Raju's Resource Hub
Google ad
\[\color {purple} {Example\ 1:}\ \color{red}{Find\ the\ inverse\ of}\ \begin{bmatrix}
1 & -1 \\
-2 & 0 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix}
1 & -1 \\
-2 & 0 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 1(0)\ -\ (-2) \hspace{13cm}\]
\[= 0 – 2 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = – 2\ \neq {0}\ \hspace{13cm}\]
\[\therefore\ A^{-1}\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ (0)\ \hspace{15cm}\]
\[= (-1)^2 (0)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 0\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ (-2)\ \hspace{15cm}\]
\[= (-1)^3 (-2)\ \hspace{15cm}\]
\[cofactor\ of\ – 1 = 2\ \hspace{15cm}\]
\[cofactor\ of\ -2 = (-1)^{2\ +\ 1}\ (-1)\ \hspace{15cm}\]
\[= (-1)^3 (-1)\ \hspace{15cm}\]
\[cofactor\ of\ – 2 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 0 = (-1)^{2\ +\ 2}\ (1)\ \hspace{15cm}\]
\[= (-1)^4 (1)\ \hspace{15cm}\]
\[cofactor\ of\ 0 = 1\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{bmatrix}
0 & 2 \\
1 & 1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{bmatrix}
0 & 1 \\
2 & 1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{-2}\ \begin{bmatrix}
0 & 1 \\
2 & 1 \\
\end{bmatrix}\ \hspace{2cm}\]
\[\color {purple} {Example\ 2:}\ \color{red}{Find\ the\ inverse\ of}\ \begin{bmatrix}
2 & -1 \\
4 & 5 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {purple} {Example\ 3:}\ \color{red}{Find\ the\ inverse\ of}\ \begin{bmatrix}
1 & -1 & 1 \\
2 & -3 & -3 \\
6 & -2 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix}
1 & -1 & 1 \\
2 & -3 & -3 \\
6 & -2 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =1\begin{vmatrix}
-3 & -3 \\
-2 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -3 \\
6 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -3\\
6 & -2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[ =1(3\ -\ 6)\ + 1 (-2\ +\ 18) + 1(-4\ +\ 18)\
\hspace{9cm}\]
\[ =1(-3)\ + 1 (16) + 1(14)\
\hspace{13cm}\]
\[ = -3\ + 16 + 14\
\hspace{14cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 27\ \neq\ 0\
\hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ \begin{vmatrix}
-3 & -3 \\
-2 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (3 – 6)\ \hspace{15cm}\]
\[= (1) (-3)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ \begin{vmatrix}
2 & -3 \\
6 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (-2 + 18)\ \hspace{15cm}\]
\[= (-1) (16)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -16\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 3}\ \begin{vmatrix}
2 & -3 \\
6 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-4 + 18)\ \hspace{15cm}\]
\[= (1) (14)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 14\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 1}\ \begin{vmatrix}
-1 & 1 \\
-2 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (1+ 2)\ \hspace{15cm}\]
\[= (-1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -3 = (-1)^{2\ +\ 2}\ \begin{vmatrix}
1 & 1 \\
6 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-1- 6)\ \hspace{15cm}\]
\[= (1) (-7)\ \hspace{15cm}\]
\[cofactor\ of\ -3 = -7\ \hspace{15cm}\]
\[cofactor\ of\ -3 = (-1)^{2\ +\ 3}\ \begin{vmatrix}
1 & -1 \\
6 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-2+ 6)\ \hspace{15cm}\]
\[= (-1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ -3 = -4\ \hspace{15cm}\]
\[cofactor\ of\ 6 = (-1)^{3\ +\ 1}\ \begin{vmatrix}
-1 & 1 \\
-3 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (3 + 3)\ \hspace{15cm}\]
\[= (1) (6)\ \hspace{15cm}\]
\[cofactor\ of\ 6 = 6\ \hspace{15cm}\]
\[cofactor\ of\ -2 = (-1)^{3\ +\ 2}\ \begin{vmatrix}
1 & 1 \\
2 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-3- 2)\ \hspace{15cm}\]
\[= (-1) (-5)\ \hspace{15cm}\]
\[cofactor\ of\ -2 = 5\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{3\ +\ 3}\ \begin{vmatrix}
1 & -1 \\
2 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (-3+ 2)\ \hspace{15cm}\]
\[= (1) (-1)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -1\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
-3 & -16 & 14 \\
-3 & -7 & -4 \\
6 & 5 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix}
-3 & -3 & 6 \\
-16 & -7 & 5 \\
14 & -4 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{27}\ \begin{bmatrix}
-3 & -3 & 6 \\
-16 & -7 & 5 \\
14 & -4 & -1 \\
\end{bmatrix}\ \hspace{2cm}\]
\[\color {purple} {Example\ 4:}\ \color{red}{Find\ the\ inverse\ of\ the\ matrix}\ \begin{bmatrix}
3 & 4 & 1 \\
0 & -1 & 2 \\
5 & -2 & 6 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix}
3 & 4 & 1 \\
0 & -1 & 2 \\
5 & -2 & 6 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 3\begin{vmatrix}
-1 & 2 \\
-2 & 6 \\
\end{vmatrix}\ -\ 4\begin{vmatrix}
0 & 2 \\
5 & 6 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
0 & -1\\
5 & -2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[ =\ 3(- 6\ +\ 4)\ -\ 4 (0\ -\ 10)\ +\ 1(0\ +\ 5)\
\hspace{9cm}\]
\[ =\ 3(-2)\ -\ 4 (-10)\ +\ 1(5)\
\hspace{13cm}\]
\[ =\ -6\ +\ 40\ + 5\
\hspace{14cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 39\ \neq\ 0\
\hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 3\ = (-1)^{1\ +\ 1}\ \begin{vmatrix}
-1 & 2 \\
-2 & 6 \\
\end{vmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 3\ = (-1)^2\ (-6\ +\ 4)\ \hspace{15cm}\]
\[= (1) (-\ 2)\ \hspace{15cm}\]
\[cofactor\ of\ 3\ =\ -\ 2\ \hspace{15cm}\]
\[cofactor\ of\ 4\ = (-1)^{1\ +\ 2}\ \begin{vmatrix}
0 & 2 \\
5 & 6 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (0\ -\ 10)\ \hspace{15cm}\]
\[= (-1) (-\ 10)\ \hspace{15cm}\]
\[cofactor\ of\ 1 =\ 10\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 3}\ \begin{vmatrix}
0 & -1 \\
5 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (0\ +\ 5)\ \hspace{15cm}\]
\[= (1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 1 =\ 5\ \hspace{15cm}\]
\[cofactor\ of\ 0\ =\ (-1)^{2\ +\ 1}\ \begin{vmatrix}
4 & 1 \\
-2 & 6 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (24\ +\ 2)\ \hspace{15cm}\]
\[= (-1) (26)\ \hspace{15cm}\]
\[cofactor\ of\ 0\ =\ -\ 26\ \hspace{15cm}\]
\[cofactor\ of\ -\ 1\ =\ (-1)^{2\ +\ 2}\ \begin{vmatrix}
3 & 1 \\
5 & 6 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (18\ -\ 5)\ \hspace{15cm}\]
\[= (1) (13)\ \hspace{13cm}\]
\[cofactor\ of\ -\ 1\ =\ 13\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 3}\ \begin{vmatrix}
3 & 4 \\
5 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-\ 6 -\ 20)\ \hspace{15cm}\]
\[= (-1) (-\ 26)\ \hspace{15cm}\]
\[cofactor\ of\ 2 =\ 26\ \hspace{15cm}\]
\[cofactor\ of\ 5\ =\ (-1)^{3\ +\ 1}\ \begin{vmatrix}
4 & 1 \\
-1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (8\ +\ 1)\ \hspace{15cm}\]
\[= (1) (9)\ \hspace{15cm}\]
\[cofactor\ of\ 5\ =\ 9\ \hspace{15cm}\]
\[cofactor\ of\ -\ 2\ =\ (-1)^{3\ +\ 2}\ \begin{vmatrix}
3 & 1 \\
0 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (6\ -\ 0)\ \hspace{15cm}\]
\[= (-1) (6)\ \hspace{15cm}\]
\[cofactor\ of\ -\ 2\ =\ -\ 6\ \hspace{15cm}\]
\[cofactor\ of\ 6\ =\ (-1)^{3\ +\ 3}\ \begin{vmatrix}
3 & 4 \\
0 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (-3\ -\ 0)\ \hspace{15cm}\]
\[= (1) (-3)\ \hspace{15cm}\]
\[cofactor\ of\ 6\ =\ -\ 3\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
-2 & 10 & 5 \\
-26 & 13 & 26 \\
9 & -6 & -3 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix}
-2 & -26 & 9 \\
10 & 13 & – 6 \\
5 & 26 & -3 \\
\end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{39}\ \begin{bmatrix}
-2 & -26 & 9 \\
10 & 13 & -6 \\
5 & 26 & -3 \\
\end{bmatrix}\ \hspace{2cm}\]
\[\color {purple} {Example\ 5:}\ \color{red}{Find\ the\ inverse\ of\ the\ matrix}\ \begin{bmatrix}
2 & 3 & 4 \\
4 & 3 & 1 \\
1 & 2 & 4 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix}
2 & 3 & 4 \\
4 & 3 & 1 \\
1 & 2 & 4 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =2\begin{vmatrix}
3 & 1 \\
2 & 4 \\
\end{vmatrix}\ -\ 3\begin{vmatrix}
4 & 1 \\
1 & 4 \\
\end{vmatrix}\ +\ 4\begin{vmatrix}
4 & 3\\
1 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[ =2(12\ -\ 2)\ – 3 (16\ -\ 1) + 4(8\ -\ 3)\
\hspace{9cm}\]
\[ =2(10)\ – 3 (15) + 4(5)\
\hspace{13cm}\]
\[ =20\ -45 + 20\
\hspace{14cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =-5\ \neq\ 0\
\hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ \begin{vmatrix}
3 & 1 \\
2 & 4 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (12 – 2)\ \hspace{15cm}\]
\[= (1) (10)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 10\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{1\ +\ 2}\ \begin{vmatrix}
4 & 1 \\
1 & 4 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (16 – 1)\ \hspace{15cm}\]
\[= (-1) (15)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -15\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{1\ +\ 3}\ \begin{vmatrix}
4 & 3 \\
1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (8- 3)\ \hspace{15cm}\]
\[= (1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = 5\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{2\ +\ 1}\ \begin{vmatrix}
3 & 4 \\
2 & 4 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (12- 8)\ \hspace{15cm}\]
\[= (-1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = -4\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{2\ +\ 2}\ \begin{vmatrix}
2 & 4 \\
1 & 4 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (8- 4)\ \hspace{15cm}\]
\[= (1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = 4\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{2\ +\ 3}\ \begin{vmatrix}
2 & 3 \\
1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (4- 3)\ \hspace{15cm}\]
\[= (-1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -1\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{3\ +\ 1}\ \begin{vmatrix}
3 & 4 \\
3 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (3- 12)\ \hspace{15cm}\]
\[= (1) (-9)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -9\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{3\ +\ 2}\ \begin{vmatrix}
2 & 4 \\
4 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (2- 16)\ \hspace{15cm}\]
\[= (-1) (-14)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 14\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{3\ +\ 3}\ \begin{vmatrix}
2 & 3 \\
4 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (6- 12)\ \hspace{15cm}\]
\[= (1) (-6)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = -6\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
10 & -15 & 5 \\
-4 & 4 & -1 \\
-9 & 14 & -6 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix}
10 & -4 & -9 \\
-15 & 4 & 14 \\
5 & -1 & -6 \\
\end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{-5}\ \begin{bmatrix}
10 & -4 & -9 \\
-15 & 4 & 14 \\
5 & -1 & -6 \\
\end{bmatrix}\ \hspace{2cm}\]
\[\color {purple} {Example\ 5:}\ \color{red}{Find\ the\ inverse\ of\ the\ matrix}\ \begin{bmatrix}
2 & 3 & 4 \\
4 & 3 & 1 \\
1 & 2 & 4 \\
\end{bmatrix}\ \hspace{15cm}\]
📚 Related Posts
Google ad