1 - Determinant (Matrix)

In this post, you can check the solutions to the exercises of Determinant of 2x2 square 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. Find the value of the following:

a)  $\left | \displaylines{cosA & -sinA \\ sinA & cosA} \right|$
Solution:

| A | = cosA × cos A - (- sinA) × sinA
= cos²A + sin²A
= 1


c) $\left | \displaylines{cosecA & -cotA \\ cotA & cosecA} \right |$
Solution:

Determinant = cosecA × cosecA - cotA × cotA
= cosec²A - cot²A
= 1

2. Evaluate the determinant of the following matrices.

a) $\left ( \displaylines{4 &3 \\ 1 &2} \right )$
Solution:

Determinant = (4 × 2) - (3 × 1)
= 8 - 3
= 5

c) $\left ( \displaylines{-1 & -3 \\ -4 & - 20} \right )$
Solution:

Determinant = (-1)×(-20) - (-3)×(-4)
= 20 - 12
= 8

e) $\left( \displaylines{-1 &-18 \\ 18&44} \right )$
Solution:

Determinant = (-1) × 44 - (-18)×18
= -44 + 324
= 280


g) $\left ( \displaylines{3 & -2 \\ 2&8} \right )$
Solution:

Determinant = (3×8) - (-2)×2
= 24 + 4
= 28

3. Find the value of x, when:

a) $\left | \displaylines{-3 &x \\ 5&2 }\right | = 9$
Solution:

Determinant = (-3)×(2) - (x)×(5)
or, 9 = -6 - 5x
or, 5x = - 6 - 9
or, 5x = -15
So, x = - 3

c) $\left | \displaylines{x &2 \\8 &x } \right | = 0$
Solution:

Determinant = (x)(x) - (2)(8)
or, 0 = x² - 16
or, x² = 16
So, x = ±4

4. If A = $\left ( \displaylines{2 &3\\ 0&4} \right)$, find the value of:


a) |3A + 5I|
Solution:

3A = 3$\left ( \displaylines{2&3\\0&4} \right )$

$= \left ( \displaylines{ 6 &9\\ 0 & 12}\right )$

We know,
I = $\left ( \displaylines{ 1&0\\0&1} \right )$
So,
5I = $\left ( \displaylines{5 &0\\0&5} \right )$

Now,
3A + 5I = $\left( \displaylines{6 &9\\0 &12}\right ) + \left( \displaylines{5&0\\0&5} \right )$

$= \left ( \displaylines{ 6+5 & 9+0\\ 0+0&12+5}\right )$

$= \left ( \displaylines{ 11 & 9\\ 0 & 17} \right)$

And,
$\left | \displaylines{11 &9\\0&17}\right|$

$= 11(17) - 9(0)$
$= 187 - 0$
$= 187$


5 a) If A = $\left ( \displaylines{3&4\\2&5} \right )$ and B = $\left ( \displaylines{2 & 1\\1 & 3} \right )$, find the determinant of 3A - 4B.
Solution:

3A - 4B = $ 3 \left ( \displaylines{3&4\\2&5} \right ) - 4 \left ( \displaylines{2&1\\1&3} \right)$

$= \left ( \displaylines{9&12\\6&15} \right ) - \left ( \displaylines{8 & 4\\4&12}\right )$

$= \left ( \displaylines{9 - 8 & 12-4\\ 6-4&15-12} \right )$

$= \left ( \displaylines{1 & 8\\2 & 3} \right )$


Now,
$\left | \displaylines{1 &8\\2 & 3}\right |$

$= (1)3 - 8(2)$
$= 3 - 16$
$= -13$

6 a) If P = $\left ( \displaylines{-4&5\\7&8} \right )$, Q = $\left ( \displaylines{4&6\\x&3} \right )$ and the determinant of P - Q - 5I is 14 where I is a 2x2 identity matrix, find the value of x.

Solution:
We know,
5I = $\left ( \displaylines{5 &0\\0&5} \right )$

And,
P - Q - 5I = $\left ( \displaylines{-4&5\\7&8}\right ) - \left ( \displaylines{4&6 \\x&3} \right) - \left ( \displaylines{5&0\\0&5} \right )$

$= \left ( \displaylines{-4 -4 -5 & 5-6-0\\7-x-0 & 8-3-5} \right )$

$= \left ( \displaylines{-13 & -1\\7-x & 0} \right )$


Now,
$\left | \displaylines{-13 & -1 \\7-x &0} \right ) = 14$

$or, (-13)(0) - (-1)(7-x) = 14$

$or, 0 +(7-x) = 14$
$or, 7 - x = 14$
$or, 7 - 14 = x$
$\therefore x = -7$

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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Determinant of Matrix- Class 10 Solved Exercises | Readmore Optional Mathematics is a collection of the solutions related to exercises of Determinant of 2x2 square matrix from the Matrix chapter for Nepal's Secondary Education Examination (SEE) appearing students.

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