Class 12 Examples Applications of Matrices and Determinants

Dear students, here are the Class 12 Business Maths Examples in Chapter 1 Applications of Matrices and Determinants for your reference.

$Example \;1.1: \;Find \;the \;rank \;of \;the \;matrix \;\large \begin{pmatrix} 1 &5 \\ 3& 9 \end{pmatrix}\\[.5 cm]$

Ans:

$Let \;A = \large \begin{pmatrix} 1 &5 \\ 3& 9 \end{pmatrix}\\[.5 cm]$

Order of A is 2 × 2

$\therefore \rho(A)\leq 2\\[.5 cm]$

Consider the second order minor

$\large \begin{vmatrix} 1 & 5\\ 3 & 9 \end{vmatrix}=6\neq 0 \\[.5 cm]$

There is a minor of order 2, which is not zero. $\therefore \rho(A)= 2\\[.5 cm]$

$Example \; 1.2: \; Find \; the \; rank\; of\; the \;matrix \; \large \begin{pmatrix} -5 &-7 \\ 5& 7 \end{pmatrix}\\[.5 cm]$

$Let \; A =\large \begin{pmatrix} -5 &-7 \\ 5& 7 \end{pmatrix}\\[.5 cm]$

Order of A is 2 × 2

$\therefore \rho(A)\leq 2\\[.5 cm]$

Consider the second order minor

$\large \begin{vmatrix} -5 & -7\\ 5 & 7 \end{vmatrix}=0 \\[.5 cm]$

Since the second order minor vanishes, $\rho(A)\leq 2\\[.5 cm]$

Consider a first order minor $\large \left | -5 \right |\neq 0 \\[.5 cm]$

There is a minor of order 1, which is not zero. $\therefore \rho(A)= 1\\[.5 cm]$

$Example\; 1.3: \;Find\; the \;rank \;of\; the \;matrix \;\large \begin{pmatrix} 0 &-1 & 5\\ 2 & 4 &-6 \\ 1& 1 & 5 \end{pmatrix} \\[.5 cm]$

$Let \; A=\large \begin{pmatrix} 0 &-1 & 5\\ 2 & 4 &-6 \\ 1& 1 & 5 \end{pmatrix} \\[.5 cm]$

Order of A is 3 × 3

$\therefore \rho(A)\leq 3\\[.5 cm]$

Consider the third order minor

$\large \begin{vmatrix} 0 & -1 & 5\\ 2 & 4 &-6 \\ 1& 1 & 5 \end{vmatrix}=6\neq 0 \\[.5 cm]$

There is a minor of order 3, which is not zero. $\therefore \rho(A)= 3\\[.5 cm]$

$Example \; 1.4:\; Find \;the \;rank \;of \;the \;matrix \; \large \begin{pmatrix} 5 &3 & 0\\ 1 & 2 &-4 \\ -2& -4 & 8 \end{pmatrix} \\[.5 cm]$

$Let \; A =\large \begin{pmatrix} 5 &3 & 0\\ 1 & 2 &-4 \\ -2& -4 & 8 \end{pmatrix} \\[.5 cm]$

Order of A is 3 × 3

$\therefore \rho(A)\leq 3\\[.5 cm]$

Consider the third order minor

$\large \begin{vmatrix} 5 & 3 & 0\\ 1 & 2 &-4 \\ -2& -4 & 8 \end{vmatrix}=0 \\[.5 cm]$

Since the third order minor vanishes, $\rho(A)\leq 3\\[.5 cm]$

Consider a second order minor $\large \begin{vmatrix} 5 & 3\\ 1 & 2 \end{vmatrix}=7\neq 0 \\[.5 cm]$

There is a minor of order 2, which is not zero. $\therefore \rho(A)= 2\\[.5 cm]$

$Example \; 1.5: \;Find \;the \;rank \;of \;the \;matrix\; \begin{pmatrix} 1 & 2 & -1 & 3\\ 2 & 4 & 1&-2\\ 3 & 6 & 3 &-7 \end{pmatrix} \\[.5 cm]$

$Let \; A = \begin{pmatrix} 1 & 2 & -1 & 3\\ 2 & 4 & 1& -2\\ 3 & 6 & 3 &-7 \end{pmatrix} \\[.5 cm]$

Order of A is 3 × 4

$\therefore \rho(A)\leq 3\\[.5 cm]$

Consider the third order minors

$\begin{vmatrix} 1 & 2 & -1 \\ 2 & 4 & 1\\ 3 & 6 & 3 \end{vmatrix}=0 \\[.5 cm]$

$\begin{vmatrix} 1 & -1 & 3 \\ 2 & 1 & -2\\ 3 & 3 & -7 \end{vmatrix}=0 \\[.5 cm]$

$\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & -2\\ 3 & 6 & -7 \end{vmatrix}=0 \\[.5 cm]$

$\begin{vmatrix} 2 & -1 & 3 \\ 4 & 1 & -2\\ 6 & 3 & -7 \end{vmatrix}=0 \\[.5 cm]$

Since all third order minors vanish, $\rho(A)\leq 3\\[.5 cm]$

Now, let us consider the second order minors,

Consider a second order minor $\large \begin{vmatrix} 2 & -1\\ 4 & 1 \end{vmatrix}=6\neq 0 \\[.5 cm]$

There is a minor of order 2, which is not zero. $\therefore \rho(A)= 2\\[.5 cm]$

$Example\; 1.6:\; Find\; the \;rank \;of \;the \;matrix \;\begin{bmatrix} 1 & 2 & 3\\ 2 & 3 & 4\\ 3 & 5 & 7 \end{bmatrix} \\[.5 cm]$

$Let \;A =\begin{bmatrix} 1 & 2 & 3\\ 2 & 3 & 4\\ 3 & 5 & 7 \end{bmatrix} \\[.5 cm]$

Order of $\therefore \rho(A)\leq 3\\[.5 cm]$

The above matrix is in echelon form. The number of non zero rows is 2.

$\therefore \rho(A)=2\\[.5 cm]$

$Example\; 1.7:\; Find \;the \;rank \;of \;the \;matrix \; \begin{bmatrix} 0 & 1 & 2 & 1 \\ 1 & 2 & 3 & 2 \\ 3 & 1 & 1 & 3 \end{bmatrix} \\[.5 cm]$

$Let \;A = \begin{bmatrix} 0 & 1 & 2 & 1 \\ 1 & 2 & 3 & 2 \\ 3 & 1 & 1 & 3 \end{bmatrix} \\[.5 cm]$

Order of matrix is 3 x 4. $\therefore \rho(A)\leq 3\\[.5 cm]$

The above matrix is in echelon form. The number of non zero rows is 3.

$\therefore \rho(A)=3\\[.5 cm]$

Example 1.8: Find the rank of the matrix $\begin{bmatrix} 1 & 1 & 1 & 1 \\ 3 & 4 & 5 & 2 \\ 2 & 3 & 4 & 0 \end{bmatrix} \\[.5 cm]$

Let A = $\begin{bmatrix} 1 & 1 & 1 & 1 \\ 3 & 4 & 5 & 2 \\ 2 & 3 & 4 & 0 \end{bmatrix} \\[.5 cm]$

Order of matrix is 3 x 4. $\therefore \rho(A)\leq 3\\[.5 cm]$

\sim \begin{bmatrix}1 & 1 & 1 & 1 \\0 & 1 & 2 & -1 \\0 & 1 & 2 & -2\end{bmatrix}\\[.5 cm]

The above matrix is in echelon form. The number of non zero rows is 3.

$\therefore \rho(A)=3\\[.5 cm]$

Example 1.9: Show that the equations x + y = 5, 2x + y = 8 are consistent and solve them.

$\begin{pmatrix} 1&1 \\ 2&1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} 5 \\ 8 \end{pmatrix} \\[.5 cm]$

A X =B

\begin{bmatrix}1 & 1 & 5 \\2 & 1 & 8\end{bmatrix}\\[.5 cm]

Number of non-zero rows is 2.

$\rho(A)=\rho([A,B]) =2 \\[.5 cm]$

The given system is consistent and has a unique solution.

$\begin{pmatrix} 1&1 \\ 0&-1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} 5 \\ -2 \end{pmatrix} \\[.5 cm]$

x + y =5

y =2

x + 2= 5

x =3

Solution is x = 3, y = 2

Example 1.10 Show that the equations 2x + y = 5, 4x + 2y = 10 are consistent and solve them.

$\begin{pmatrix} 2&1 \\ 4&2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} 5\\ 10 \end{pmatrix} \\[.5 cm]$

A X =B

\begin{pmatrix}2&1 \4&2\end{pmatrix}\\[.5 cm]

\begin{bmatrix}2 & 1 & 5 \\4 & 2 & 10\end{bmatrix}\\[.5 cm]

$\rho(A)=\rho([A,B]) =1 \\[.5 cm]$

Therefore, the given system is consistent and has infinitely many solutions.

$\begin{pmatrix} 2&1 \\ 0&0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} 5\\ 10 \end{pmatrix} \\[.5 cm]$

$=2x+y=5$

$Let\; us \;take\; y = k, \; k\in\; R$

$2x+k=5$

$x = \dfrac{5-k}{2} \\[.5 cm]$

$x = \dfrac{5-k}{2}; y=k for \: all\: k\in \;R \\[.5 cm]$

By giving different values for k, we get different solutions. Hence the system has infinite number of solutions.

Example 1.11 Show that the equations 3x − 2y = 6, 6x − 4y = 10 are inconsistent.

$\begin{pmatrix} 3&-2 \\ 6&-4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} 6\\ 10 \end{pmatrix} \\[.5 cm]$

A X =B

\begin{pmatrix}3&-2 \\6&-4\end{pmatrix}\\[.5 cm]

\begin{bmatrix}3 & -2 & 6 \\6 & -4 & 10\end{bmatrix}\\[.5 cm]

$\rho(A)=1 \: \rho([A,B]) =2 \\[.5 cm]$

$\rho(A)=1 \: \rho([A,B]) =2$
$\therefore \rho(A)\neq \rho([A,B])$

Therefore, the given system is inconsistent and has no solution.

Example 1.12 Show that the equations 2x + y + z = 5, x + y + z = 4, x − y + 2z = 1 are consistent and hence solve them.

$\begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \begin{bmatrix} 5 \\ 4 \\ 1 \end{bmatrix} \\[.5cm]$

A X=B

\begin{bmatrix}2 & 1 & 1 & 5 \\1 & 1 & 1 & 4 \\1 & -1 & 2 & 1\end{bmatrix}\\[.25 cm]

$\therefore$ the given system is consistent and has a unique solution.

$\begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & -1 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \begin{bmatrix} 4 \\ -3 \\ 3 \end{bmatrix} \\[.5cm]$

$x+y+z=4\cdots (1)\\[.25 cm] y+z=3\cdots (2)\\[.25 cm] 3z=3\cdots (3)\\[.25 cm] z=1\\[.25 cm] y=3-z =3-1 =2\\[.25 cm] y=2\\[.25 cm] x=4-y-z\\[.25 cm] =4-2-1\\[.25 cm] x=1\\[.25 cm] \therefore x=1 \: y=2 \: z=1\\[.25 cm] \\[.5cm]$

Example 1.13 Show that the equations x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30 are consistent and solve them.

$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \begin{bmatrix} 6 \\ 14 \\ 30 \end{bmatrix} \\[.5cm]$

\begin{bmatrix}1 & 1 & 1 & 6 \\1 & 2 & 3 & 14 \\1 & 4 & 7 & 30\end{bmatrix}\\[.25 cm]

\sim \begin{bmatrix}1 & 1 & 1 & 6 \\0 & 1 & 2 & 8 \\0 & 2 & 4 & 16\end{bmatrix}\\[.25 cm]

The given system is consistent and has infinitely many solutions.

$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \begin{bmatrix} 6 \\ 8 \\ 0 \end{bmatrix} \\[.5cm]$

$x+y+z=6\cdots (1)\\[.25 cm] y+2z=8\cdots (2)\\[.25 cm] y=8-2z \\[.25 cm] x=6-y-z\\[.25 cm] =6-(8-2z)-z =6-8-2z-z\\[.25 cm] =z-2\\[.25 cm] \\[.5cm]$

$Take\: z = k,\: k \in R\\[.25 cm] x=k-2\\[.25 cm] y=8-2k \\[.5cm]$

Therefore, by giving different values for k we get different solutions. Hence the given system has infinitely many solutions.

Example 1.14 Show that the equations are inconsistent: x − 4y + 7z = 14, 3x + 8y − 2z = 13, 7x − 8y + 26z = 5

$\begin{bmatrix} 1 & -4 & 7 \\ 3 & 8 & -2 \\ 7 & -8 & 26 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \begin{bmatrix} 14 \\ 13 \\ 5 \end{bmatrix} \\[.5cm]$

A X = B

\begin{bmatrix}1 & -4 & 7 & 14 \\3 & 8 & -2 & 13 \\7 & -8 & 26 & 5\end{bmatrix}\\[.25 cm]

R_{2}\rightarrow R_{2}-3R_{1}\\R_{3}\rightarrow R_{3}-7R_{1}

$\rho (A)\leq \rho ([A,B]) \\[.5cm]$

$\rho (A)\neq \rho ([A,B]) \\[.5cm]$

Therefore, the system is inconsistent and has no solution.

Example 1.15 Find k, if the equations x + 2y − 3z = −2, 3x − y − 2z = 1, 2x + 3y − 5z = k are consistent.

$\begin{bmatrix} 1 & 2 & -3 \\ 3 & -1 & -2 \\ 2 & 3 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \begin{bmatrix} -2\\ 1 \\ k \end{bmatrix} \\[.5cm]$

A X= B

\begin{bmatrix}1 & 2 & -3 & -2 \\3 & -1 & -2 & 1 \\2 & 3 & -5 & k\end{bmatrix}\\[.25 cm]

\sim \begin{bmatrix}1 & 2 & -3 & -2 \\0 & -7 & 7 & 7 \\0 & -1 & 1 & 4+k\end{bmatrix}\\[.25 cm]

For the equations to be consistent,

$\rho (A)=\rho ([A,B]) = 2 \\$
$21+7k =0 \\$
$21+7k =0 \\$
$7k=-21 \\$
$k=-3$

Example 1.16 Find k, if the equations x + y + z = 7, x + 2y + 3z = 18, y + kz = 6 are inconsistent.

$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & k \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \begin{bmatrix} 7\\ 18 \\ 6 \end{bmatrix} \\[.5cm]$

A X =B

\begin{bmatrix}1 & 1 & 1 & 7 \\1 & 2 & 3 & 18 \\0 & 1 & k & 6\end{bmatrix}\\[.25 cm]

\sim \begin{bmatrix}1 & 1 & 1 & 7 \\0 & 1 & 2 & 11 \\0 & 1 & k & 6\end{bmatrix}\\[.25 cm]

For the equations to be inconsistent $\rho (A)\neq \rho ([A,B])\\[.25cm]$

It is possible if $k-2=0 \\[.25cm]$

$\therefore k=2\\[.25cm]$

Example 1.17 Investigate for what values of ‘a’ and ‘b’ the following system of equations
x + y + z = 6, x + 2y + 3z = 10, x + 2y + az = b have (i) no solution (ii) a unique solution (iii) an infinite number of solutions.

$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & k \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \begin{bmatrix} 6\\ 10 \\ b \end{bmatrix} \\[.5cm]$

A X =B

\begin{bmatrix}1 & 1 & 1 & 6 \\1 & 2 & 3 & 10 \\1 & 2 & a & b\end{bmatrix}\\[.25 cm]

\sim \begin{bmatrix}1 & 1 & 1 & 6 \\0 & 1 & 2 & 4 \\0 & 1 & a-1 & b-6\end{bmatrix}\\[.25 cm]

Case (i) For no solution:

The system possesses no solution only when $\rho (A)\neq \rho ([A,B])$ which is possible only when $a - 3 = 0 \:and \: b -10 \neq 0.$ Hence for $a = 3, b \neq 10 ,$ the system possesses no solution.

Case (ii) For a unique solution:

The system possesses a unique solution only when $\rho (A)= \rho ([A,B])=3.$ When $a-3\neq 0$ and b is any real number (b \in R), the system possesses a unique solution.

Case (iii) For an infinite number of solutions:

The system possesses an infinite number of solutions only when $\rho (A)= \rho ([A,B])=2$ . It is possible only when $a − 3 = 0, b −10 = 0$ Hence for Hence for a = 3, b =10, the system possesses infinite number of solutions.

Example 1.18 The total number of units produced (P) is a linear function of amount of over times in labour (in hours) (l), amount of additional machine time (m) and fixed finishing time (a) i.e, P = a + bl + cm. From the data given below, find the values of constants a, b and c.

Day	Production (in Units P)	Labour (in Hrs l)	Additional Machine Time (in Hrs m)
Monday	6,950	40	10
Tuesday	6,725	35	9
Wednesday	7,100	40	12

Estimate the production when overtime in labour is 50 hrs and additional machine time is 15 hrs.

$P = a + bl + cm \\[.25 cm] 6,950 = a + 40b + 10c\\[.25 cm] 6,725 = a + 35b + 9c\\[.25 cm] 7,100 = a + 40b + 12c \\[.25 cm]$

$\begin{bmatrix} 1 & 40 & 10 \\ 1 & 35 & 9 \\ 1 & 40 & 12 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} \begin{bmatrix} 6950\\ 6725 \\ 7100 \end{bmatrix} \\[.5cm]$

\begin{bmatrix}1 & 40 & 10 & 6950 \\1 & 35 & 9 & 6725 \\1 & 40 & 12 & 7100\end{bmatrix}\\[.25 cm]

\sim \begin{bmatrix}1 & 40 & 10 & 6950 \\0 & -5 & -1 & -225 \\0 & 0 & 2 & 150\end{bmatrix}\\[.25 cm]

$\begin{bmatrix} 1 & 40 & 10 \\ 0 & -5 & -1 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} \begin{bmatrix} 6950\\ -225 \\ 150 \end{bmatrix} \\[.5cm]$

$a+40b+10c=6950\cdots (1)\\[.25 cm] -5b-c=-225\\[.25 cm] 2c=150 \\[.25 cm] c=75 \\[.25cm]$

$(2) -5b-75=-225\\[.25cm] b=30 \\[.25cm]$

$(1) a+1200+750=6950\\[.25cm] a=5000\\[.25cm] a = 5000,\: b = 30, \:c = 75 \\[.25cm]$

The production equation is P = 5000 + 30l + 75m

P at l=50 and m =15 =5000 + 30 (50) +75 (15) =7625

Therefore, the production is 7625 units.

Example 1.19 Solve the equations 2x + 3y = 7, 3x + 5y= 9 by Cramer’s rule.

$\Delta = \begin{vmatrix} 2&3 \\ 3&5 \end{vmatrix}\\[.25cm] =10-9 \\[.25cm] =1 \neq 0\\[.25cm]$

$\Delta_{x} = \begin{vmatrix} 7&3 \\ 9&5 \end{vmatrix}\\[.25cm] =35-27 \\[.25cm] =8 \\[.25cm]$

$\Delta_{y} = \begin{vmatrix} 2&7 \\ 3&9 \end{vmatrix}\\[.25cm] =18-21 \\[.25cm] =-3 \\[.25cm]$

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{8}{1}\\$
$=8$

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{-3}{1}\\$
$=-3 \\$

Therefore x = 8; y = -3

Example 1.20 The following table represents the number of shares of two companies A and B during the month of January and February and it also gives the amount in rupees invested by Ravi during these two months for the purchase of shares of two companies. Find the the price per share of A and B purchased during both the months.

Months	Number of shares of the company		Amount invested by Ravi in Rs.
	A	B
January	10	5	125
February	9	12	150

Let the price of one share of A be x and the price of one share of B be y.
The equations are: 10x+ 5y= 125 ; 9x+ 12y= 150

$\Delta = \begin{vmatrix} 10&5 \\ 9&12 \end{vmatrix}\\[.25cm] =120-45 \\[.25cm] =75 \neq 0\\[.25cm]$

$\Delta_{x} = \begin{vmatrix} 125&5 \\ 150&12 \end{vmatrix}\\[.25cm] =1500-750 \\[.25cm] =750 \\[.25cm]$

$\Delta_{y} = \begin{vmatrix} 10&125 \\ 9&150 \end{vmatrix}\\[.25cm] =1500-1125\\[.25cm] =375 \\[.25cm]$

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{750}{75} \\$
$=10 \\$

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{375}{75} \\$
$=5$

Therefore, the price of the share A is Rs. 10 and the price of the share B is Rs. 5.

Example 1.21 The total cost of 11 pencils and 3 erasers is Rs. 64 and the total cost of 8 pencils and 3 erasers is Rs. 49. Find the cost of each pencil and each eraser by Cramer’s rule.

Let x be the cost of a pencil and y be the cost of an eraser. The equations are: 11x+3y=64 and 8x+3y=49.

$\Delta = \begin{vmatrix} 11&3 \\ 8&3 \end{vmatrix}\\[.25cm] =33-24\\[.25cm] =9\neq 0\\[.25cm]$

$\Delta_{x} = \begin{vmatrix} 64&3 \\ 49&3 \end{vmatrix}\\[.25cm] =192-147\\[.25cm] =45\\[.25cm]$

$\Delta_{y} = \begin{vmatrix} 11&64\\ 8&49 \end{vmatrix}\\[.25cm] =539-512\\[.25cm] =27\\[.25cm]$

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{45}{9} \\$
$=5\\$

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{27}{9} \\$
$=3 \\$

Therefore, The cost of a pencil is Rs. 5 and the cost of an eraser is Rs. 3.

Example 1.22 Solve by Cramer’s rule x + y+ z = 4; 2x – y + 3z =1; 3x +2y -z =1.

$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{vmatrix}\\[.25cm] = 1 (1 - 6) - 1 (-2 - 9) + 1 (4 + 3) \\[.25cm] = 1 (-5) -1 (-11) +1 (7) \\[.25cm] = -5 + 11 + 7 \\[.25cm] =13 \neq 0\\[.25cm]$

$\Delta_{x} = \begin{vmatrix} 4 & 1 & 1 \\ 1 & -1 & 3 \\ 1 & 2 & -1 \end{vmatrix}\\[.25cm] = 4 (1 - 6) - 1 (-1 - 3) + 1 (2 + 1) \\[.25cm] = 4 (-5) -1 (-4) +1 ( 3) \\[.25cm] = -20 +4 + 3 \\[.25cm] =-13 \\[.25cm]$

$\Delta_{y} = \begin{vmatrix} 1 & 4 & 1 \\ 2 & 1 & 3 \\ 3 & 1 & -1 \end{vmatrix}\\[.25cm] = 1 (- 1 - 3) -4 (-2 - 9) +1 (2 - 3) \\[.25cm] = 1 (-4) - 4 (-11) + 1 (-1) \\[.25cm] = -4 + 44 - 1 \\[.25cm] =39\\[.25cm]$

$\Delta_{z} = \begin{vmatrix} 1 & 1 & 4 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{vmatrix}\\[.25cm] = 1 (-1 -2) - 1 (2 - 3) + 4 (4 + 3) \\[.25cm] = 1 (-3) - 1(-1) + 4 (7) \\[.25cm] = -3 +1 +28 \\[.25cm] =26 \\[.25cm]$

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{-13}{13} \\$
$=-1 \\[.25cm]$

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{39}{13} \\$
$=3 \\[.25cm]$

$y =\dfrac{\Delta_{z}}{\Delta} \\$
$=\dfrac{26}{13} \\$
$=2 \\[.25cm]$

Therefore x = -1; y = 3; z = 2

Example 1.23 The price of 3 Business Mathematics books, 2 Accountancy books and one Commerce book is Rs. 840. The price of 2 Business Mathematics books, one Accountancy book and one Commerce book is Rs. 570. The price of one Business Mathematics book, one Accountancy book and 2 Commerce books is Rs. 630. Find the cost of each book by using Cramer’s rule.

Let x be the cost of a Business Mathematics book, y be the cost of a Accountancy book and z be the cost of a Commerce book. The equations are 3x +2y +z =840, 2x +y +z =570, and x +y +z =630.

$\Delta = \begin{vmatrix} 3 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{vmatrix}\\[.25cm] = 3 (2 - 1) - 2 (4 - 1) + 1 (2 - 1) \\[.25cm] = 3 (1) -2 (3) +1 (1) \\[.25cm] = 3 -6 + 1 \\[.25cm] =-2 \neq 0\\[.25cm]$

$\Delta_{x} = \begin{vmatrix} 840 & 2 & 1 \\ 570 & 1 & 1 \\ 630 & 1 & 2 \end{vmatrix}\\[.25cm] = 840 (2 - 1) - 2 (1140 - 630) + 1 (570-630) \\[.25cm] = 840 (1) -2 (510) +1 ( -60) \\[.25cm] = 840 -1020 -60 \\[.25cm] =-240 \\[.25cm]$

$\Delta_{y} = \begin{vmatrix} 3 & 840& 1 \\ 2 & 570 & 1 \\ 1 & 630 & 2 \end{vmatrix}\\[.25cm] = 3 (1140-630) -840 (4 - 1) +1 (1260-570) \\[.25cm] = 3 (510) - 840 (3) + 1 (690) \\[.25cm] = 1530 -2520 +690 \\[.25cm] =-300\\[.25cm]$

$\Delta_{z} = \begin{vmatrix} 3 & 2 & 840 \\ 2 & 1 & 570 \\ 1 & 1 & 630 \end{vmatrix}\\[.25cm] = 3 (630-570) - 2 (1260-570) + 840 (2 - 1) \\[.25cm] = 3 (60) - 2(690) + 840 (1) \\[.25cm] = 180 -1380 + 840 \\[.25cm] =-360\\[.25cm]$

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{-240}{-2} \\$
$=120 \\[.25cm]$

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{-300}{-2} \\$
$=150 \\[.25cm]$

$y =\dfrac{\Delta_{z}}{\Delta} \\$
$=\dfrac{-360}{-2} \\$
$=180 \\[.25cm]$

Therefore, the cost of a Business Mathematics book is Rs. 120, the cost of a Accountancy book is Rs. 150 and the cost of a Commerce book is Rs. 180.

Example 1.24 An automobile company uses three types of Steel S1, S2 and S3 for providing three different types of Cars C1, C2 and C3. Steel requirement R (in tonnes) for each type of car and total available steel of all the three types are summarized in the following table.

Types of Steel	Types of car			Total Steel available
	C1	C2	C3
S1	3	2	4	28
S2	1	1	2	13
S3	2	2	1	14

Determine the number of Cars of each type which can be produced by Cramer’s rule.

Let x be the number of cars of type $C_{1}$ , y be the number of cars of type $C_{2}$ and z be the number of cars of type $C_{3}$ .

The equations are: 3x + 2y + 4z =28; x + y + 2z =13; and 2x + 2y + z =14.

$\Delta = \begin{vmatrix} 3 & 2 & 4 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{vmatrix}\\[.25cm] = 3 (1 - 4) - 2 (1 - 4) + 4 (2 - 2) \\[.25cm] = 3 (-3) -2 (-3) \\[.25cm] = -9 +6 \\[.25cm] =-3 \neq 0\\[.25cm]$

$\Delta_{x} = \begin{vmatrix} 28& 2 & 4 \\ 13& 1 & 2 \\ 14& 2 & 1 \end{vmatrix}\\[.25cm] = 28(1-4) - 2 (13-28) + 4 (26-14) \\[.25cm] = 28(-3) -2 (-15) +4 (12) \\[.25cm] = -84+30+48 \\[.25cm] =-6\\[.25cm]$

$\Delta_{y} = \begin{vmatrix} 3 & 28& 4 \\ 1 & 13& 2 \\ 2 & 14& 1 \end{vmatrix}\\[.25cm] = 3 (13-28) -28(1-4) +4 (14-26) \\[.25cm] = 3 (-15) - 28(-3) + 4 (-12) \\[.25cm] = -45+84-48\\[.25cm] =-9\\[.25cm]$

$\Delta_{z} = \begin{vmatrix} 3 & 2 & 28\\ 1 & 1 & 13\\ 2 & 2 & 14 \end{vmatrix}\\[.25cm] = 3 (14-26) - 2 (14-26) + 28(2 - 2) \\[.25cm] = 3 (-12) - 2(-12) \\[.25cm] = -36+24\\[.25cm] =-12\\[.25cm]$

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{-6}{-3} \\$
$=2\\[.25cm]$

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{-9}{-3} \\$
$=3\\[.25cm]$

$y =\dfrac{\Delta_{z}}{\Delta} \\$
$=\dfrac{-12}{-3} \\$
$=4\\[.25cm]$

Therefore, the number of cars of each type which can be produced are 2, 3 and 4.

Example 1.25 Consider the matrix of transition probabilities of a product available in the market in two brands A and B.

$\begin{matrix} & \\ A & \\ B& \end{matrix} \begin{pmatrix} A & B\\ .9& .1\\ .3& .7 \end{pmatrix} \\[.25cm]$

Determine the market share of each brand in equilibrium position.

Transition Probability Matrix

$T = \begin{matrix} & \\ A & \\ B& \end{matrix} \begin{pmatrix} A & B\\ .9& .1\\ .3& .7 \end{pmatrix} \\[.25cm]$

At Equilibrium:

$( A,B) =\begin{pmatrix} .9 &.1 \\ .3 & .7 \end{pmatrix} = ( A,B) \\[.25cm]$ We know that A + B =1 ; $\therefore B =1-A \\[.25cm]$

$0.9A+ 0.3B= A \\[.25cm] = 0.9A+ 0.3 (1 - A) = A \\[.25cm] = 0.9A+ 0.3 - 0.3A = A \\[.25cm] = 0.6A + 0.3 = A \\[.25cm] = 0.3 = A - 0.6A \\[.25cm] = 0.4A= 0.3 \\[.25cm] = \frac{.3}{.4} \\[.25cm] =.75 \\[.25cm]$

Hence the market share of brand A is 75% and the market share of brand B is 25.

Example 1.26 Parithi is either sad (S) or happy (H) each day. If he is happy in one day, he is sad on the next day by four times out of five. If he is sad on one day, he is happy on the next day by two times out of three. Over a long run, what are the chances that Parithi is happy on any given day?

Transition Probability Matrix

$T= \begin{matrix} & \\ A & \\ B& \end{matrix} \begin{pmatrix} A & B\\ 1/3 & 2/3 \\ 4/5& 1/5 \end{pmatrix} \\[.25cm]$

At Equilibrium:

$( S,H) =\begin{pmatrix} 1/3&2/3 \\ 4/5 & 1/5 \end{pmatrix} = ( S,H) \\[.25cm]$ We know that S + H =1 ; $\therefore H =1-S \\[.25cm]$

$1/3S +4/5H = S\\[.25cm] = 1/3S+ 4/5 (1 - S) = S \\[.25cm] = 1/3S+ 4/5 - 4/5S = S \\[.25cm] = 1/3S -4/5S -S = -4/5 \\[.25cm] = -22S/15S = -4/5 \\[.25cm] = -22S=-12 \\[.25cm] = \frac{-12}{-22} \\[.25cm] =6/11 \\[.25cm]$

Therefore, S = 6/11 H = 5/11. In the long run, on a randomly selected day, his chances of being happy is 5/11.

Example 1.27 Akash bats according to the following traits. If he makes a hit (S), there is a 25% chance that he will make a hit his next time at bat. If he fails to hit (F), there is a 35% chance that he will make a hit his next time at bat. Find the transition probability matrix for the data and determine Akash’s long- range batting average.

Transition Probability Matrix

$T= \begin{pmatrix} A & B\\ .25 & .75 \\ .35& .65 \end{pmatrix} \\[.25cm]$

At Equilibrium:

$( S,F) =\begin{pmatrix} .25 &.75 \\ .35 & .65 \end{pmatrix} = ( S,F) \\[.25cm]$ We know that S + F =1 ; $\therefore F =1-S \\[.25cm]$

$0.25S+ 0.35F= S \\[.25cm] = 0.25S+ 0.35 (1 - S) = S \\[.25cm] = 0.25S+ 0.35 - 0.35S = S \\[.25cm] = -0.1S + 0.35 = S \\[.25cm] = 0.35 = S +.1S \\[.25cm] = 1.1S= 0.35 \\[.25cm] = \frac{.35}{1.1} \\[.25cm] =0.318 \\[.25cm]$

Therefore, S = 0.318 and F = 0.682. Akash’s batting average is 31.8%

Example 1.28 80% of students who do maths work during one study period, will do the maths work at the next study period. 30% of students who do English work during one study period, will do the English work at the next study period.

Initially there were 60 students do maths work and 40 students do English work. Calculate,
(i) The transition probability matrix
(ii) The number of students who do maths work, English work for the next subsequent 2 study periods.

Transition Probability Matrix

$T= \begin{matrix} & \\ M & \\ E& \end{matrix} \begin{pmatrix} M & E\\ .8& .2\\ .7& .3 \end{pmatrix} \\[.25cm]$

After 1 study period,

$\begin{pmatrix} M&E \\ .60& .40 \end{pmatrix} \begin{matrix} & \\ M & \\ E& \end{matrix} \begin{pmatrix} M & E\\ .8& .2\\ .7& .3 \end{pmatrix}\\[.25cm]$

$\begin{pmatrix} 0.48+0.28 & 1.12+0.12\\ \end{pmatrix}\\[.25cm]$

$\begin{pmatrix} M&E \\ .76& .24 \end{pmatrix}\\[.25cm]$

So in the next study period, there will be 76 students do maths work and 24 students do the
English work.

After two study periods,

$\begin{pmatrix} M&E \\ .76& .24 \end{pmatrix} \begin{matrix} & \\ M & \\ E& \end{matrix} \begin{pmatrix} M & E\\ .8& .2\\ .7& .3 \end{pmatrix}\\[.25cm]$

$\begin{pmatrix} 60.8+16.8 & 15.2+7.2\\ \end{pmatrix}\\[.25cm]$

$\begin{pmatrix} M&E \\ 77.6& 22.4 \end{pmatrix}\\[.25cm]$

After two study periods there will be approximately 78 students who do maths work and approximately 22 students who do English work.

Matrix A	Elementary Transformation
$\sim$ $\begin{bmatrix}1 & 2 & 3\\2 & 3 & 4\\3 & 5 & 7\end{bmatrix}\\[.5 cm]$	$R_{2}\rightarrow R_{2}- 2R_{1}$ $R_{3}\rightarrow R_{3}- 2R_{1}$
$\sim$ $\begin{bmatrix}1 & 2 & 3\\0 & -1 & -2\\0 & -1 & -2\end{bmatrix}\\[.5 cm]$	$R_{3}\rightarrow R_{3}- R_{2}$

Matrix A	Elementary Transformation
A= $\begin{bmatrix}0 & 1 & 2 & 1 \\1 & 2 & 3 & 2 \\3 & 1 & 1 & 3\end{bmatrix}\\[.5 cm]$	$R_{1}\leftrightarrow R_{2}$
$\sim$ $\begin{bmatrix}0 & 1 & 2 & 1 \\1 & 2 & 3 & 2 \\3 & 1 & 1 & 3\end{bmatrix}\\[.5 cm]$	$R_{3}\rightarrow R_{3}- 3R_{1}$
$\sim$ $\begin{bmatrix}1 & 2 & 3 & 2 \\0 & 1 & 2 & 1 \\3 & -5 & -8 & -3\end{bmatrix}\\[.5 cm]$	$R_{3}\rightarrow R_{3}+5R_{2}$

Matrix A	Elementary Transformation
$A=\begin{bmatrix}1 & 1 & 1 & 1 \\3 & 4 & 5 & 2 \\2 & 3 & 4 & 0\end{bmatrix}\\[.5 cm]$
$\sim \begin{bmatrix}1 & 1 & 1 & 1 \\0 & 1 & 2 & -1 \\0 & 1 & 2 & -2\end{bmatrix}\\[.5 cm]$	$R_{2}\rightarrow R_{2}-3R_{1}$ $R_{3}\rightarrow R_{3}- 2R_{1}$
$\sim \begin{bmatrix}1 & 1 & 1 & 1 \\0& 1 & 2 & -1 \\0 & 0 & 0 & -1\end{bmatrix}\\[.5 cm]$	$R_{3}\rightarrow R_{3}- 3_{2}$

Matrix A	Augmented matrix	Elementary Transformation
$\begin{pmatrix}1&1 \\2&1\end{pmatrix}\\[.5 cm]$	$\begin{bmatrix}1 & 1 & 5 \\2 & 1 & 8\end{bmatrix}\\[.5 cm]$
$\sim \begin{pmatrix}1&1 \\0&-1\end{pmatrix}\\[.5 cm]$	$\begin{bmatrix}1 & 1 & 5 \\0 & -1 & -2\end{bmatrix}\\[.5 cm]$	$R_{2}\rightarrow R_{2}- 2R_{1}$
$\rho(A)=2\\[.5 cm]$	$\rho([A,B]) =2\\[.5 cm]$

Matrix A	Augmented matrix	Elementary Transformation
$\begin{pmatrix}2&1 \4&2\end{pmatrix}\\[.5 cm]$	$\begin{bmatrix}2 & 1 & 5 \\4 & 2 & 10\end{bmatrix}\\[.5 cm]$	$R_{2}\rightarrow R_{2}- 2R_{1}$
$\sim \begin{pmatrix}2&1 \\0&0\end{pmatrix}\\[.5 cm]$	$\sim \begin{bmatrix}2 & 1 & 5 \\0 & 0 & 0\end{bmatrix}\\[.5 cm]$
$\rho(A)=1\\[.5 cm]$	$\rho([A,B]) =1\\[.5 cm]$

Class 12 Business Maths Examples Applications of Matrices and Determinants

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