Exercise 1.2 Solutions – Applications of Matrices and Determinants (12th Maths)

IMPORTANT QUESTIONS

2 Marks : Q1 (Very Important)
3 Marks : Q2 (Very Important ) Q3(Rare)
5 Marks : Nill

Q1 (i) : Find the rank by minor method:

$A =\begin{bmatrix}2 & -4 \\-1 & 2\end{bmatrix}$

$\text{Given matrix is a } 2 \times 2 \text{ matrix}$

$\therefore \; P(A) \le 2$

$|A| =\begin{vmatrix}2 & -4 \\-1 & 2\end{vmatrix}= (4 - 4)= 0$

$\text{Since it is equal to zero, } P(A) \ne 2$

$\text{hence } P(A) = 1$

Q1 (ii) : Find the rank by minor method:

$A =\begin{bmatrix}-1 & 3 \\4 & -7 \\3 & -4\end{bmatrix}$

$\text{Given matrix is } 3 \times 2 \text{ matrix}$

$\therefore \; P(A) \le 2$

$\text{Second order minor}$

$A =\begin{vmatrix}-1 & 3 \\4 & -7\end{vmatrix}$

$|A| = 7 - 12 = -5 \ne 0$

$\therefore \; P(A) = 2$

Q1 (iii) : Find the rank by minor method:

$A =\begin{bmatrix}1 & -2 & -1 & 0 \\3 & -6 & -3 & 1\end{bmatrix}$

$\text{Given matrix is } 2 \times 4 \text{ matrix}$

$\therefore \; P(A) \le 2$

$\text{Second order minor}$

$A =\begin{vmatrix}1 & -2 \\3 & -6\end{vmatrix}\quad |A| = -6 + 6 = 0$

$A =\begin{vmatrix}-2 & -1 \\-6 & -3\end{vmatrix}\quad |A| = 6 - 6 = 0$

$A =\begin{vmatrix}-1 & 0 \\-3 & 1\end{vmatrix}\quad |A| = -1 \ne 0$

$\therefore \; P(A) = 2$

Q1 (iv) : Find the rank by minor method:

$A =\begin{bmatrix}1 & -2 & 3 \\2 & 4 & -6 \\5 & 1 & -1\end{bmatrix}$

$\text{Given matrix is } 3 \times 3 \text{ matrix}$

$\therefore \; P(A) \le 3$

$|A| = 1(-4 + 6) + 2(-2 + 30) + 3(2 - 20)$

$= 1(2) + 2(28) + 3(-18)$

$= 2 + 56 - 54$

$= 4 \ne 0$

$\therefore \; P(A) = 3$

Q1 (v) : Find the rank by minor method:

$A =\begin{bmatrix}0 & 1 & 2 & 1 \\0 & 2 & 4 & 3 \\8 & 1 & 0 & 2\end{bmatrix}$

$\text{Given matrix is } 3 \times 4 \text{ matrix}$

$\therefore \; P(A) \le 3$

$\text{Lets take the minor matrix of order } 3 \times 3$

$A =\begin{vmatrix}0 & 1 & 1 \\0 & 2 & 3 \\8 & 1 & 2\end{vmatrix}$

$|A| = 0(4-3) - 1(0-24) + 1(0-16)$

$= -(-24) + (-16)$

$= 24 - 16 = 8 \ne 0$

$\therefore \; P(A) = 3$

Q2 (i) : Find the rank by Row reduction

$\textbf{2(i). Find the rank of the matrix}\begin{pmatrix}1 & 1 & 1 & 3\\2 & -1 & 3 & 4\\5 & -1 & 7 & 11\end{pmatrix}$

$R_2 \to R_2 - 2R_1,\quad R_3 \to R_3 - 5R_1$

$=\begin{pmatrix}1 & 1 & 1 & 3\\0 & -3 & 1 & -2\\0 & -6 & 2 & -4\end{pmatrix}$

$R_3 \to R_3 - 2R_2$

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

$\therefore\ \text{Rank}=2$

Q2 (ii) : Find the rank by Row reduction

$\textbf{2(ii). Find the rank of the matrix}\begin{pmatrix}1 & 2 & -1\\3 & -1 & 2\\1 & -2 & 3\\1 & -1 & 1\end{pmatrix}$

$R_2 \to R_2 - 3R_1,\quad R_3 \to R_3 - R_1,\quad R_4 \to R_4 - R_1$

$=\begin{pmatrix}1 & 2 & -1\\0 & -7 & 5\\0 & -4 & 4\\0 & -3 & 2\end{pmatrix}$

$R_3 \to 7R_3 - 4R_2,\quad R_4 \to 7R_4 - 3R_2$

$=\begin{pmatrix}1 & 2 & -1\\0 & -7 & 5\\0 & 0 & 8\\0 & 0 & -1\end{pmatrix}$

$R_4 \to 8R_4 + R_3$

$=\begin{pmatrix}1 & 2 & -1\\0 & -7 & 5\\0 & 0 & 8\\0 & 0 & 0\end{pmatrix}$

$\therefore\ \text{Rank}=3$

Q3 (iii) : Find the rank by Row reduction

$\textbf{2(iii). Find the rank of the matrix}\begin{pmatrix}3 & -8 & 5 & 2\\2 & -5 & 1 & 4\\-1 & 2 & 3 & -2\end{pmatrix}$

$R_1 \leftrightarrow R_3$

$=\begin{pmatrix}-1 & 2 & 3 & -2\\2 & -5 & 1 & 4\\3 & -8 & 5 & 2\end{pmatrix}$

$R_2 \to R_2 + 2R_1,\quad R_3 \to R_3 + 3R_1$

$=\begin{pmatrix}-1 & 2 & 3 & -2\\0 & -1 & 7 & 0\\0 & -2 & 14 & -4\end{pmatrix}$

$R_3 \to R_3 - 2R_2$

$=\begin{pmatrix}-1 & 2 & 3 & -2\\0 & -1 & 7 & 0\\0 & 0 & 0 & -4\end{pmatrix}$

$\therefore\ \text{Rank}=3$

Q3 (i)

$\textbf{3(i). Find the inverse by Gauss-Jordan method}\quad A=\begin{pmatrix}2 & -1\\ 5 & -2\end{pmatrix}$

$\left[\begin{array}{cc|cc}2 & -1 & 1 & 0\\5 & -2 & 0 & 1\end{array}\right]$

$R_1 \to \frac{1}{2}R_1$

$\left[\begin{array}{cc|cc}1 & -\frac{1}{2} & \frac{1}{2} & 0\\5 & -2 & 0 & 1\end{array}\right]$

$R_2 \to R_2 - 5R_1$

$\left[\begin{array}{cc|cc}1 & -\frac{1}{2} & \frac{1}{2} & 0\\0 & \frac{1}{2} & -\frac{5}{2} & 1\end{array}\right]$

$R_2 \to 2R_2$

$\left[\begin{array}{cc|cc}1 & -\frac{1}{2} & \frac{1}{2} & 0\\0 & 1 & -5 & 2\end{array}\right]$

$R_1 \to R_1 + \frac{1}{2}R_2$

$\left[\begin{array}{cc|cc}1 & 0 & -2 & 1\\0 & 1 & -5 & 2\end{array}\right]$

$A^{-1}=\begin{pmatrix}-2 & 1\\ -5 & 2\end{pmatrix}$

Q3 (ii)

$\textbf{3(ii). Find the inverse by Gauss-Jordan method}\quad A=\begin{pmatrix}1 & -1 & 0\\1 & 0 & -1\\6 & -2 & -3\end{pmatrix}$

$\left[\begin{array}{ccc|ccc}1 & -1 & 0 & 1 & 0 & 0\\1 & 0 & -1 & 0 & 1 & 0\\6 & -2 & -3 & 0 & 0 & 1\end{array}\right]$

$R_2 \to R_2 - R_1,\quad R_3 \to R_3 - 6R_1$

$\left[\begin{array}{ccc|ccc}1 & -1 & 0 & 1 & 0 & 0\\0 & 1 & -1 & -1 & 1 & 0\\0 & 4 & -3 & -6 & 0 & 1\end{array}\right]$

$R_3 \to R_3 - 4R_2$

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

$R_2 \to R_2 + R_3$

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

$R_1 \to R_1 + R_2$

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

$A^{-1}=\begin{pmatrix}-2 & -3 & 1\\-3 & -3 & 1\\-2 & -4 & 1\end{pmatrix}$

Q3 (iii)

$\textbf{3(iii). Find the inverse by Gauss-Jordan method}\quad A=\begin{pmatrix}1 & 2 & 3\\2 & 5 & 3\\1 & 0 & 8\end{pmatrix}$

$\left[\begin{array}{ccc|ccc}1 & 2 & 3 & 1 & 0 & 0\\2 & 5 & 3 & 0 & 1 & 0\\1 & 0 & 8 & 0 & 0 & 1\end{array}\right]$

$R_2 \to R_2 - 2R_1,\quad R_3 \to R_3 - R_1$

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

$R_3 \to R_3 + 2R_2$

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

$R_3 \to -R_3$

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

$R_2 \to R_2 + 3R_3$

$\left[\begin{array}{ccc|ccc}1 & 2 & 3 & 1 & 0 & 0\\0 & 1 & 0 & 13 & -5 & -3\\0 & 0 & 1 & 5 & -2 & -1\end{array}\right]$

$R_1 \to R_1 - 3R_3$

$\left[\begin{array}{ccc|ccc}1 & 2 & 0 & -14 & 6 & 3\\0 & 1 & 0 & 13 & -5 & -3\\0 & 0 & 1 & 5 & -2 & -1\end{array}\right]$

$R_1 \to R_1 - 2R_2$

$\left[\begin{array}{ccc|ccc}1 & 0 & 0 & -40 & 16 & 9\\0 & 1 & 0 & 13 & -5 & -3\\0 & 0 & 1 & 5 & -2 & -1\end{array}\right]$

$A^{-1}=\begin{pmatrix}-40 & 16 & 9\\13 & -5 & -3\\5 & -2 & -1\end{pmatrix}$