4 - Cramer's Rule (Matrix)

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

Before starting, make sure to check the following note of Determinant.
  1. Cramer's Rule - Matrix Class 10


Important Points:
  1. Value of x = $\frac{D_x}{D}$
  2. Value of y = $\frac{D_y}{D}$
  3. Determinant (D) = $\left |\displaylines{cof. x_1 &cof. y_1\\cof.x_2 & cof.y_2} \right | $



1 a) If $D = \left | \displaylines{ 1&1\\2&1} \right |$, $D_x = \left | \displaylines{4&1 \\6&1} \right |$ and $D_y = \left | \displaylines{1&4 \\ 2&6} \right |$ then, find the values of x and y.

Solution:

$D = \left | \displaylines{1 & 1 \\ 2& 1} \right|$
$= 1×1- 2×1$
$=  1- 2$
$= -1$
$D_x = \left |\displaylines{4&1\\6&1} \right |$
$= 4×1 - 6×1$
$= 4 - 6$
$= -2$
$D_y = \left | \displaylines{1&4\\2&6} \right |$
$= 1×6 - 4×2$
$= 6 - 8$
$= - 2$

We know,
x = $\dfrac{ D_x}{D}$
$or, x = \frac{-2}{-1}$

$\therefore x = 2$
And,
y = $\dfrac{D_y }{D}$
$or, y = \frac{-2}{-1}$
$\therefore y = 2$
Hence, the required values of x and y are 2 and 2, respectively.


1 b) If $D = \left | \displaylines{ 1&1\\1&2} \right |$, $D_x = \left | \displaylines{2&1 \\3&2} \right |$ and $D_y = \left | \displaylines{1&2\\1&3} \right |$ then, find the values of x and y.

Solution:

$D = \left | \displaylines{1 & 1 \\ 1& 2} \right|$
$= 1×2- 1×1$
$=  2-1$
$= 1$

$D_x = \left |\displaylines{2&1\\3&2} \right |$
$= 2×2 - 1×3$
$= 4 - 3$
$= 1$

$D_y = \left | \displaylines{1&2\\1&3} \right |$
$= 1×3 - 1×2$
$= 3 - 2$
$= 1$


We know,
x = $\dfrac{ D_x}{D}$

$or, x = \frac{1}{1}$

$\therefore x = 1$

And,
y = $\dfrac{D_y }{D}$

$or, y = \frac{1}{1}$

$\therefore y = 1$

Hence, the required values of x and y are 1 and 1, respectively.
2. Use Cramer's rule to solve the following equations if the solutions exists:

a) x + y = 7, x - y = 3
Solution:

Arranging the given data in a table,
Coefficient of x = 1, 1
Coefficient of y = 1, -1
Constant terms = 7, 3
We know,

The actual determinant can be obtained by the values of coefficient of x and y,
D = $\left | \displaylines{ 1 &1\\1&-1} \right |$

$or, D = 1(-1) - (1)(1)$
$or, D = -1 - 1$
$\therefore D = -2$

And,
Determinant value of x can be obtained by the values of constant terms c and coefficient of y ,
$D_x = \left | \displaylines{7&1\\3&-1} \right |$

$or, D_x = 7(-1) - (3)1$
$or, D_x = -7-3$
$\therefore D_x =-10$

Also,
Determinant value of y can be obtained by the values of coefficient of x and constant terms c ,
$D_y = \left | \displaylines{ 1&7 \\1&3} \right | $

$or, D_y = 1(3) - 1(7)$
$or, D_y = 3 - 7$
$or, D_y = -4$

Now,
x = $\dfrac{D_x}{D}$
$= \frac{-10}{-2}$
$= 5$
y = $\dfrac{D_y}{D}$
$= \frac{-4}{-2}$
$= 2$
Hence, the required values of x and y are 5 and 2, respectively.


c) 12x + 3y = 15, 2x - 3y = 13
Solution:

Arranging the given data in a table,
Coefficient of x = 12, 2
Coefficient of y = 3, -3
Constant terms = 15, 13

We know,

The actual determinant can be obtained by the values of coefficient of x and y,
D = $\left | \displaylines{ 12 &3\\2&-3} \right |$

$or, D = 12(-3) - (3)(2)$

$or, D = -36 - 6$

$\therefore D = -42$


And,
Determinant value of x can be obtained by the values of constant terms c and coefficient of y ,
$D_x = \left | \displaylines{15&3\\13&-3} \right |$

$or, D_x = 15(-3) - (13)3$

$or, D_x = -45-39$

$\therefore D_x =-84$


Also,
Determinant value of y can be obtained by the values of coefficient of x and constant terms c ,
$D_y = \left | \displaylines{ 12&15 \\2&13} \right | $

$or, D_y = 12(13) - 15(2)$

$or, D_y = 156 - 30$

$or, D_y = 126$


Now,
x = $\dfrac{D_x}{D}$

$= \frac{84}{42}$

$= 2$

y = $\dfrac{D_y}{D}$

$= \frac{126}{42}$

$= 3$

Hence, the required values of x and y are 2 and 3, respectively.


3. Use Cramer's rule to solve the following equations:

a) $\frac{x}{2} + \frac{y}{2} = 7$, $\frac{x}{2} + \frac{y}{3} = 3$
Solution:

Arranging the data, we get,
Coefficient of x = $\frac{1}{2}$, $\frac{1}{2}$
Coefficient of y = $\frac{1}{2}$, $\frac{1}{3}$
Constant terms = $7$, $3$
We know,
D = $\left | \displaylines{ \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{3}} \right |$

$= \frac{1}{2} × \frac{1}{3} - \frac{1}{2} × \frac{1}{2}$

$= \frac{1}{6} - \frac{1}{4}$
$= \frac{2 - 3}{12}$
$= -\frac{1}{12}$
And,
$D_x$ = $\left | \displaylines{ 7 & \frac{1}{2} \\ 3 & \frac{1}{3}} \right |$

$= 7 × \frac{1}{3} - 3× \frac{1}{2}$

$= \frac{7}{3} - \frac{3}{2}$
$= \frac{14-9}{6}$
$= \frac{5}{6}$
Also,
$D_y$ = $\left | \displaylines{ \frac{1}{2} & 7\\ \frac{1}{2} & 3} \right |$

$= \frac{1}{2} × 3 - \frac{1}{2} × 7$

$= \frac{3}{2} - \frac{7}{2}$
$= \frac{3-7}{2}$
$= -2$
Now,
x = $\dfrac{D_x}{D}$
$= \dfrac{\frac{5}{6}}{-\frac{1}{12}}$

$=- \dfrac{ 5 × 12}{6×1}$
$=- 10$
y = $\dfrac{D_y}{D}$
$= \dfrac{-2}{\frac{-1}{12}}$
$= (-2) × (-12)$
$= 24$
Hence, the required values of x and y are -10 and 24, respectively.

About Readmore Optional Mathematics Book 10

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

Readmore Publishers and Distributors
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Cramer's Rule - Class 10 Solved Exercises | Readmore Optional Mathematics is a collection of the solutions related to exercises of Creamer's Rule for solving simultaneous equations from the Matrix chapter for Nepal's Secondary Education Examination (SEE) appearing students.

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