc|c}1 & 1 & 3 & 0\\0 & -1 & -4 & 0\\0 & 0 & \lambda - 8 & 0\end{array}\right]$

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$\text{The matrix is in echelon form}$

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$\text{Case 1: For unique solution}$

$\lambda \ne 8,\quad \rho(A)=\rho(A|B)=3=n$

$\therefore\ \text{Unique (trivial) solution}$

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$\text{Case 2: For non-trivial solution}$

$\lambda = 8$

$\rho(A)=\rho(A|B)=2<3$

$\therefore\ \text{System has non-trivial solutions}$

Q3

$\textbf{3.By using Gaussian elimination method, balance the chemical reaction equation:}\]\[C_2H_6 + O_2 \rightarrow H_2O + CO_2$

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$\text{Let } x_1, x_2, x_3, x_4 \text{ be the number of moles respectively}$

$x_1 C_2H_6 + x_2 O_2 \rightarrow x_3 H_2O + x_4 CO_2\quad \rightarrow (1)$

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$\textbf{Equating atoms}$

$\text{Carbon: } 2x_1 = x_4 \;\Rightarrow\; 2x_1 - x_4 = 0 \quad (1)$

$\text{Hydrogen: } 6x_1 = 2x_3 \;\Rightarrow\; 6x_1 - 2x_3 = 0 \quad (2)$

$\text{Oxygen: } 2x_2 = x_3 + 2x_4 \;\Rightarrow\; 2x_2 - x_3 - 2x_4 = 0 \quad (3)$

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$\text{Augmented matrix}$

$\left[\begin{array}{cccc|c}2 & 0 & 0 & -1 & 0\\6 & 0 & -2 & 0 & 0\\0 & 2 & -1 & -2 & 0\end{array}\right]$

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$R_2 \rightarrow R_2 - 3R_1$

$\left[\begin{array}{cccc|c}2 & 0 & 0 & -1 & 0\\0 & 0 & -2 & 3 & 0\\0 & 2 & -1 & -2 & 0\end{array}\right]$

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$R_2 \leftrightarrow R_3$

$\left[\begin{array}{cccc|c}2 & 0 & 0 & -1 & 0\\0 & 2 & -1 & -2 & 0\\0 & 0 & -2 & 3 & 0\end{array}\right]$

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$\textbf{Equivalent equations}$

$2x_1 - x_4 = 0 \quad (1)$

$2x_2 - x_3 - 2x_4 = 0 \quad (2)$

$-2x_3 + 3x_4 = 0 \quad (3)$

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$\text{Sub } x_4 = t \text{ in (1)}$

$2x_1 - t = 0$

$2x_1 = t$

$x_1 = \frac{t}{2}$

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$\text{Sub } x_4 = t \text{ in (3)}$

$-2x_3 + 3x_4 = 0$

$-2x_3 + 3t = 0$

$-2x_3 = -3t$

$x_3 = \frac{3t}{2}$

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$\text{Sub } x_3, x_4 \text{ in (2)}$

$2x_2 - \frac{3t}{2} - 2t = 0$

$2x_2 = \frac{3t}{2} + 2t$

$2x_2 = \frac{3t + 4t}{2}$

$2x_2 = \frac{7t}{2}$

$x_2 = \frac{7t}{4}$

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$\textbf{Solution}$

$x_1 = \frac{t}{2}, \quad x_2 = \frac{7t}{4}, \quad x_3 = \frac{3t}{2}, \quad x_4 = t$

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$\textbf{Final simplification}$

$\text{Take } t = 4 \text{ (LCM of denominators)}$

$x_1 = \frac{4}{2} = 2,\;x_2 = \frac{7(4)}{4} = 7,\;x_3 = \frac{3(4)}{2} = 6,\;x_4 = 4$

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$\textbf{Balanced equation}$

$2C_2H_6 + 7O_2 \rightarrow 6H_2O + 4CO_2$