3 - Solutions of a System of Linear Equations (Matrix)

In this post, you can check the solutions to the exercises of solution of a system of linear equations of Matrix chapter of Readmore Optional Mathematics by DR Simkhada. It consists of complete answers to all odd questions mentioned in the textbook.

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Before starting, make sure to check the following note of Determinant.
  1. Solutions of a system of Linear Equations - Matrix Class 10

Important Points:
  1. When the determinant of a matrix formed by a system of linear equations is zero (singular matrix), the equations have no unique solutions.
  2. The general form of matrix is AX = B.
  3. When AX = B, X = A$^{-1}$B.
  4. When XA = B, X = BA$^{-1}$.
  5. Always arrange both the equations in the form of ax + by = c.
  6. If the equation is ax + by - c = 0 then arrange it as ax + by = c to get correct solution.


Let,
Equation i: ax + by = c
Equation ii: gx + fy = z
Matrix form of above equations is:
$\left ( \displaylines{a & b\\g & f}\right) \left ( \displaylines{x \\y}\right ) = \left ( \displaylines{c \\ z} \right )$


1. Check whether the following system of equations contain unique solutions or not.


a) 3x + 2y = 5 and 6x + 4y = 10
Solution:
Arranging in matrix form,
$\left ( \displaylines { 3&6\\2&4}\right ) \left ( \displaylines{x \\y} \right ) = \left ( \displaylines{5 \\10} \right )$
i.e. AX = B
To check whether the system of equations contains unique solution or not, we need to determine whether matrix A is singular or non-singular matrix.

We have,

A = $\left ( \displaylines{ 3&6\\2&4} \right )$

$or, |A| = \left | \displaylines{3&6\\2&4}  \right )$

$= 3(4) - 6(2)$
$= 12 - 12$
$= 0$
Since |A| = 0, the system of solutions does not have unique solutions.


c) 3x - 5y = 7 and 5y - 3x + 2 = 0
Solution:
We need each equation in the form of ax + by = c

So,
Required equations are:
3x - 5y = 7
3x - 5y = 2
Arranging in matrix form, we get,
$\left ( \displaylines{3 & -5\\3&-5} \right ) \left ( \displaylines{x\\y} \right ) \left ( \displaylines{7\\2}\right )$

i.e. AX = B
Now,

|A| = $\left | \displaylines{3&-5\\3&-5}\right |$

$= 3(-5) - (-5)(3)$
$= -15 +15$
$= 0$
Since, matrix A is a singular matrix, given system of equations has no unique solutions.

2. Find $\left ( \displaylines{x\\y}\right )$ when:


a) $\left ( \displaylines{-2&1\\-3&1} \right ) \left ( \displaylines{ x \\y} \right )\left ( \displaylines{11\\1} \right )$

Solution:
$\left ( \displaylines{-2&1\\-3&1} \right ) \left ( \displaylines{x \\y} \right ) \left ( \displaylines{ 11\\1} \right )$

We have,
AX = B
So,
A = $\left ( \displaylines{ -2&1\\-3&1} \right )$

Now,
|A| = $\left | \displaylines{ -2&1\\-3&1} \right |$

$= 2(1) - (-3)1$
$= 2 + 3$
$= 5$
Since |A| ≠ 0, the system of equations have unique solutions.

And,
A$^{-1}$ = $\dfrac{1}{|A|} (adjoint of A)$

$= \dfrac{1}{5} \left ( \displaylines{ 1&-1\\3&2} \right )$

Also,
X = $A^{-1}B$
$or, \left ( \displaylines{ x\\y} \right )= \dfrac{1}{5} \left ( \displaylines{1&-1\\3&2} \right ) \left ( \displaylines{11\\1} \right )$

$= \dfrac{1}{5} \left ( \displaylines{ 1(11) -1(1)\\3(11)+2(1) } \right )$

$= \dfrac{1}{5} \left ( \displaylines{11-1\\33+2} \right )$

$= \dfrac{1}{5} \left ( \displaylines{10\\35} \right )$

$= \left ( \displaylines{ \frac{10}{5} \\ \frac{35}{5} } \right )$

$= \left ( \displaylines{ 2 \\7} \right )$

Hence, x = 2 and y = 7.

c) $\left ( \displaylines{2&5\\3&-5} \right ) \left ( \displaylines{x\\y} \right ) = \left ( \displaylines{2\\3} \right )$

Solution:
Given,
AX = B

So, A = $\left ( \displaylines{2&5\\3&-5} \right )$

|A| = $\left | \displaylines{2&5\\3&-5} \right )$

$= 2(-5) - 3(5)$
$= -10 - 15$
$= -25$
Since |A| ≠ 0, the system of equations has unique solutions.

We know,
X = A$^{-1}$B
$or, \left ( \displaylines{x\\y} \right ) = \dfrac{1}{|A|} \left ( adjoint of A) \right ) × \left ( \displaylines{2\\3} \right )$

$= \dfrac{1}{-25} \left ( \displaylines{-5&-5\\-3&2} \right )\left ( \displaylines{2\\3} \right )$

$= \dfrac{1}{-25} \left ( \displaylines{ -5(2)-5(3)\\-3(2)+2(3) } \right )$

$= \dfrac{1}{-25} \left ( \displaylines{-10-15\\-6+6} \right )$

$= \dfrac{1}{-25} \left (\displaylines{-25\\0} \right )$

$= \left ( \displaylines { \frac{-25}{-25} \\ \frac{0}{-25} } \right )$

$= \left ( \displaylines{1\\0} \right )$

Hence, the required values of x and y are 1 and 0, respectively.


3. Solve the following system of equations with the help of matrix.


a) $3x + 4y = 5$ and $x - y = -3$

Solution:

Arranging the given equations in matrix form, we get,

$\left ( \displaylines{3&4\\1&-1} \right ) \left ( \displaylines{x \\y} \right ) = \left ( \displaylines{5\\-3} \right )$

i.e. AX = B
So,
A = $\left ( \displaylines{3&4\\1&-1} \right )$

|A| = $\left | \displaylines{3&4\\1&-1} \right )$

$= 3(-1) - 4(1)$
$= -3 - 4$

$= -7$
Since |A| ≠ 0, the system of linear equations have unique solutions.

We know,
X = A$^{-1}$B
$or, \left ( \displaylines{x\\y} \right ) = \dfrac{1}{|A|} \left ( Adjoint of A \right ) \left ( \displaylines{5\\-3} \right )$

$= \dfrac{1}{-7} \left ( \displaylines{-1&-4\\-1&3} \right ) \left ( \displaylines{5\\-3} \right)$

$= \dfrac{1}{-7} \left ( \displaylines{-1(5)-4(-3)\\-1(5)+3(-3)} \right )$

$= \dfrac{1}{-7} \left ( \displaylines{ -5+12\\-5-9} \right )$

$= \dfrac{1}{-7} \left ( \displaylines{7\\-14} \right )$

$= \left ( \displaylines{\frac{7}{-7} \\ \frac{-14}{-7} } \right )$

$= \left ( \displaylines{-1\\2} \right )$

Hence, the required values of x and y are -1 and 2, respectively.


c) $2x + 5y = 7$ and $x + 10y = -4$
Solution:

Arranging the given equations in matrix form, we get,

$\left ( \displaylines{2&5\\1&10} \right ) \left ( \displaylines{x \\y} \right ) = \left ( \displaylines{7\\-4} \right )$

i.e. AX = B
Now, solve and get the answer.


e) $4y - 5 = 0$ and $x - y + 3 = 0$
Solution:

Arranging the given equations in the form of ax + by = c,

(i) $4y - 5 = 0$
$or, 0x + 4y = 5$
(ii) $x - y + 3 = 0$
$or, x - y = -3$
Now,
Arranging the given system of equations in matrix form, we get,

$\left ( \displaylines{0&4\\1&-2} \right ) \left ( \displaylines{x \\y} \right ) = \left ( \displaylines{5\\-3} \right )$

Solve and get the answer.

g) $2x - 5y = 4$ and $4x + y = 30$
Solution:

Arranging the given system of solutions in matrix form, we get,

$\left ( \displaylines{ 2&-5\\4&1} \right ) \left ( \displaylines{x \\y} \right ) = \left ( \displaylines{4 \\30} \right )$

Now solve and get the answer.

i) $\frac{x}{7} - \frac{2y}{7} = -1 $ and $\frac{3x}{5} + \frac{7y}{5} = 1$
Solution:
Arranging the given system of equations in matrix form, we get,

$\left ( \displaylines { \frac{1}{7} & \frac{-2}{7} \\ \frac{3x}{5} & \frac{7y}{5} } \right ) \left ( \displaylines{x \\y} \right ) = \left ( \displaylines { -1 \\1} \right )$

Now, solve and get the answer.


k) $\frac{1}{x} - \frac{1}{2y} = 8$ and $\frac{1}{2x} - \frac{1}{y} = -1$
Solution:
Arranging the given system of equations in matrix form, we get,

$\left ( \displaylines{ 1 & \frac{-1}{2} \\ \frac{1}{2} & -1 } \right ) \left ( \displaylines{\frac{1}{x} \\ \frac{1}{y}} \right ) = \left ( \displaylines{8 \\-1 } \right )$

Now solve and get the answer.

m) $x + \frac{6}{y} - 7 = 0$ and $2xy + 4 = 10y$
Solution:

Arranging the equations,
(i) $x+ \frac{6}{y} = 7$
(ii) $2xy + 4 = 10y$
$or, 2(xy +2) = 2(5y)$
$or, xy + 2 = 5y$
Dividing by y
$or, \frac{xy}{y} + \frac{2}{y} = \frac{5y}{y}$

$or, x + \frac{2}{y} = 5$

Arranging the given equations in matrix form, we get,

$\left ( \displaylines{1&6\\1&2} \right ) \left ( \displaylines{x \\ \frac{1}{y}}\right ) = \left ( \displaylines{7\\5} \right )$

Now perform the operation, solve and get your answer.

About Readmore Optional Mathematics Book 10

Author: D. R. Simkhada
Editor: I. R. Simkhada

Readmore Publishers and Distributors
T.U. Road, Kuleshwor - 14, Kathmandu, Nepal
Phone: 4672071, 5187211, 5187226
readmorenepal6@gmail.com

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Solution of a System of Linear Equations - Class 10 | Readmore Optional Mathematics is a collection of the solutions related to exercises of solving simultaneous equations from the Matrix chapter using matrix method for Nepal's Secondary Education Examination (SEE) appearing students.

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