SYLLABUS
Definition and expansion of second and third order determinants – Solution of simultaneous equations using Cramer’s rule for 2 and 3 unknowns – Types of matrices – Algebra of matrices – Equality, addition, subtraction, scalar multiplication and multiplication of matrices– Cofactor matrix – Adjoint matrix – Singular and non-singular matrices – Inverse of a matrix – Simple problems – Engineering applications of Determinants and Matrices.
\[\color {purple} { DETERMINANTS}\ \hspace{20cm}\]
\[\color {green} {Definition\ of\ Determinant}\ \hspace{20cm}\]
\[A\ system\ of\ Linear\ equations\ like\ \hspace{15cm}\]
\[a_1x + b_1y = c_1\ \hspace{15cm}\]
\[a_2x + b_2y = c_2\ \hspace{15cm}\]
Determinant is a square arrangement of numbers within two vertical lines.
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\[\color {black} {Eg:}\ \Delta =\begin{vmatrix}
a_1 & b_1 \\
a_2 & b_2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\color {green} {Order\ and\ Value\ of\ the\ Determinant}\ \hspace{20cm}\]
\[\color {green} {Determinant\ of\ Second\ order:}\ \hspace{20cm}\]
\[A =\begin{vmatrix}
a_1 & b_1 \\
a_2 & b_2 \\
\end{vmatrix}\ \hspace{15cm}\]
consisting of two rows and two columns is called a determinant of second order.
Value of the Determinant is
\[\Delta =\begin{vmatrix}
a_1 & b_1 \\
a_2 & b_2 \\
\end{vmatrix}\ = a_1b_2\ -\ a_2b_1\ \hspace{15cm}\]
\[\color {purple} {Example\ 1:}\ \color {red}{Find\ the\ determinant\ of}\ A =\begin{vmatrix}
2 & -5 \\
1 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ A = \begin{vmatrix}
2 & -5 \\
1 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 2(3) – (-5)(1) \hspace{13cm}\]
\[= 6 + 5 \hspace{12cm}\]
\[\Delta= 11 \hspace{12cm}\]
\[\color {purple} {Example\ 2:}\ \color {red}{Find\ the\ value\ of\ x\ if}\ \begin{vmatrix}
2 & 3 \\
4 & x \\
\end{vmatrix}\ = 0\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue}{Solution:}\ \begin{vmatrix}
2 & 3 \\
4 & x \\
\end{vmatrix}\ = 0 \hspace{15cm}\]
\[2(x)\ -\ 4 (3) = 0\ \hspace{13cm}\]
\[2x\ -\ 12\ =\ 0\ \hspace{13cm}\]
\[2x\ =\ 12\ \hspace{13cm}\]
\[x\ =\ 6\ \hspace{13cm}\]
\[\color {purple} {Example\ 3:}\ \color {red}{Find\ the\ value\ of\ x\ if}\ \begin{vmatrix}
3 & 3 \\
4 & x \\
\end{vmatrix}\ = 0\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue}{Solution:}\ \begin{vmatrix}
3 & 3 \\
4 & x \\
\end{vmatrix}\ = 0 \hspace{15cm}\]
\[3(x)\ -\ 3 (3) = 0\ \hspace{13cm}\]
\[3x\ -\ 9\ =\ 0\ \hspace{13cm}\]
\[3x\ =\ 9\ \hspace{13cm}\]
\[x\ =\ 3\ \hspace{13cm}\]
\[\color {purple} {Example\ 4:}\ \color {red}{Find\ the\ value\ of\ x\ if}\ \begin{vmatrix}
x & 1 \\
4 & x \\
\end{vmatrix}\ = 0\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue}{Solution:}\ \begin{vmatrix}
x & 1 \\
4 & x \\
\end{vmatrix}\ = 0 \hspace{15cm}\]
\[x(x) – 4 (1) = 0\ \hspace{13cm}\]
\[x^2 – 4 = 0\ \hspace{13cm}\]
\[x^2 = 4\ \hspace{13cm}\]
\[x = \pm\sqrt{4}\ \hspace{13cm}\]
\[x = \pm 2\ \hspace{13cm}\]
\[\color{green}{\boxed{x\ =\ 2\ or\ x\ =\ -\ 2}}\]
\[\color {green} {Determinant\ of\ Third\ order:}\ \hspace{20cm}\]
\[The\ expression\ \begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3 \\
\end{vmatrix}\ \hspace{20cm}\]
consisting of three rows and three columns is called a determinant of third order.
Value of the Determinant is :
\[\Delta =a_1\begin{vmatrix}
b_2 & b_3 \\
c_2 & c_3 \\
\end{vmatrix}\ -\ a_2\begin{vmatrix}
b_1 & b_3 \\
c_1 & c_3 \\
\end{vmatrix}\ +\ a_3\begin{vmatrix}
b_1 & b_2 \\
c_1 & c_2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =a_1(b_2c_3\ -\ b_3c_2)\ – a_2 (b_1c_3\ -\ b_3c_1) + a_3(b_1c_2\ -\ b_2c_1)\
\hspace{10cm}\]
\[\color {purple} {Example\ 5:}\ \color {red}{Find\ the\ determinant\ of}\ A =\begin{bmatrix}
3 & 1 & -1 \\
2 & -1 & 2 \\
2 & 1 & -2 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ A =\begin{bmatrix}
3 & 1 & -1 \\
2 & -1 & 2 \\
2 & 1 & -2 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\Delta =3\begin{vmatrix}
-1 & 2 \\
1 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 2 \\
2 & -2 \\
\end{vmatrix}\ +\ -1\begin{vmatrix}
2 & -1 \\
2 & 1 \\
\end{vmatrix}\ \hspace{4cm}\]
\[ =3(2\ -\ 2)\ – 1 (-4\ -\ 4) – 1(2\ +\ 2)\
\hspace{10cm}\]
\[ =3(0)\ – 1 (-8) – 1(4)\
\hspace{14cm}\]
\[ =0\ +8 – 4\
\hspace{15cm}\]
\[\Delta =4\
\hspace{16cm}\]
\[\color {purple} {Example\ 6:}\ \color {red}{Find\ the\ value\ of\ m\ if}\ \begin{vmatrix}
3 & 4 & -2 \\
-3 & 6 & 2 \\
4 & 1 & m \\
\end{vmatrix}\ =\ 0\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \begin{vmatrix}
3 & 4 & -2 \\
-3 & 6 & 2 \\
4 & 1 & m \\
\end{vmatrix}\ =\ 0\ \hspace{15cm}\]
\[3\begin{vmatrix}
6 & 2 \\
1 & m \\
\end{vmatrix}\ -\ 4\begin{vmatrix}
-3 & 2 \\
4 & m \\
\end{vmatrix}\ -2\begin{vmatrix}
-3 & 6 \\
4 & 1 \\
\end{vmatrix}\ =\ 0\ \hspace{4cm}\]
\[3(6m\ -\ 2)\ -\ 4 (-3m\ -\ 8)\ -\ 2(-3\ -\ 24)\ =\ 0\
\hspace{10cm}\]
\[ 18m\ -\ 6\ +\ 12m\ +\ 32\ – 2(-27)\ =\ 0\
\hspace{14cm}\]
\[ 30m\ +\ 26\ +\ 54\ =\ 0\
\hspace{14cm}\]
\[ 30m\ +\ 80\ =\ 0\
\hspace{14cm}\]
\[ 30m\ =\ -\ 80\
\hspace{14cm}\]
\[ \boxed{m\ =\ \frac{-8}{3}}\
\hspace{14cm}\]
\[\color {purple} {SOLUTION\ OF\ SIMULTANEOUS\ EQUATIONS\ USING\ CRAMERS\ RULE}\ \hspace{20cm}\]
\[a_1x\ + b_1y\ +\ c_1z\ = d_1\ —– (1)\ \hspace{20cm}\]
\[a_2x\ + b_2y\ +\ c_2z\ = d_2\ \hspace{20cm}\]
\[a_3x\ + b_3y\ +\ c_3z\ = d_3\ \hspace{20cm}\]
\[\color {black}{Solution:}\ to\ find\ x,\ y,\ z\ \hspace{20cm}\]
\[Step\ 1:\ \Delta = \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \\
\end{vmatrix}\ \hspace{20cm}\]
\[Step\ 2:\ \Delta_x = \begin{vmatrix}
d_1 & b_1 & c_1 \\
d_2 & b_2 & c_2 \\
d_3 & b_3 & c_3 \\
\end{vmatrix}\ \hspace{20cm}\]
\[Step\ 3:\ \Delta_y= \begin{vmatrix}
a_1 & d_1 & c_1 \\
a_2 & d_2 & c_2 \\
a_3 & d_3 & c_3 \\
\end{vmatrix}\ \hspace{20cm}\]
\[Step\ 4:\ \Delta_z = \begin{vmatrix}
a_1 & b_1 & d_1 \\
a_2 & b_2 & d_2 \\
a_3 & b_3 & d_3 \\
\end{vmatrix}\ \hspace{20cm}\]
\[Solution\ is\ x=\ \frac{\Delta_x}{\Delta}.\ y=\ \frac{\Delta_y}{\Delta},\ z=\ \frac{\Delta_z}{\Delta}\ \hspace{15cm}\]
\[\color {purple}{Example\ 7:}\ \color{red}{Solve}\ the\ following\ equations\ x + y\ +\ z\ =\ 3,\ 2x\ -\ y\ +\ z\ =\ 2\ and\ \hspace{15cm}\\ 3\ x\ +\ 2\ y\ -\ 2\ z\ =\ 3\ \color{red}{using\ Cramers\ Rule}\ \hspace{18cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[x + y\ +\ z\ =\ 3\ ————— (1) \hspace{7cm}\]
\[2x\ -\ y\ +\ z\ =\ 2\ \hspace{15cm}\]
\[3\ x\ +\ 2\ y\ -\ 2\ z\ =\ 3\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
1 & 1 & 1 \\
2 & – 1 & 1 \\
3 & 2 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =1\begin{vmatrix}
-1 & 1 \\
2 & – 2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
3 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =\ 1(2\ -\ 2)\ – 1 (-4\ -\ 3)\ +\ 1(4\ +\ 3)\
\hspace{9cm}\]
\[\Delta =1(0)\ – 1\ (-\ 7)\ +\ 1(7)\
\hspace{13cm}\]
\[\Delta =\ 0\ +\ 7\ +\ 7\
\hspace{14cm}\]
\[\boxed{\Delta =\ 14}\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
3 & 1 & 1 \\
2 & -1 & 1 \\
3 & 2 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =\ 3\begin{vmatrix}
-1 & 1 \\
2 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
3 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x\ =\ 3(2\ -\ 2)\ -\ 1 (-4\ -\ 3)\ + \ 1(4\ +\ 3)\
\hspace{9cm}\]
\[\Delta_x =\ -3(0)\ – 1 (-7)\ +\ 1 (7)\
\hspace{13cm}\]
\[\Delta_x\ =\ 0\ +\ 7\ +\ 7\
\hspace{14cm}\]
\[\boxed{\Delta_x\ =\ 14}\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
1 & 3 & 1 \\
2 & 2 & 1 \\
3 & 3 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y =1\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ -\ 3\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & 2\\
3 & 3 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y =1(-\ 4\ -\ 3)\ -\ 3\ (-\ 4\ -\ 3)\ +\ 1(6\ -\ 6)\
\hspace{9cm}\]
\[\Delta_y =1(-7)\ -\ 3\ (-\ 7)\ +\ 1(0)\
\hspace{13cm}\]
\[\Delta_y\ =\ -\ 7\ +\ 21\ +\ 0\
\hspace{14cm}\]
\[\boxed{\Delta_y\ =\ 14}\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
1 & 1 & 3 \\
2 & -1 & 2 \\
3 & 2 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z =1\begin{vmatrix}
-1 & 2 \\
2 & 3 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 2 \\
3 & 3 \\
\end{vmatrix}\ +\ 3\begin{vmatrix}
2 & -1\\
3 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z\ =\ 1(-\ 3\ -\ 4)\ -\ 1(6\ -\ 6)\ +\ 3(4\ +\ 3)\
\hspace{9cm}\]
\[\Delta_z\ =\ 1(-\ 7)\ -\ 1 (0)\ +\ 3(7)\
\hspace{13cm}\]
\[\Delta_z =\ -\ 7\ -\ 0\ +\ 21\
\hspace{14cm}\]
\[\boxed{\Delta_z\ =\ 14}\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{14}{14}\ =\ 1\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x\ =\ 1\ y\ =\ 1\ and\ z\ =\ 1\ in\ equation (1)\ \hspace{18cm}\]
\[LHS\ =\ 1 +\ 1\ +\ 1\]\[ =\ 3\]\[ = RHS\]
\[\color {violet}{Example\ 8:}\ \color {red} {Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}\]
\[4\ x\ +\ y\ +\ z\ =\ 6,\ 2\ x\ -\ y\ -\ 2\ z\ =\ -\ 6\ and\ x\ +\ y\ +\ z\ =\ 3\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2024\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[4\ x\ +\ y\ +\ z\ =\ 6\ ——————-(1)\ \hspace{6cm}\]
\[2\ x\ -\ y\ -\ 2\ z\ =\ -\ 6\ \hspace{15cm}\]
\[x\ +\ y\ +\ z\ =\ 3\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
4 & 1 & 1 \\
2 & -1 & -2\\
1 & 1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =4\begin{vmatrix}
-1 & -2 \\
1 & 1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -2 \\
1 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =4(-1\ +\ 2)\ – 1 (2\ +\ 2)\ +\ 1(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta\ =\ 4(1)\ – 1 (4)\ +\ 1(7)\
\hspace{13cm}\]
\[\Delta =4\ -\ 4\ +\ 3\
\hspace{14cm}\]
\[\boxed{\Delta\ =\ 3}\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
6 & 1 & 1 \\
-6 & -1 & -2 \\
3 & 1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =6\begin{vmatrix}
-1 & -2 \\
1 & 1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
-6 & -2 \\
3 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
-6 & -1\\
3 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x\ =\ 6(-1\ +\ 2) – 1 (\ -6\ +\ 6)\ +\ 1(-6\ +\ 3)\
\hspace{9cm}\]
\[\Delta_x\ =\ 6(1)\ – 1 (0)\ +\ 1(-3)\
\hspace{13cm}\]
\[\Delta_x = 6\ +\ 0\ -3\
\hspace{14cm}\]
\[\boxed{\Delta_x\ =\ 3}\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
4 & 6 & 1 \\
2 & -6 & -2 \\
1 & 3 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y\ =\ 4\begin{vmatrix}
-6 & -2 \\
3 & 1\\
\end{vmatrix}\ -\ 6\begin{vmatrix}
2 & -2 \\
1 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -6\\
1 & 3 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y\ =\ 4(-\ 6\ +\ 6)\ -\ 6 (2\ +\ 2)\ +\ 1(6\ +\ 6)\
\hspace{9cm}\]
\[\Delta_y\ =\ 4(0)\ -\ 6 (4)\ +\ 1(12)\
\hspace{13cm}\]
\[\Delta_y\ =\ -\ 0\ -\ 24\ +\ 12\
\hspace{14cm}\]
\[\boxed{\Delta_y\ =\ -12}\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
4 & 1 & 6 \\
2 & -1 & -6 \\
1 & 1 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z\ =\ 4\begin{vmatrix}
-1 & -6 \\
1 & 3 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -6 \\
1 & 3 \\
\end{vmatrix}\ +\ 6\begin{vmatrix}
2 & – 1\\
1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z\ =\ 4(-\ 3\ +\ 6)\ -\ 1 (6\ +\ 6)\ +\ 6(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta_z\ =\ 4(3)\ -\ 1 (12)\ +\ 6(3)\
\hspace{13cm}\]
\[\Delta_z\ =\ 12\ -\ 12\ +\ 18\
\hspace{14cm}\]
\[\boxed{\Delta_z\ =\ 18}\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{-12}{3} =\ -4\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{18}{3} =\ 6\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x\ =\ 1,\ y\ =\ -4\ and\ z = 6\ in\ equation (1)\ \hspace{18cm}\]
\[LHS\ =\ 4(1) – 4 + 6\]\[ = 4 – 4 + 6 = 6\]\[ = RHS\]
\[\color {violet}{Example\ 9:}\ \color {red} {Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}\]
\[3\ x\ +\ y\ -\ z\ =\ 2,\ 2\ x\ -\ y\ +\ 2\ z\ =\ 6\ and\ 2\ x\ +\ y\ -\ 2\ z\ =\ -\ 2\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2023\ April\ 2025\ June\ 2025(Supp)\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[3\ x\ +\ y\ -\ z\ =\ 2\ ——————-(1)\ \hspace{6cm}\]
\[2\ x\ -\ y\ +\ 2\ z\ =\ 6\ \hspace{15cm}\]
\[2\ x\ +\ y\ -\ 2\ z\ =\ -\ 2\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
3 & 1 & -1 \\
2 & -1 & 2\\
2 & 1 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =\ 3\begin{vmatrix}
-1 & 2 \\
1 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 2 \\
2 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -1\\
2 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =\ 3(2\ -\ 2)\ – 1 (-\ 4\ -\ 4)\ -\ 1(2\ +\ 2)\
\hspace{9cm}\]
\[\Delta\ =\ 3(0)\ – 1 (-8)\ -\ 1(4)\
\hspace{13cm}\]
\[\Delta =\ 0\ +\ 8\ -\ 4\
\hspace{14cm}\]
\[\boxed{\Delta\ =\ 4}\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
2 & 1 & -1 \\
6 & -1 & 2 \\
-2 & 1 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x\ =\ 2\begin{vmatrix}
-1 & 2 \\
1 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
6 & 2 \\
-2 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
6 & -1\\
-2 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x\ =\ 2(2\ -\ 2) – 1 (-12\ +\ 4)\ -\ 1(6\ -\ 2)\
\hspace{9cm}\]
\[\Delta_x\ =\ 2(0)\ – 1 (-8)\ -\ 1(4)\
\hspace{13cm}\]
\[\Delta_x =\ 0\ +\ 8\ -\ 4\
\hspace{14cm}\]
\[\boxed{\Delta_x\ =\ 4}\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
3 & 2 & -1 \\
2 & 6 & 2 \\
2 & -2 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y\ =\ 3\begin{vmatrix}
6 & 2 \\
-2 & -2\\
\end{vmatrix}\ -\ 2\begin{vmatrix}
2 & 2 \\ 2 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 6\\
2 & -2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y\ =\ 3(-\ 12\ +\ 4)\ -\ 2 (-4\ -\ 4)\ -\ 1(-4\ -\ 12)\
\hspace{9cm}\]
\[\Delta_y\ =\ 3(-8)\ -\ 2 (-8)\ -\ 1(-16)\
\hspace{13cm}\]
\[\Delta_y\ =\ -\ 24\ +\ 16\ +\ 16\
\hspace{14cm}\]
\[\boxed{\Delta_y\ =\ 8}\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
3 & 1 & 2 \\
2 & -1 & 6 \\
2 & 1 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z\ =\ 3\begin{vmatrix}
-1 & 6 \\
1 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 6 \\
2 & -2 \\
\end{vmatrix}\ +\ 2\begin{vmatrix}
2 & – 1\\
2 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z\ =\ 3(2\ -\ 6)\ -\ 1 (-\ 4\ -\ 12)\ +\ 2(2\ +\ 2)\
\hspace{9cm}\]
\[\Delta_z\ =\ 3(-4)\ -\ 1 (-16)\ +\ 2(4)\
\hspace{13cm}\]
\[\Delta_z\ =\ -\ 12\ +\ 16\ +\ 8\
\hspace{14cm}\]
\[\boxed{\Delta_z\ =\ 12}\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{4}{4} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{8}{4} =\ 2\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{12}{4} =\ 3\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x\ =\ 1,\ y\ =\ 2\ and\ z = 3\ in\ equation (1)\ \hspace{18cm}\]
\[LHS\ =\ 3(1)\ +\ 2 -\ 3\]\[ =\ 3\ +\ 2\ -\ 3=\ 2\]\[ = RHS\]
\[\color {royalblue} {Definition\ of\ a\ Matrix}:\ \hspace{20cm}\]
A matrix is a rectangular array of numbers arranged in rows and columns enclosed by brackets.
\[\color {royalblue} {Ex:}\ 1)\ A =\begin{bmatrix}
3 & 6 \\
1 & 2\\
\end{bmatrix}\ \hspace{2cm}\ 2)\ B =\begin{bmatrix}
1 & -1 & 2\\
2 & -2 & 4\\
3 & -3 & 6\\
\end{bmatrix}\ \hspace{10cm}\]
\[\color {royalblue} {Order\ of\ a\ Matrix}:\ \hspace{20cm}\]
If there are m rows and n columns in a matrix, then the order of the matrix is m × n.
\[\color {royalblue} {Ex:}\ A =\begin{bmatrix}
1 & 2 & -1 & 3\\
2 & 4 & -4 & 7\\
-1 & -2 & -2 & -2\\
\end{bmatrix}\ is\ a\ matrix\ of\ order\ 3×4 \hspace{10cm}\]
\[\color {green} {Square\ Matrix}:\ \hspace{20cm}\]
A matrix which has equal number of rows and columns is called a square matrix.
\[\color {black} {Eg:}\ 1)\ A =\begin{bmatrix}
3 & 6 \\
1 & 4 \\
\end{bmatrix}\ is\ a\ square\ matrix\ of\ order\ 2×2 \hspace{20cm}\]
\[\color {black} {Eg:}\ 2)\ B =\begin{bmatrix}
1 & -1 & 2 \\
2 & -2 & 4 \\
3 & -3 & 6 \\
\end{bmatrix}\ is\ a\ square\ matrix\ of\ order\ 3×3 \hspace{20cm}\]
\[\color {green} {Transpose\ of\ Matrix}:\ \hspace{20cm}\]
Let A be a square matrix. The transpose of A is obtained by changing rows into columns and vise-versa and is denoted by AT
\[\color {black} {Eg:}\ 1)\ A =\begin{bmatrix}
3 & 6 \\
1 & 4 \\
\end{bmatrix}\ \hspace{10cm}\]\[A^T =\begin{bmatrix}
3 & 1 \\
6 & 4 \\
\end{bmatrix}\ \hspace{8cm}\]
\[\color {black} {Eg:}\ 2)\ A =\begin{bmatrix}
3 & 4 & 1 \\
0 & -1 & 2 \\
5 & -2 & 6 \\
\end{bmatrix}\ \hspace{10cm}\]\[A^T =\begin{bmatrix}
3 & 0 & 5 \\
4 & -1 & -2 \\
1 & 2 & 6 \\
\end{bmatrix}\ \hspace{8cm}\]
\[\color {green} {6.\ Unit\ Matrix}:\ \hspace{20cm}\]
Unit matrix is a square matrix in which the diagonal elements are all ones and all the other elements are zeros.
\[\color {black} {Eg:}\ 1)\ I =\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}\ is\ a\ unit\ matrix\ of\ order\ 2×2 \hspace{10cm}\]
\[2)\ I =\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\ is\ a\ unit\ matrix\ of\ order\ 3×3 \hspace{8cm}\]
\[\color {purple} {Operation\ on\ Matrices:}\ \hspace{20cm}\]
i) Addition and subtraction of matrices
ii) Multiplication of matrix by a scalar
iii) Multiplication of matrices
\[\color {purple} {i)\ Addition\ and\ Subtraction\ of\ Matrices:}\ \hspace{20cm}\]
Two Matrices can be added (or) subtracted if they have the same order. We can add (or) subtract two matrices by the corresponding element by element.
\[\color {purple} {Example\ 10:}\ \color{red}{If}\ A =\begin{bmatrix}
1 & 2 & 7 \\
0 & 4 & 5 \\
3 & 1 & 6 \\
\end{bmatrix}\ ,\ B =\begin{bmatrix}
1 & 3 & 1 \\
2 & 4 & 0 \\
1 & 7 & 5 \\
\end{bmatrix}\ \hspace{5cm}\]
\[\color {red}{Find\ A + B}\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ A + B=\begin{bmatrix}
1 + 1 & 2 + 3 & 7 + 1 \\
0 + 2 & 4 + 4 & 5 + 0 \\
3 + 1 & 1 + 7 & 6 + 5 \\
\end{bmatrix}\ \hspace{6cm}\]
\[A + B=\begin{bmatrix}
2 & 5 & 8 \\
2 & 8 & 5 \\
4 & 8 & 11 \\
\end{bmatrix}\ \hspace{7cm}\]
\[\color {purple} {Example\ 11:}\ \color {red}{If}\ A =\begin{bmatrix}
1 & 3 & 5 \\
2 & 0 & 7 \\
1 & 5 & 2 \\
\end{bmatrix}\ ,\ B =\begin{bmatrix}
7 & 3 & 4 \\
1 & -1 & 5 \\
0 & 2 & 4 \\
\end{bmatrix}\ \hspace{5cm}\]
\[\color {red}{Find\ A – B}\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ A – B=\begin{bmatrix}
1 – 7 & 3 – 3 & 5 – 4 \\
2 – 1 & 0 + 1 & 7 – 5 \\
1 – 0 & 5 – 2 & 2 – 4 \\
\end{bmatrix}\ \hspace{6cm}\]
\[A – B=\begin{bmatrix}
-6 & 0 & 1 \\
1 & 1& 2 \\
1 & 3 & -2 \\
\end{bmatrix}\ \hspace{7cm}\]
\[\color {purple} {Example\ 12:}\ \color {red}{If}\ A =\ \begin{pmatrix}
4 & 6 & 2 \\
0 & 1 & 5 \\
0 & 3 & 2 \\
\end{pmatrix}\ ,\ B =\begin{pmatrix}
0 & 1 & – 1 \\
3 & – 1 & 4 \\
– 1 & 2 & 1 \\
\end{pmatrix}\ \hspace{7cm}\]\[\color {red}{Prove\ that\ (A\ +\ B)^T\ =\ A^T\ +\ B^T}\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2024\]
\[\color {blue}{Solution:}\ Given\ A =\ \begin{pmatrix}
4 & 6 & 2 \\
0 & 1 & 5 \\
0 & 3 & 2 \\
\end{pmatrix}\ ,\ B =\begin{pmatrix}
0 & 1 & – 1 \\
3 & – 1 & 4 \\
– 1 & 2 & 1 \\
\end{pmatrix}\ \hspace{5cm}\]
\[A + B=\begin{pmatrix}
4 + 0 & 6 + 1 & 2 – 1 \\
0 + 3 & 1 – 1 & 5 + 4 \\
0 – 1 & 3 + 2 & 2 + 1 \\
\end{pmatrix}\ \hspace{6cm}\]
\[A + B=\ \begin{pmatrix}
4 & 7 & 1 \\
3 & 0 & 9\\
-1 & 5 & 3\\
\end{pmatrix}\ \hspace{6cm}\]
\[(A + B)^T\ =\ \begin{pmatrix}
4 & 3 & – 1 \\
7 & 0 & 5 \\
1 & 9 & 3\\
\end{pmatrix}\ \hspace{2cm}\ ——– (1)\]
\[A^T =\begin{pmatrix}
4 & 0 & 0 \\
6 & 1 & 3 \\
2 & 5 & 2 \\
\end{pmatrix}\ \hspace{10cm}\]
\[B^T =\begin{pmatrix}
0 & 3 & – 1 \\
1 & – 1 & 2 \\
– 1 & 4 & 1 \\
\end{pmatrix}\ \hspace{10cm}\]
\[A^T\ +\ B^T\ =\begin{pmatrix}
4 + 0 & 0\ +\ 3 & 0\ -\ 1\\
6\ +\ 1 & 1 – 1 & 3 + 2 \\
2 – 1 & 5 + 4 & 2 + 1 \\
\end{pmatrix}\ \hspace{6cm}\]
\[A^T\ +\ B^T\ =\begin{pmatrix}
4 & 3 & – 1\\
7 & 0 & 5 \\
1 & 9 & 3 \\
\end{pmatrix}\ \hspace{2cm}\ ———- (2)\]
\[From\ (1)\ and\ (2), It\ is\ concluded\ that (A\ +\ B)^T\ =\ A^T\ +\ B^T\]
\[\color {purple} {ii)\ Matrix\ Multiplication\ by\ a\ scalar}\ \hspace{20cm}\]
We can multiply the matrix by any non-zero scalar [(value) number] obtain we get the matrix whose all the elements are multiplied by that same scalar.
\[\color {purple} {Example\ 13:}\ \color {red}{If}\ A =\begin{pmatrix}
5 & – 6 \\
3 & 7\\
\end{pmatrix}, \color {red} {Find\ 3A}\ \hspace{12cm}\]
\[\color {blue}{Solution:}\ \ A =\begin{pmatrix}
5 & – 6 \\
3 & 7\\
\end{pmatrix}\ \hspace{15cm}\]
\[ 3A = 3\begin{pmatrix}
5 & – 6 \\
3 & 7\\
\end{pmatrix}\ \hspace{13cm}\]
\[3A = \begin{pmatrix}
15 & – 18 \\
9 & 21\\
\end{pmatrix}\ \hspace{13cm}\]
\[\color {purple} {Example\ 14:}\ \color {red}{If}\ A =\begin{bmatrix}
9 & 10 \\
13 & 20\\
\end{bmatrix}\ and\ B =\begin{bmatrix}
3 & 5 \\
8 & 9\\
\end{bmatrix},\ \hspace{15cm}\]
\[\color {red}{\ find}\ 2A\ +\ B\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2023,\ June\ 2025(Supp)\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[ 2A = 2\begin{bmatrix}
9 & 10 \\
13 & 20\\
\end{bmatrix}\ \hspace{13cm}\]
\[2A = \begin{bmatrix}
18 & 20 \\
26 & 40\\
\end{bmatrix}\ ———- (1)\ \hspace{8cm}\]
\[B = \begin{bmatrix}
3 & 5 \\
8 & 9\\
\end{bmatrix}\ ———- (2)\ \hspace{8cm}\]
\[2A\ +\ B = \begin{bmatrix}
18 & 20 \\
26 & 40\\
\end{bmatrix}\ +\ \begin{bmatrix}
3 & 5 \\
8 & 9\\
\end{bmatrix}\ \hspace{8cm}\]
\[=\begin{bmatrix}
18\ +\ 3 & 20\ +\ 5 \\
26\ +\ 8 & 40\ +\ 9\\
\end{bmatrix}\ \hspace{8cm}\]
\[2A\ +\ B = \begin{bmatrix}
21 & 25 \\
34 & 49\\
\end{bmatrix}\ \hspace{10cm}\]
\[\color {purple} {Example\ 15:}\ \color {red}{If}\ A =\begin{pmatrix}
1 & 2 \\
3 & 5\\
\end{pmatrix}\ and\ B =\begin{pmatrix}
– 5 & 7 \\
0 & 4\\
\end{pmatrix},\ \hspace{15cm}\]
\[\color {red}{\ find}\ 2A\ +\ B\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2024\ October\ 25\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[ 2A = 2\begin{pmatrix}
1 & 2 \\
3 & 5\\
\end{pmatrix}\ \hspace{13cm}\]
\[2A = \begin{pmatrix}
2 & 4 \\
6 & 10\\
\end{pmatrix}\ ———- (1)\ \hspace{8cm}\]
\[B = \begin{pmatrix}
-5 & 7 \\
0 & 4\\
\end{pmatrix}\ ———- (2)\ \hspace{8cm}\]
\[2A\ +\ B = \begin{pmatrix}
2 & 4 \\
6 & 10\\
\end{pmatrix}\ +\ \begin{pmatrix}
-5 & 7 \\
0 & 4\\
\end{pmatrix}\ \hspace{8cm}\]
\[=\begin{pmatrix}
2\ -\ 5 & 4\ +\ 7 \\
6\ +\ 0 & 10\ +\ 4\\
\end{pmatrix}\ \hspace{8cm}\]
\[2A\ +\ B = \begin{pmatrix}
-3 & 11 \\
6 & 14\\
\end{pmatrix}\ \hspace{10cm}\]
\[\color {purple} {Example\ 16:}\ \color {red}{If}\ A =\begin{pmatrix}
5 & 2 \\
16 & 15\\
\end{pmatrix}\ and\ B =\begin{pmatrix}
5 & 2 \\
4 & 6\\
\end{pmatrix},\ \hspace{15cm}\]
\[\color {red}{\ find}\ 3A\ +\ 2B\ \hspace{10cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[ 3A = 3\begin{pmatrix}
5 & 2 \\
16 & 15\\
\end{pmatrix}\ \hspace{13cm}\]
\[3A = \begin{pmatrix}
15 & 6 \\
48 & 45\\
\end{pmatrix}\ ———- (1)\ \hspace{8cm}\]
\[B = \begin{pmatrix}
5 & 2 \\
4 & 6\\
\end{pmatrix}\ \hspace{13cm}\]
\[2B = \begin{pmatrix}
10 & 4 \\
8 & 12\\
\end{pmatrix}\ ———- (2)\ \hspace{8cm}\]
\[3A\ +\ 2B = \begin{pmatrix}
15 & 6 \\
48 & 45\\
\end{pmatrix}\ +\ \begin{pmatrix}
10 & 4\\
4 & 8\\
\end{pmatrix}\ \hspace{8cm}\]
\[=\begin{pmatrix}
15\ +\ 10 & 6\ +\ 4 \\
48\ +\ 8& 45\ +\ 12\\
\end{pmatrix}\ \hspace{8cm}\]
\[3A\ +\ 2B = \begin{pmatrix}
25 & 10 \\
56 & 57\\
\end{pmatrix}\ \hspace{10cm}\]
\[\color {purple} {Example\ 17:}\ \color {red}{If}\ A =\begin{bmatrix}
1 & 2 \\
3 & 5\\
\end{bmatrix}\ and\ B =\begin{bmatrix}
-5 & 7 \\
0 & 4\\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {red}{Then\ Find}\ 4A\ -\ 2B\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[ 4A = 4\begin{bmatrix}
1 & 2 \\
3 & 5\\
\end{bmatrix}\ \hspace{13cm}\]
\[4A = \begin{bmatrix}
4 & 8 \\
12 & 20\\
\end{bmatrix}\ ———- (1)\ \hspace{8cm}\]
\[ 2B = 2\begin{bmatrix}
-5 & 7 \\
0 & 4\\
\end{bmatrix}\ \hspace{13cm}\]
\[2B = \begin{bmatrix}
-10 & 14 \\
0 & 8\\
\end{bmatrix}\ ———- (2)\ \hspace{8cm}\]
\[4A\ – 2B = \begin{bmatrix}
4 & 8 \\
12 & 20\\
\end{bmatrix}\ -\ \begin{bmatrix}
-10 & 14 \\
0 & 8\\
\end{bmatrix}\ \hspace{8cm}\]
\[=\begin{bmatrix}
4 + 10 & 8 -14 \\
12 – 0 & 20 -8\\
\end{bmatrix}\ \hspace{8cm}\]
\[4A\ – 2B = \begin{bmatrix}
14 & -6 \\
12 & 12\\
\end{bmatrix}\ \hspace{10cm}\]
\[\color {purple} {Example\ 18:}\ If\ f(x)\ =\ 3\ x\ +\ 2\ and\ A =\begin{pmatrix}
1 & 0 \\
2 & -1\\
\end{pmatrix},\ \color {red}{\ find\ f(A)}.\ \hspace{10cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[\hspace{3cm}\ Given\ \hspace{15cm}\]
\[\hspace{3cm}\ A =\begin{pmatrix}
1 & 0 \\
2 & -1\\
\end{pmatrix}\ and\ \hspace{15cm}\]
\[\hspace{3cm}\ f(x)\ =\ 3\ x\ +\ 2\ \hspace{15cm}\]
\[f(A) = \ 3A\ +\ 2\ I, Where\ I\ is\ a\ Identity\ matrix\ of\ order\ 2\ ———- (1)\]
\[3A\ =\ 3\begin{pmatrix}
1 & 0 \\
2 & -1\\
\end{pmatrix}\ \hspace{13cm}\]
\[3A\ =\ \begin{pmatrix}
3 & 0 \\
6 & -3\\
\end{pmatrix}\ \hspace{13cm}\]
\[I\ =\ \begin{pmatrix}
1 & 0 \\
0 & 1\\
\end{pmatrix}\ \hspace{13cm}\]
\[2I\ =\ \begin{pmatrix}
2 & 0 \\
0 & 2\\
\end{pmatrix}\ \hspace{13cm}\]
\[3A\ +\ 2I\ = \begin{pmatrix}
3 & 0 \\
6 & -3\\
\end{pmatrix}\ +\ \begin{pmatrix}
2 & 0 \\
0 & 2\\
\end{pmatrix}\ \hspace{8cm}\]
\[=\begin{pmatrix}
3\ +\ 2 & 0\ +\ 0 \\
6\ +\ 0\ & -3\ +\ 2\\
\end{pmatrix}\ \hspace{8cm}\]
\[3A\ +\ 2I = \begin{pmatrix}
5 & 0 \\
6 & -1\\
\end{pmatrix}\ \hspace{10cm}\]
\[\boxed{\color{green}{\therefore\ f(A)\ = \begin{pmatrix}
5 & 0 \\
6 & -1\\
\end{pmatrix}}}\ \hspace{10cm}\]
\[\color {purple} {iii)\ Multiplication\ of\ Matrices}\ \hspace{20cm}\]
The condition of multiplication of two matrices A and B is the number of columns in A is equal to the number of rows in B.
\[\color {purple} {Example\ 19:}\ \color {red}{If}\ A =\begin{bmatrix}
1 & 2 \\
3 & 4 \\
\end{bmatrix}\ ,\ B =\begin{bmatrix}
5 & 6 \\
7 & 8 \\
\end{bmatrix}\ ,\ \color {red} {Find\ AB}\ \hspace{12cm}\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[AB =\begin{bmatrix}
1 & 2 \\
3 & 4 \\
\end{bmatrix}\ \begin{bmatrix}
5 & 6 \\
7 & 8 \\
\end{bmatrix}\ \hspace{12cm}\]
\[ = \begin{bmatrix}
1 × 5 + 2 × 7 & 1 × 6 + 2 × 8 \\
3 × 5 + 4 × 7 & 3 × 6 + 4 × 8\\
\end{bmatrix}\ \hspace{9cm}\]
\[ = \begin{bmatrix}
5 + 14 & 6 + 16 \\
15 + 28 & 18 + 32\\
\end{bmatrix}\ \hspace{10cm}\]
\[\boxed{AB = \begin{bmatrix}
19 & 22 \\
43 & 50\\
\end{bmatrix}}\ \hspace{12cm}\]
\[\color {purple} {Example\ 20:}\ \color {Red}{If}\ A =\begin{bmatrix}
1 & 1 \\
-1 & 1 \\
\end{bmatrix}\ and\ B =\begin{bmatrix}
1 & 7 &; 0\\
4 & 3 & – 2; \\
\end{bmatrix}\ ,\ \color {red} {Find\ AB}\ \hspace{12cm}\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[AB =\begin{bmatrix}
1 & 1 \\
– 1 & 1 \\
\end{bmatrix}\ \begin{bmatrix}
1 & 7 &; 0\\
4 & 3 & – 2; \\
\end{bmatrix}\ \hspace{12cm}\]
\[ = \begin{bmatrix}
1 × 1 + 1 × 4 & 1 × 7 + 1× 3 & 1 × 0 + 1× -2 \\
-1 × 4 + 1 × 4 & – 1 × 7 + 1× 3 & -1 × 0 + 1× – 2 \\
\end{bmatrix}\ \hspace{9cm}\]
\[ = \begin{bmatrix}
1 + 4 & 7 + 3 &; 0 – 2 \\
– 4 + 4 & – 7 + 3 &; 0 – 2 \\
\end{bmatrix}\ \hspace{9cm}\]
\[\boxed{AB\ =\ \begin{bmatrix}
5 & 10 & – 2\\
0& – 4 & – 2\\
\end{bmatrix}}\ \hspace{12cm}\]
\[\color {purple} {Example\ 21:}\ \color {red}{Verify\ (AB)^T\ =\ B^T\ A^T}\ If\ A =\begin{bmatrix}
1 & 0 & 3 \\
2 & 1 & – 1 \\
1 & -1 & 1 \\
\end{bmatrix}\ and\ \ B =\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0 \\
\end{bmatrix}\ \hspace{12cm}\]
\[\hspace{5cm}\ October\ 2023,\ June\ 2025\ (Supp)\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[Let\ A =\begin{bmatrix}
1 & 0 & 3 \\
2 & 1 & – 1 \\
1 & -1 & 1 \\
\end{bmatrix}\ and\ \ B =\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0 \\
\end{bmatrix}\ \hspace{12cm}\]
\[AB =\begin{bmatrix}
1 & 0 & 3 \\
2 & 1 & – 1 \\
1 & -1 & 1 \\
\end{bmatrix}\ \begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0 \\
\end{bmatrix}\ \hspace{12cm}\]
\[ = \begin{bmatrix}
1\ ×\ 1\ +\ 0 ×\ 0\ +\ 3\ ×\ 1& 1\ ×\ 0\ +\ 0 ×\ 1\ +\ 3\ ×\ 2 & 1\ ×\ 2\ +\ 0 ×\ 2\ +\ 3\ ×\ 0\\
2\ ×\ 1\ +\ 1 ×\ 0\ +\ -1\ ×\ 1\ & 2\ ×\ 0\ +\ 1 ×\ 1\ +\ -1\ ×\ 2\ & 2\ ×\ 2\ +\ 1 ×\ 2\ +\ -1\ ×\ 0\\
1\ ×\ 1\ +\ -1 ×\ 0\ +\ 1\ ×\ 1\ & 1\ ×\ 0\ +\ -1 ×\ 1\ +\ 1\ ×\ 2\ & 1\ ×\ 2\ +\ -1 ×\ 2\ +\ 1\ ×\ 0\\
\end{bmatrix}\ \hspace{9cm}\]
\[ = \begin{bmatrix}
1\ +\ 0\ +\ 3\ & 0\ +\ 0\ +\ 6 & 2\ +\ 0 +\ 0\\
2\ +\ 0\ +\ -1\ & 0\ +\ 1\ -\ 2\ & 4\ +\ 2\ +\ 0\\
1\ +\ 0\ +\ 1\ & 0\ -\ 1\ +\ 2\ & 2\ -\ 2\ +\ 0\\
\end{bmatrix}\ \hspace{9cm}\]
\[ AB = \begin{bmatrix}
4 & 6 & 2 \\
1 & -1 & 6 \\
2 & 1 & 0 \\
\end{bmatrix}\ \hspace{12cm}\]
\[(AB)^T\ = \begin{bmatrix}
4 & 1 & 2 \\
6 & -1 & 1 \\
2 & 6 & 0 \\
\end{bmatrix}\ ————– (1)\ \hspace{10cm}\]
\[Let\ B^T =\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 2 \\
2 & 2 & 0 \\
\end{bmatrix}\ and\ \ A^T\ =\begin{bmatrix}
1 & 2 & 1 \\
0 & 1 & – 1 \\
3 & -1 & 1 \\
\end{bmatrix}\ \hspace{12cm}\]
\[B^T\ A^T\ =\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 2 \\
2 & 2 & 0 \\
\end{bmatrix}\ \begin{bmatrix}
1 & 2 & 1 \\
0 & 1 & – 1 \\
3 & -1 & 1 \\
\end{bmatrix}\ \hspace{12cm}\]
\[ = \begin{bmatrix}
1\ ×\ 1\ +\ 0 ×\ 0\ +\ 1\ ×\ 3& 1\ ×\ 2\ +\ 0 ×\ 1\ +\ 1\ ×\ -1 & 1\ ×\ 1\ +\ 0 ×\ -1\ +\ 1\ ×\ 1\\
0\ ×\ 1\ +\ 1 ×\ 0\ +\ 2\ ×\ 3\ & 0\ ×\ 2\ +\ 1 ×\ 1\ +\ 2\ ×\ -1\ & 0\ ×\ 1\ +\ 1 ×\ -1\ +\ 2\ ×\ 1\\
2\ ×\ 1\ +\ 2 ×\ 0\ +\ 0\ ×\ 3\ & 2\ ×\ 2\ +\ 2 ×\ 1\ +\ 0\ ×\ -1\ & 2\ ×\ 1\ +\ 2 ×\ -1\ +\ 0\ ×\ 1\\
\end{bmatrix}\ \hspace{9cm}\]
\[ = \begin{bmatrix}
1\ +\ 0\ +\ 3\ & 2\ +\ 0\ -\ 1 & 1\ +\ 0 +\ 1\\
0\ +\ 0\ +\ 6\ & 0\ +\ 1\ -\ 2\ & 0\ -\ 1\ +\ 2\\
2\ +\ 0\ +\ 0\ & 4\ +\ 2\ +\ 0\ & 2\ -\ 2\ +\ 0\\
\end{bmatrix}\ \hspace{9cm}\]
\[B^T\ A^T\ = \begin{bmatrix}
4 & 1 & 2 \\
6 & -1 & 1 \\
2 & 6 & 0 \\
\end{bmatrix}\ ————– (2)\ \hspace{10cm}\]
\[From\ (1)\ and\ (2)\ (AB)^T\ =\ B^T\ A^T\ \hspace{10cm}\]
\[\color {purple} {Example\ 22:}\ \color {red}{Verify\ (AB)^T\ =\ B^T\ A^T}\ If\ A =\begin{pmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{pmatrix}\ and\ \ B =\begin{pmatrix}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0 \\
\end{pmatrix}\ \hspace{12cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[Let\ A =\begin{pmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{pmatrix}\ and\ \ B =\begin{pmatrix}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0 \\
\end{pmatrix}\ \hspace{12cm}\]
\[AB =\begin{pmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{pmatrix}\ \begin{pmatrix}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0 \\
\end{pmatrix}\ \hspace{12cm}\]
\[ = \begin{pmatrix}
1\ ×\ 1\ +\ 2 ×\ 0\ +\ 2\ ×\ 1& 1\ ×\ 0\ +\ 2 ×\ 1\ +\ 2\ ×\ 2 & 1\ ×\ 2\ +\ 2 ×\ 2\ +\ 2\ ×\ 0\\
2\ ×\ 1\ +\ 1 ×\ 0\ +\ 2\ ×\ 1\ & 2\ ×\ 0\ +\ 1 ×\ 1\ +\ 2\ ×\ 2\ & 2\ ×\ 2\ +\ 1 ×\ 2\ +\ 2\ ×\ 0\\
2\ ×\ 1\ +\ 2×\ 0\ +\ 1\ ×\ 1\ & 1\ ×\ 0\ +\ 2 ×\ 1\ +\ 1\ ×\ 2\ & 2\ ×\ 2\ +\ 2 ×\ 2\ +\ 1\ ×\ 0\\
\end{pmatrix}\ \hspace{9cm}\]
\[ = \begin{pmatrix}
1\ +\ 0\ +\ 2\ & 0\ +\ 2\ +\ 4 & 2\ +\ 4 +\ 0\\
2\ +\ 0\ +\ 2\ & 0\ +\ 1\ +\ 4\ & 4\ +\ 2\ +\ 0\\ 2\ +\ 0\ +\ 1\ & 0\ +\ 2\ +\ 2\ & 4\ +\ 4\ +\ 0\\
\end{pmatrix}\ \hspace{9cm}\]
\[ AB = \begin{pmatrix}
3 & 6 & 6 \\
4 & 5 & 6 \\
3 & 4 & 8 \\
\end{pmatrix}\ \hspace{12cm}\]
\[(AB)^T\ = \begin{pmatrix}
3 & 4 & 3 \\
6 & 5 & 4 \\
6 & 6 & 8\\
\end{pmatrix}\ ————– (1)\ \hspace{10cm}\]
\[Let\ B^T =\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 2 \\
2 & 2 & 0 \\
\end{pmatrix}\ and\ \ A^T\ =\begin{pmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{pmatrix}\ \hspace{12cm}\]
\[B^T\ A^T\ =\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 2 \\
2 & 2 & 0 \\
\end{pmatrix}\ \begin{pmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{pmatrix}\ \hspace{12cm}\]
\[ = \begin{pmatrix}
1\ ×\ 1\ +\ 0 ×\ 2\ +\ 1\ ×\ 2& 1\ ×\ 2\ +\ 0 ×\ 1\ +\ 1\ ×\ 2 & 1\ ×\ 2\ +\ 0 ×\ 2\ +\ 1\ ×\ 1\\
0\ ×\ 1\ +\ 1 ×\ 2\ +\ 2\ ×\ 2\ & 0\ ×\ 2\ +\ 1 ×\ 1\ +\ 2\ ×\ 2\ & 0\ ×\ 2\ +\ 1 ×\ 2\ +\ 2\ ×\ 1\\
2\ ×\ 1\ +\ 2 ×\ 2\ +\ 0\ ×\ 2\ & 2\ ×\ 2\ +\ 2 ×\ 1\ +\ 0\ ×\ 2\ & 2\ ×\ 2\ +\ 2 ×\ 2\ +\ 0\ ×\ 1\\
\end{pmatrix}\ \hspace{9cm}\]
\[ = \begin{pmatrix}
1\ +\ 0\ +\ 2\ & 2\ +\ 0\ +\ 2 & 2\ +\ 0 +\ 1\\
0\ +\ 2\ +\ 4\ & 0\ +\ 1\ +\ 4\ & 0\ +\ 2\ +\ 2\\
2\ +\ 4\ +\ 0\ & 4\ +\ 2\ +\ 0\ & 4\ +\ 4\ +\ 0\\
\end{pmatrix}\ \hspace{9cm}\]
\[B^T\ A^T\ = \begin{pmatrix}
3 & 4 & 3 \\
6 & 5 & 4 \\
6 & 6 & 8 \\
\end{pmatrix}\ ————– (2)\ \hspace{10cm}\]
\[From\ (1)\ and\ (2)\ (AB)^T\ =\ B^T\ A^T\ \hspace{10cm}\]
\[\color {purple} {Example\ 23:}\ If\ A =\begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{bmatrix}\ then\ \color{red}{show\ that\ A^2\ -\ 4\ A\ -\ 5I\ =0}\ \hspace{12cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[Let\ A =\begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{bmatrix}\ \hspace{10cm}\]
\[A^2 =\begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{bmatrix}\ \begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{bmatrix}\ \hspace{12cm}\]
\[ = \begin{bmatrix}
1\ ×\ 1\ +\ 2 ×\ 2\ +\ 2\ ×\ 2 & 1\ ×\ 2\ +\ 2 ×\ 1\ +\ 2\ ×\ 2 & 1\ ×\ 2\ +\ 2\ ×\ 2\ +\ 2\ ×\ 1\\
2\ ×\ 1\ +\ 1 ×\ 2\ +\ 2\ ×\ 2\ & 2\ ×\ 2\ +\ 1 ×\ 1\ +\ 2\ ×\ 2\ & 2\ ×\ 2\ +\ 1 ×\ 2\ +\ 2\ ×\ 1\\
2\ ×\ 1\ +\ 2 ×\ 2\ +\ 1\ ×\ 2\ & 2\ ×\ 2\ +\ 2 ×\ 1\ +\ 1\ ×\ 2\ & 2\ ×\ 2\ +\ 2 ×\ 2\ +\ 1\ ×\ 1\\
\end{bmatrix}\ \hspace{9cm}\]
\[ = \begin{bmatrix}
1\ +\ 4\ +\ 4\ & 2\ +\ 2\ +\ 4 & 2\ +\ 4 +\ 2\\
2\ +\ 2\ +\ 4\ & 4\ +\ 1\ +\ 4\ & 4\ +\ 2\ +\ 2\\
2\ +\ 4\ +\ 2\ & 4\ +\ 2\ +\ 2\ & 4\ +\ 4\ +\ 1\\
\end{bmatrix}\ \hspace{9cm}\]
\[A^2 = \begin{bmatrix}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9 \\
\end{bmatrix}\ \hspace{12cm}\]
\[4\ A\ =\ 4\ \begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{bmatrix}\ \hspace{12cm}\]
\[4\ A\ = \begin{bmatrix}
4 & 8 & 8 \\
8 & 4 & 8 \\
8 & 8 & 4 \\
\end{bmatrix}\ \hspace{10cm}\]
\[5\ I\ =\ 5\ \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\ \hspace{12cm}\]
\[5\ I\ = \begin{bmatrix}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5 \\
\end{bmatrix}\ \hspace{10cm}\]
\[A^2\ -\ 4\ A\ -\ 5I\ =\ \begin{bmatrix}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9 \\
\end{bmatrix}\ -\ \begin{bmatrix}
4 & 8 & 8 \\
8 & 4 & 8 \\
8 & 8 & 4 \\
\end{bmatrix}\ -\ \begin{bmatrix}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5 \\
\end{bmatrix}\]
\[ = \begin{bmatrix}
9\ -\ 4\ -\ 5\ & 8\ -\ 8\ -\ 0 & 8\ -\ 8 -\ 0\\
8\ -\ 8\ -\ 0\ & 9\ -\ 4\ -\ 5\ & 8\ -\ 8\ -\ 0\\
8\ -\ 8\ -\ 0\ & 8\ -\ 8\ -\ 0\ & 9\ -\ 4\ -\ 5\\
\end{bmatrix}\ \hspace{9cm}\]
\[\color{green}{\boxed{\therefore\ A^2\ -\ 4\ A\ -\ 5I\ =\ 0}}\]
\[\color {green} {Singular\ and\ Non-Singular\ Matrix}:\ \hspace{20cm}\]
A square matrix A is called a singular matrix
\[if\ \begin{vmatrix} A \\ \end{vmatrix}\ = 0\ and\ non\ –\ singular\ matrix\ if\ \begin{vmatrix} A \\ \end{vmatrix}\ \neq {0}\ \hspace{10cm}\]
\[\color {purple} {Example\ 24:}\ \color {red}{Prove\ that}\ \begin{pmatrix}
1 & 3 \\
2 & 6 \\
\end{pmatrix}\ \color {red} {is\ a\ singular\ matrix}\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue}{Solution\ :}\ \hspace{18cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 1(6)\ -\ 2(3) \hspace{13cm}\]
\[=\ 6\ -\ 6 \hspace{12cm}\]
\[= 0 \hspace{13cm}\]
\[A\ is\ a\ singular\ matrix\ \hspace{10cm}\]
\[\color {purple} {Example\ 25:}\ Prove\ that\ the\ matrix \begin{pmatrix}
1 & – 2 \\
– 2 & 4\\
\end{pmatrix}\ \color {red} {is\ a\ singular\ matrix}\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2024\ June\ 2025\ (Supp)\]
\[\color {blue}{Solution:}\ Given\ A =\begin{pmatrix}
1 & – 2 \\
– 2 & 4\\
\end{pmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 1(4) – (-2)(- 2) \hspace{13cm}\]
\[=\ 4\ -\ 4 \hspace{12cm}\]
\[= 0 \hspace{13cm}\]
\[A\ is\ a\ singular\ matrix\ \hspace{10cm}\]
\[\color {purple} {Example\ 26:}\ \color {red}{Prove\ that}\ A =\begin{bmatrix}
2 & 3 \\
4 & 5 \\
\end{bmatrix}\ \color {red} {is\ Non\ -\ singular}\ \hspace{15cm}\]
\[\color {blue}{Solution\ :}\ \hspace{18cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 2(5) – 3(4) \hspace{13cm}\]
\[= 10 – 12 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = -2\ \neq {0}\ \hspace{13cm}\]
\[A\ is\ a\ Non\ – \ Singular\ matrix\ \hspace{10cm}\]
\[\color {purple}{Example\ 27:}\ \color {red}{Prove\ that}\ \begin{bmatrix}
1 & -1 & 2 \\
2 & -2 & 4 \\
3 & -3 & 6 \\
\end{bmatrix}\ \color {red} {is\ a\ singular\ matrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ A\ =\begin{bmatrix}
1 & -1 & 2 \\
2 & -2 & 4 \\
3 & -3 & 6 \\
\end{bmatrix}\ \hspace{12cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =1\begin{vmatrix}
-2 & 4 \\
-3 & 6 \\
\end{vmatrix}\ – \ 1\begin{vmatrix}
2 & 2 \\
2 & -2 \\
\end{vmatrix}\ -\ 1 \begin{vmatrix}
2 & -1 \\
2 & 1 \\
\end{vmatrix}\ \hspace{4cm}\]
\[ =1(-12\ +\ 12)\ + 1 (12\ -\ 12) + 2(-6\ +\ 6)\
\hspace{10cm}\]
\[ =1(0)\ + 1 (0) + 2(0)\
\hspace{10cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = 0\ \hspace{13cm}\]
\[A\ is\ a\ Singular\ matrix\ \hspace{10cm}\]
\[\color {purple}{Example\ 28:}\ \color {red}{Prove\ that\ the\ matrix}\ \begin{pmatrix}
2 & 3 & – 1 \\
4 & 6 & 5 \\
6 & 2 & 1 \\
\end{pmatrix}\ \color {red} {is\ non\ -\ singular}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ A\ =\begin{pmatrix}
2 & 3 & -1 \\
4 & 6 & 5 \\
6 & 2 & 1 \\
\end{pmatrix}\ \hspace{12cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 2\begin{vmatrix}
6 & 5 \\
2 & 1 \\
\end{vmatrix}\ – \ 3\begin{vmatrix}
4 & 5 \\
6 & 1 \\
\end{vmatrix}\ -\ 1 \begin{vmatrix}
4 & 6 \\
6 & 2 \\
\end{vmatrix}\ \hspace{4cm}\]
\[ =\ 2(6\ -\ 10)\ -\ 3 (4\ -\ 30)\ -\ 1(8\ -\ 36)\
\hspace{10cm}\]
\[ =\ 2(- 4)\ -\ 3 (-26)\ -\ 1(-28)\
\hspace{10cm}\]
\[ =\ -\ 8\ +\ 78\ +\ 28\
\hspace{10cm}\]
\[\boxed{\begin{vmatrix} A \\ \end{vmatrix}\ = 98\ \neq {0}}\ \hspace{13cm}\]
\[A\ is\ a\ Non\ – \ Singular\ matrix\ \hspace{10cm}\]
\[\color {purple} {Minor\ of\ an\ element\ of\ a\ Matrix}\ \hspace{20cm}\]
\[Minor\ of\ an\ element\ is\ a\ determinant\ obtained\ by\ deleting\ the\ row\ and\ column\ in\ which\ the\ element\ occurs\]
\[\color{purple}{Example:\ 29}\ \color{red}{Find\ the\ Minor\ of\ a_1\ in\ the\ matrix}\ \hspace{15cm}\]
\[
A
\ = \begin{bmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[Minor\ of\ a_1 =\begin{vmatrix}
b_2 & c_2 \\
b_3 & c_3 \\
\end{vmatrix}\ = b_2c_3\ -\ b_3c_2\ \hspace{15cm}\]
\[\color {purple}{Example\ 30:}\ \color{red}{Find\ the\ Minor\ of\ 2\ in\ the\ following\ matrix}\ \hspace{15cm}\]
\[
A
\ = \begin{bmatrix}
1 & 0 & -1 \\
2 & 3 & 4 \\
7 & 8 & -2 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[\therefore\ Minor\ of\ 2 =\begin{vmatrix}
0 & -1 \\
8 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= 0 + 8\ \hspace{15cm}\]
\[\boxed{Minor\ of\ 2= 8}\ \hspace{15cm}\]
\[\color {purple} {Cofactor\ of\ an\ element of\ a\ Matrix}\ \hspace{20cm}\]
\[Cofactor\ of\ an\ element\ is\ a\ signed\ minor\ of\ that\ element\]
\[\therefore\ cofactor\ of\ a_{ij} = (-1)^{i\ +\ j}\ minor\ of\ a_{ij}\]
\[\color {purple}{Example\ 31 :}\ \color {red}{Find\ the\ cofactor\ of\ ‘2’\ in}\ \begin{pmatrix}
3 & 0 & 2 \\
5 & 1 & 7 \\
4 & 5 & -3 \\
\end{pmatrix}\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[cofactor\ of\ 2\ =\ (-1)^{1\ +\ 3}\ \begin{vmatrix}
5 & 1 \\
4 & 5 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (25\ -\ 4)\ \hspace{15cm}\]
\[=\ 1(21)\ \hspace{15cm}\]
\[\color{green}{\boxed{cofactor\ of\ 2\ =\ 21}}\ \hspace{15cm}\]
\[\color {purple}{Example\ 32 :}\ \color {red}{Find\ the\ cofactor\ of\ ‘2’\ in}\ \begin{pmatrix}
1 & 0 & – 1 \\
2 & 3 & 4 \\
7 & 8 & – 2 \\
\end{pmatrix}\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2024\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[cofactor\ of\ 2\ =\ (-1)^{2\ +\ 1}\ \begin{vmatrix}
0 & – 1 \\
8 & – 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (0\ -\ (- 8)\ \hspace{15cm}\]
\[=\ – 1(8)\ \hspace{15cm}\]
\[\color{green}{\boxed{cofactor\ of\ 2\ =\ – 8}}\ \hspace{15cm}\]
\[\color {purple}{Example\ 33 :}\ \color {red}{Find\ the\ cofactor\ of\ ‘2’\ in}\ \begin{pmatrix}
1 & – 1 & 1 \\
2 & 3 & – 3 \\
6 & – 2 & – 1 \\
\end{pmatrix}\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[cofactor\ of\ 2\ =\ (-1)^{2\ +\ 1}\ \begin{vmatrix}
– 1 & 1 \\
– 2 & – 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (1 -\ (- 2)\ \hspace{15cm}\]
\[=\ – 1(3)\ \hspace{15cm}\]
\[\color{green}{\boxed{cofactor\ of\ 2\ =\ – 3}}\ \hspace{15cm}\]
\[\color {purple}{Example\ 34 :}\ \color {red}{Find\ the\ cofactor\ of\ -\ 2\ in}\ \hspace{20cm}\]
\[\hspace{5cm} \begin{vmatrix}
1 & – 1 & 1 \\
2 & 3 & -\ 3 \\
6 & -\ 2 & -\ 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[cofactor\ of\ -\ 2\ =\ (-1)^{3\ +\ 2}\ \begin{vmatrix}
1 & 1 \\
2 & -\ 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-\ 3\ -\ 2)\ \hspace{15cm}\]
\[= (-1) (-\ 5)\ \hspace{15cm}\]
\[\boxed{cofactor\ of\ -\ 2 =\ 5}\ \hspace{15cm}\]
\[\color {purple} {Example\ 35:} \color {red}{Find\ the\ cofactor\ matrix\ of}\ \begin{bmatrix}
2 & 3 & 4 \\
1 & 2 & 3 \\
-1 & 1 & 2 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ A\ =\begin{bmatrix}
2 & 3 & 4 \\
1 & 2 & 3 \\
-1 & 1 & 2 \\
\end{bmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ \begin{vmatrix}
2 & 3 \\
1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (4 -3)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{1\ +\ 2}\ \begin{vmatrix}
1 & 3 \\
-1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (2 + 3)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -5\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{1\ +\ 3}\ \begin{vmatrix}
1 & 2 \\
-1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (1 + 2)\ \hspace{15cm}\]
\[= (1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = 3\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{2\ +\ 1}\ \begin{vmatrix}
3 & 4 \\
1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (6 – 4)\ \hspace{15cm}\]
\[= (-1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -2\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 2}\ \begin{vmatrix}
2 & 4 \\
-1 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (4 + 4)\ \hspace{15cm}\]
\[= (1) (8)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 8\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{2\ +\ 3}\ \begin{vmatrix}
2 & 3 \\
-1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (2 + 3)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -5\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{3\ +\ 1}\ \begin{vmatrix}
3 & 4 \\
2 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (9 – 8)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{3\ +\ 2}\ \begin{vmatrix}
2 & 4 \\
1 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (6 – 4)\ \hspace{15cm}\]
\[= (-1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -2\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{3\ +\ 3}\ \begin{vmatrix}
2 & 3 \\
1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (4 – 3)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 1\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
1 & -5 & 3 \\
-2 & 8 & -5 \\
1 & -2 & 1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {purple} {Adjoint\ of\ Matrix}\ \hspace{20cm}\]
\[The\ Adjoint\ of\ a\ square\ matrix\ A\ is\ the\ transpose\ of\ the\ matrix\ which\ is\ formed\ by\]\[the\ elements\ which\ are\ the\ cofactors\ of\ the\ corresponding\ elements\ of\ the\ determinant\ of\ the\ matrix\ A\]
\[\color {purple}{Method\ for\ to\ find\ adjoint\ of\ Matrix\ of\ order\ 3\ (order 2)}\ \hspace{20cm}\]
\[i)\ A\ is\ square\ Matrix\ of\ order\ 3\ (order 2)\ \hspace{5cm}\]
\[ii)\ Find\ the\ co-factor\ of\ all\ the\ elements\ of\ det\ A\ \hspace{8cm}\]
\[iii)\ Form\ the\ matrix\ by\ replacing\ all\ the\ elements\ of\ A\ by\ the\ corresponding\ cofactor\ in\ \begin{vmatrix}
A \\
\end{vmatrix}\]
\[iv)\ Then\ take\ the\ Transpose\ of\ that\ matrix,\ then\ we\ get\ adj. A.\ \hspace{8cm}\]
\[\color {purple} {Example\ 36:}\ \color{red}{Find\ the\ Adjoint\ of}\ \begin{pmatrix}
1 & 5 \\
3 & 6 \\
\end{pmatrix}\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2024\ June\ 2025\ (Supp)\]
\[\color {blue}{Solution:}\ A\ =\begin{pmatrix}
1 & 5 \\
3 & 6 \\
\end{pmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ (6)\ \hspace{15cm}\]
\[= (1) (6)\ \hspace{15cm}\]
\[cofactor\ of\ 1\ = \ 6\ \hspace{15cm}\]
\[cofactor\ of\ 5 = (-1)^{1\ +\ 2}\ (3)\ \hspace{15cm}\]
\[= (-1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ -1 =\ -\ 3\ \hspace{15cm}\]
\[cofactor\ of\ 3\ = (-1)^{2\ +\ 1}\ (5)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3\ =\ -\ 5\ \hspace{15cm}\]
\[cofactor\ of\ 6\ = (-1)^{2\ +\ 2}\ (1)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 6\ = 1\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{pmatrix}
6 & -\ 3 \\
-\ 5 & 1 \\
\end{pmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{pmatrix}
6 & -\ 5 \\
-\ 3 & 1 \\
\end{pmatrix}\ \hspace{15cm}\]
\[\color {purple} {Example\ 37:}\ \color{red}{Find\ the\ Adjoint\ Matrix\ of}\ \begin{bmatrix}
5 & -6 \\
3 & 2 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix}
5 & -6 \\
3 & 2 \\
\end{bmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 5 = (-1)^{1\ +\ 1}\ (2)\ \hspace{15cm}\]
\[= (1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 5 = \ 2\ \hspace{15cm}\]
\[cofactor\ of\ -6 = (-1)^{1\ +\ 2}\ (3)\ \hspace{15cm}\]
\[= (-1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ -6 = -3\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{2\ +\ 1}\ (-6)\ \hspace{15cm}\]
\[= (-1) (-6)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = 6\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 2}\ (5)\ \hspace{15cm}\]
\[= (1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 5\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{bmatrix}
2 & -3 \\
6 & 5 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{bmatrix}
2 & 6 \\
-3 & 5 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {purple} {Example\ 38:}\ \color{red}{Find\ the\ Adjoint\ of}\ A\ = \begin{bmatrix}
2 & 3 & 4 \\
1 & 2 & 3 \\
-1 & 1 & 2 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ A\ =\begin{bmatrix}
2 & 3 & 4 \\
1 & 2 & 3 \\
-1 & 1 & 2 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
1 & -5 & 3 \\
-2 & 8 & -5 \\
1 & -2 & 1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\ (Cofactor Matrix)^T\ =\ \begin{bmatrix}
1 & -2 & 1 \\
-5 & 8 & -2 \\
3 & -5 & 1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {purple} {Inverse\ of\ Matrix}\ \hspace{20cm}\]
\[Let\ A\ be\ a\ non\ singular\ matrix\ if\ there\ exists\ a\ square\ matrix\ B\, such\ that\ AB\ =\ BA\ = I\]\[Where\ I\ is\ the\ unit\ matrix\ of\ same\ order\ then\ B\ is\ called\ the\ the\ Inverse\ of\ A\ and\ it\ is\ denoted\ by\ A^{-1}.\]
\[\color {black}{Note:}\ \hspace{12cm}\]
\[i)\ inverse\ of\ matrix\ is\ unique\ \hspace{6cm}\]
\[ii)\ AB=\ BA\ =\ I\ \hspace{8cm}\]
\[iii)\ (AB)^{-1} =\ B^{-1}A^{-1}\ \hspace{8cm}\]
\[\color {black}{Formula\ for\ Inverse\ of\ Matrix:}\ \hspace{12cm}\]
\[\color {green} {\boxed {A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A}}\]
\[\color {purple} {Example\ 39:}\ \color{red}{Find\ the\ inverse\ of}\ \begin{bmatrix}
1 & -1 \\
-2 & 0 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix}
1 & -1 \\
-2 & 0 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 1(0)\ -\ (-2) \hspace{13cm}\]
\[= 0 – 2 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = – 2\ \neq {0}\ \hspace{13cm}\]
\[\therefore\ A^{-1}\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ (0)\ \hspace{15cm}\]
\[= (-1)^2 (0)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 0\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ (-2)\ \hspace{15cm}\]
\[= (-1)^3 (-2)\ \hspace{15cm}\]
\[cofactor\ of\ – 1 = 2\ \hspace{15cm}\]
\[cofactor\ of\ -2 = (-1)^{2\ +\ 1}\ (-1)\ \hspace{15cm}\]
\[= (-1)^3 (-1)\ \hspace{15cm}\]
\[cofactor\ of\ – 2 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 0 = (-1)^{2\ +\ 2}\ (1)\ \hspace{15cm}\]
\[= (-1)^4 (1)\ \hspace{15cm}\]
\[cofactor\ of\ 0 = 1\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{bmatrix}
0 & 2 \\
1 & 1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{bmatrix}
0 & 1 \\
2 & 1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{-2}\ \begin{bmatrix}
0 & 1 \\
2 & 1 \\
\end{bmatrix}\ \hspace{2cm}\]
\[\color {purple} {Example\ 40:}\ \color{red}{Find\ the\ inverse\ matrix\ of}\ \begin{pmatrix}
5 & 2 \\
-4 & 3 \\
\end{pmatrix}\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2023\ April\ 2025\]
\[\color {blue}{Solution:}\ Let\ A\ =\begin{pmatrix}
5 & 2 \\
-4 & 3 \\
\end{pmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 5(3)\ -\ 2(-4) \hspace{13cm}\]
\[=\ 15\ +\ 8 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ =\ 23\ \neq {0}\ \hspace{13cm}\]
\[\therefore\ A^{-1}\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 5 = (-1)^{1\ +\ 1}\ (3)\ \hspace{15cm}\]
\[= (-1)^2 (3)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 3\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 2}\ (-4)\ \hspace{15cm}\]
\[= (-1)^3 (-4)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 4\ \hspace{15cm}\]
\[cofactor\ of\ -4 = (-1)^{2\ +\ 1}\ (2)\ \hspace{15cm}\]
\[= (-1)^3 (2)\ \hspace{15cm}\]
\[cofactor\ of\ – 4 =\ -\ 2\ \hspace{15cm}\]
\[cofactor\ of\ 3\ = (-1)^{2\ +\ 2}\ (5)\ \hspace{15cm}\]
\[= (-1)^4 (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3 =\ 5\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{pmatrix}
3 & 4 \\
-2 & 5 \\
\end{pmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{pmatrix}
3 & -2 \\
4 & 5 \\
\end{pmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{23}\ \begin{pmatrix}
3 & -2 \\
4 & 5 \\
\end{pmatrix}\ \hspace{2cm}\]
\[\color {purple} {Example\ 41:}\ \color{red}{Find\ the\ inverse\ of}\ \begin{bmatrix}
1 & -1 & 1 \\
2 & -3 & -3 \\
6 & -2 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2023,\ April\ 2024\ June\ 2025(Supp)\]
\[\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix}
1 & -1 & 1 \\
2 & -3 & -3 \\
6 & -2 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =1\begin{vmatrix}
-3 & -3 \\
-2 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -3 \\
6 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -3\\
6 & -2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[ =1(3\ -\ 6)\ + 1 (-2\ +\ 18) + 1(-4\ +\ 18)\
\hspace{9cm}\]
\[ =1(-3)\ + 1 (16) + 1(14)\
\hspace{13cm}\]
\[ = -3\ + 16 + 14\
\hspace{14cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 27\ \neq\ 0\
\hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ \begin{vmatrix}
-3 & -3 \\
-2 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (3 – 6)\ \hspace{15cm}\]
\[= (1) (-3)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ \begin{vmatrix}
2 & -3 \\
6 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (-2 + 18)\ \hspace{15cm}\]
\[= (-1) (16)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -16\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 3}\ \begin{vmatrix}
2 & -3 \\
6 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-4 + 18)\ \hspace{15cm}\]
\[= (1) (14)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 14\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 1}\ \begin{vmatrix}
-1 & 1 \\
-2 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (1+ 2)\ \hspace{15cm}\]
\[= (-1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -3 = (-1)^{2\ +\ 2}\ \begin{vmatrix}
1 & 1 \\
6 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-1- 6)\ \hspace{15cm}\]
\[= (1) (-7)\ \hspace{15cm}\]
\[cofactor\ of\ -3 = -7\ \hspace{15cm}\]
\[cofactor\ of\ -3 = (-1)^{2\ +\ 3}\ \begin{vmatrix}
1 & -1 \\
6 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-2+ 6)\ \hspace{15cm}\]
\[= (-1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ -3 = -4\ \hspace{15cm}\]
\[cofactor\ of\ 6 = (-1)^{3\ +\ 1}\ \begin{vmatrix}
-1 & 1 \\
-3 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (3 + 3)\ \hspace{15cm}\]
\[= (1) (6)\ \hspace{15cm}\]
\[cofactor\ of\ 6 = 6\ \hspace{15cm}\]
\[cofactor\ of\ -2 = (-1)^{3\ +\ 2}\ \begin{vmatrix}
1 & 1 \\
2 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-3- 2)\ \hspace{15cm}\]
\[= (-1) (-5)\ \hspace{15cm}\]
\[cofactor\ of\ -2 = 5\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{3\ +\ 3}\ \begin{vmatrix}
1 & -1 \\
2 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (-3+ 2)\ \hspace{15cm}\]
\[= (1) (-1)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -1\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
-3 & -16 & 14 \\
-3 & -7 & -4 \\
6 & 5 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix}
-3 & -3 & 6 \\
-16 & -7 & 5 \\
14 & -4 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{27}\ \begin{bmatrix}
-3 & -3 & 6 \\
-16 & -7 & 5 \\
14 & -4 & -1 \\
\end{bmatrix}\ \hspace{2cm}\]
\[\color {purple} {Example\ 42:}\ \color{red}{Find\ the\ inverse\ of\ the\ matrix}\ \begin{bmatrix}
1 & -1 & 2 \\
4 & 0 & 6 \\
0 & 1 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix}
1 & -1 & 2 \\
4 & 0 & 6 \\
0 & 1 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 1\begin{vmatrix}
0 & 6 \\
1 & – 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
4 & 6 \\
0 & -1 \\
\end{vmatrix}\ +\ 2\begin{vmatrix}
4 & 0\\
0 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[ =\ 1(0\ -\ 6)\ +\ 1 (-4\ -\ 0)\ +\ 2(4\ -\ 0)\
\hspace{9cm}\]
\[ =\ 1(-6)\ +\ 1 (-\ 4)\ +\ 2(4)\
\hspace{13cm}\]
\[ =\ -\ 6\ -\ 4\ +\ 8\
\hspace{14cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ -\ 2\ \neq\ 0\
\hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1\ = (-1)^{1\ +\ 1}\ \begin{vmatrix}
4 & 6 \\
0 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 1\ = (-1)^2\ (-\ 4\ -\ 0)\ \hspace{15cm}\]
\[= (1) (-\ 4)\ \hspace{15cm}\]
\[cofactor\ of\ 1\ =\ -\ 4\ \hspace{15cm}\]
\[cofactor\ of\ -\ 1\ = (-1)^{1\ +\ 2}\ \begin{vmatrix}
4 & 6 \\
0 & -\ 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (-\ 4\ -\ 0)\ \hspace{15cm}\]
\[= (-1) (-\ 4)\ \hspace{15cm}\]
\[cofactor\ of\ -\ 1\ =\ 4\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 3}\ \begin{vmatrix}
4 & 0 \\
0 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (4\ -\ 0)\ \hspace{15cm}\]
\[= (1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ 2 =\ 4\ \hspace{15cm}\]
\[cofactor\ of\ 4\ =\ (-1)^{2\ +\ 1}\ \begin{vmatrix}
-\ 1 & 2 \\
1 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (1\ -\ 2)\ \hspace{15cm}\]
\[= (-1) (-\ 1)\ \hspace{15cm}\]
\[cofactor\ of\ 4\ =\ 1\ \hspace{15cm}\]
\[cofactor\ of\ 0\ =\ (-1)^{2\ +\ 2}\ \begin{vmatrix}
1 & 2 \\
0 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-\ 1\ -\ 0)\ \hspace{15cm}\]
\[= (1) (-\ 1)\ \hspace{13cm}\]
\[cofactor\ of\ 0\ =\ -\ 1\ \hspace{15cm}\]
\[cofactor\ of\ 6\ = (-1)^{2\ +\ 3}\ \begin{vmatrix}
1 & – 1 \\
0 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (1 +\ 0)\ \hspace{15cm}\]
\[= (-1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 6\ =\ -\ 1\ \hspace{15cm}\]
\[cofactor\ of\ 0\ =\ (-1)^{3\ +\ 1}\ \begin{vmatrix}
-\ 1 & 2 \\
0 & 6 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-\ 6\ -\ 0)\ \hspace{15cm}\]
\[= (1) (-\ 6)\ \hspace{15cm}\]
\[cofactor\ of\ 0\ =\ -\ 6\ \hspace{15cm}\]
\[cofactor\ of\ 1\ =\ (-1)^{3\ +\ 2}\ \begin{vmatrix}
1 & 4 \\
2 & 6 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (6\ -\ 8)\ \hspace{15cm}\]
\[= (-1) (-2)\ \hspace{15cm}\]
\[cofactor\ of\ 1\ =\ 2\ \hspace{15cm}\]
\[cofactor\ of\ -\ 1\ =\ (-1)^{3\ +\ 3}\ \begin{vmatrix}
1 & -\ 1 \\
4 & 0 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (0\ +\ 4)\ \hspace{15cm}\]
\[= (1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ -\ 1\ =\ 4\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
-4 & 4 & 4 \\
1 & – 1 & – 1 \\
– 6 & 2 & 4 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix}
-4 &1 & – 6 \\
4 & – 1 & 2 \\
4 & – 1 & 4 \\
\end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{-2}\ \begin{bmatrix}
-4 &1 & – 6 \\
4 & – 1 & 2 \\
4 & – 1 & 4 \\
\end{bmatrix}\ \hspace{2cm}\]
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