Exercise 1.6 Solutions – Applications of Matrices and Determinants (12th Maths)

IMPORTANT QUESTIONS

2 MARKS

Q1 (i) . Test for consistency and solve by rank method

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

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$\text{Matrix form}$

$\begin{pmatrix}1 & -1 & 2\\2 & 1 & 4\\4 & -1 & 1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2\\7\\4\end{pmatrix}$

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

$\left[\begin{array}{ccc|c}1 & -1 & 2 & 2\\2 & 1 & 4 & 7\\4 & -1 & 1 & 4\end{array}\right]$

$R_2 \rightarrow R_2-2R_1$

$R_3 \rightarrow R_3-4R_1$

$\left[\begin{array}{ccc|c}1 & -1 & 2 & 2\\0 & 3 & 0 & 3\\0 & 3 & -7 & -4\end{array}\right]$

$R_3 \rightarrow R_3-R_2$

$\left[\begin{array}{ccc|c}1 & -1 & 2 & 2\\0 & 3 & 0 & 3\\0 & 0 & -7 & -7\end{array}\right]$

—

$\text{It is in echelon form}$

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

$\text{The system is consistent and has a unique solution}$

—

$\text{Equivalent equations}$

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

$3y=3 \;\rightarrow\; (2)$

$-7z=-7 \;\rightarrow\; (3)$

$\text{From (2), } y=1$

$\text{From (3), } z=1$

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

$x-1+2(1)=2$

$x-1+2=2$

$x+1=2 \Rightarrow x=1$

—

$\therefore\ x=1,\; y=1,\; z=1$

Q1 (ii) . Test for consistency and solve by rank method

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

—

$\text{Matrix form}$

$\begin{pmatrix}3 & 1 & 1\\1 & -3 & 2\\7 & -1 & 4\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2\\1\\5\end{pmatrix}$

—

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

$\left[\begin{array}{ccc|c}3 & 1 & 1 & 2\\1 & -3 & 2 & 1\\7 & -1 & 4 & 5\end{array}\right]$

$R_1 \leftrightarrow R_2$

$\left[\begin{array}{ccc|c}1 & -3 & 2 & 1\\3 & 1 & 1 & 2\\7 & -1 & 4 & 5\end{array}\right]$

$R_2 \rightarrow R_2-3R_1$

$R_3 \rightarrow R_3-7R_1$

$\left[\begin{array}{ccc|c}1 & -3 & 2 & 1\\0 & 10 & -5 & -1\\0 & 20 & -10 & -2\end{array}\right]$

$R_3 \rightarrow R_3-2R_2$

$\left[\begin{array}{ccc|c}1 & -3 & 2 & 1\\0 & 10 & -5 & -1\\0 & 0 & 0 & 0\end{array}\right]$

—

$\text{It is in echelon form}$

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

$\text{Since rank } < \text{ number of unknowns, infinite solutions exist}$

—

$\text{Equivalent equations}$

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

$10y-5z=-1 \;\rightarrow\; (2)$

—

$\text{Let } z=k$

$10y-5k=-1$

$10y=5k-1$

$y=\frac{5k-1}{10}$

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

$x-3\left(\frac{5k-1}{10}\right)+2k=1$

$x = 1 + 3\left(\frac{5k-1}{10}\right) - 2k$

$= \frac{1(10)+15k-3-2k(10)}{10}$

$= \frac{10+15k-3-20k}{10}$

$x=\frac{7-5k}{10}$

—

$\therefore\ x=\frac{7-5k}{10},\quad y=\frac{5k-1}{10},\quad z=k,\; k\in\mathbb{R}$

Q1 (iii) . Test for consistency by rank method

$\textbf{1(iii). 2x+2y+z=5,\; x-y+z=1,\; 3x+y+2z=4}$

—

$\text{Matrix form}$

$\begin{pmatrix}2 & 2 & 1\\1 & -1 & 1\\3 & 1 & 2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}5\\1\\4\end{pmatrix}$

—

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

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

$R_1 \leftrightarrow R_2$

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

$R_2 \rightarrow R_2-2R_1$

$R_3 \rightarrow R_3-3R_1$

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

$R_3 \rightarrow R_3-R_2$

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

—

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

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

$\therefore\ \rho(A)\ne \rho(A|B)$

$\text{The system is inconsistent and has no solution}$

Q1 (iv) . Test for consistency by rank method

$\textbf{1(iv). 2x-y+z=2,\; 6x-3y+3z=6,\; 4x-2y+2z=4}$

—

$\text{Matrix form}$

$\begin{pmatrix}2 & -1 & 1\\6 & -3 & 3\\4 & -2 & 2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2\\6\\4\end{pmatrix}$

—

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

$\left[\begin{array}{ccc|c}2 & -1 & 1 & 2\\6 & -3 & 3 & 6\\4 & -2 & 2 & 4\end{array}\right]$

$R_2 \rightarrow R_2-3R_1$

$R_3 \rightarrow R_3-2R_1$

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

—

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

$\rho(A)=1,\quad \rho(A|B)=1$

$\text{Since rank } < \text{ number of unknowns, infinite solutions exist}$

—

$\text{Equivalent equation}$

$2x-y+z=2$

$\text{Let } y=s,\; z=t$

$2x-s+t=2$

$2x=2+s-t$

$x=\frac{1}{2}(2+s-t)$

—

$\therefore\ x=\frac{1}{2}(2+s-t),\quad y=s,\quad z=t,\; s,t\in\mathbb{R}$

Q2 . Find the value of k

$\textbf{\begin{aligned}&2.\ Find\ the\ value\ of\ } k \textbf{\ for\ which\ the\ equations}\\&\textbf{ }kx-2y+z=1,\; x-2ky+z=-2,\; x-2y+kz=1\\&\textbf{have\ (i)\ no\ solution\ \ (ii)\ unique\ solution\ \ (iii)\ infinitely\ many\ solutions.}\end{aligned}$

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$\text{Matrix form}$

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

—

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

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

$R_1 \leftrightarrow R_3$

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

—

$R_2 \rightarrow R_2 - R_1$

$R_3 \rightarrow R_3 - kR_1$

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

—

$R_3 \rightarrow R_3 + R_2$

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

—

$= \left[\begin{array}{ccc|c}1 & -2 & k & 1\\0 & -2k+2 & 1-k & -3\\0 & 0 & -(k^2+k-2) & -(k+2)\end{array}\right]$

—

$R_3 \rightarrow -R_3$

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

—

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

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

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$\text{(i) No solution}$

$k=1$

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

$\therefore\ \text{No solution}$

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$\text{(ii) Unique solution}$

$k\ne 1,\; k\ne -2$

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

$\therefore\ \text{Unique solution}$

—

$\text{(iii) Infinitely many solutions}$

$k=-2$

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

$\therefore\ \text{Infinitely many solutions}$

Q3 . Investigate the values

$\textbf{3. Investigate the values of } \lambda \textbf{ and } \mu \textbf{ for the system}$

$2x+3y+5z=9,\; 7x+3y-5z=8,\; 2x+3y+\lambda z=\mu$

$\text{(i) no solution \quad (ii) unique solution \quad (iii) infinitely many solutions}$

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$\text{Matrix form}$

$\begin{pmatrix}2 & 3 & 5\\7 & 3 & -5\\2 & 3 & \lambda\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}9\\8\\\mu\end{pmatrix}$

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

$\left[\begin{array}{ccc|c}2 & 3 & 5 & 9\\7 & 3 & -5 & 8\\2 & 3 & \lambda & \mu\end{array}\right]$

$R_2 \rightarrow 2R_2-7R_1$

$R_3 \rightarrow R_3-R_1$

$\left[\begin{array}{ccc|c}2 & 3 & 5 & 9\\0 & -15 & -45 & -47\\0 & 0 & \lambda-5 & \mu-9\end{array}\right]$

—

$\text{It is in row echelon form}$

—

$\text{(i) No solution}$

$\lambda=5,\quad \mu\ne 9$

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

$\therefore\ \text{No solution}$

—

$\text{(ii) Unique solution}$

$\lambda\ne 5$

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

$\therefore\ \text{Unique solution}$

—

$\text{(iii) Infinitely many solutions}$

$\lambda=5,\quad \mu=9$

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

$\therefore\ \text{Infinitely many solutions}$