2 - Inverse 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.

Important Definitions
  1. A square matrix whose determinant is equal to zero is said to be a singular matrix.
  2. A square matrix whose determinant is any number other than zero is said to be a non-singular matrix.
  3. Inverse of a square and non-singular matrix is only possible.
  4. When two matrices are inverse to each other, then their product is an identity matrix.


All solutions related to matrix:




1. Multiply the following matrices and show that A and B are inverse to each other.

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

Solution:

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

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

$= \left (  \displaylines{6 -5 & -2 + 2 \\ 15-15 & -5 + 6} \right )$

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


Hence, matrices A and B are inverse to each other.


2. Find the inverse of:

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

Solution:

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

First, we need to find the determinant to check whether the matrix is singular or non-singular matrix.

|A| = $\left | \displaylines{3&-5 \\-1&2} \right |$

$= 3(2) - (-5)(-1)$
$= 6 - 5$
$= 1$
Since, |A| ≠ 0, matrix A is a non-singular matrix and its inverse is possible.

Now,

Adjoint matrix of A = $\left ( \displaylines{ 2&5\\1&3} \right )$


$A ^{-1}= \frac{1}{|A|} (Adjoint matrix of A)$

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


Hence, the inverse of matrix A is found.



c) $\left ( \displaylines{2 & 15 \\ 3 & 23} \right )$

Solution:

Let A = $\left ( \displaylines{2&15 \\3&23} \right )$

First, we need to find the determinant to check whether the matrix is singular or non-singular matrix.

|A| = $\left | \displaylines{2&15 \\3&23} \right |$

$= 2(23) - (15)(3)$

$= 46 - 45$

$= 1$

Since, |A| ≠ 0, matrix A is a non-singular matrix and its inverse is possible.

Now,

Adjoint matrix of A = $\left ( \displaylines{ 23&-15\\-3&2} \right )$


$A ^{-1}= \frac{1}{|A|} (Adjoint matrix of A)$

$= \frac{1}{1} \left ( \displaylines{23&-15\\-3&2} \right )$


Hence, the inverse of matrix A is found.



e) $\left ( \displaylines{secA & tanA \\ secA & tanA} \right )$

Solution:

Let A = $\left ( \displaylines{secA&tanA \\secA&tanA} \right )$

First, we need to find the determinant to check whether the matrix is singular or non-singular matrix.

|A| = $\left | \displaylines{secA&tanA \\secA&tanA} \right |$

$= secA(secA) - (tanA)(tanA)$

$= sec^2A- tan^2A$

$= 1$

Since, |A| ≠ 0, matrix A is a non-singular matrix and its inverse is possible.

Now,

Adjoint matrix of A = $\left ( \displaylines{ secA&-tanA\\-tanA&secA} \right )$


$A ^{-1}= \frac{1}{|A|} (Adjoint matrix of A)$

$= \frac{1}{1} \left ( \displaylines{secA&-tanA\\-tanA&secA} \right )$


Hence, the inverse of matrix A is found.

3. Show that each of the following matrices is its own inverse.

a) $\left ( \displaylines{ 0&1 \\ 1&0} \right )$

Solution:

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

If A is the inverse of itself then A×A = I should be true.

LHS

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

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

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

$= I$
RHS
Hence, it is proved that the given matrix is inverse of itself.


c) $\left ( \displaylines{ 1&0 \\ 0&-1} \right )$

Solution:

Let A = $\left ( \displaylines{ 1&0 \\ 0&-1} \right )$

If A is the inverse of itself then A×A = I should be true.

LHS

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

$= \left ( \displaylines {1*1 + 0*0&1*0 + 0*(-1)\\0*1 + (-1)*0&0*0+(-1)*(-1)} \right)$

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

$= I$

RHS


Hence, it is proved that the given matrix is inverse of itself.
4. If $\text{A} = \left ( \displaylines{ 5 & -3 \\ 2 & -4 } \right) $ and  $\text{B} = \left ( \displaylines{ 4 & 2 \\ -3 & 5 } \right) $, verify that $|BA| = |A||B|$.

Solution:
Here,
$\text{A} = \left ( \displaylines{ 5 & -3 \\ 2 & -4 } \right) $
$\text{B} = \left ( \displaylines{ 4 & 2 \\ -3 & 5 } \right) $

To verify: $|BA| = |A||B|$

For $|BA|$

Let us find BA first

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

$\implies \left ( \displaylines{ 20+9 & 10 -15 \\ 8+12 & 4 -20} \right ) $

$\implies \left ( \displaylines{ 29 & -5 \\ 20 & -16} \right )$

Now,

$|BA| = 29 * (-16) - (-5)*20)$

$\implies -464 +100$

$\implies |BA| =  -364$ --- (i)

For $|A| |B|$

First, we find determinant of A,

$|A| = 5*(-4) - (-3)*2$

$\implies -20 + 6$

$\implies |A| = -14$

Second, we find determinant of B,
$|B| = 4*5 - (-3)*2$

$\implies 20 + 6$

$\implies |B| = 26$

So, $|A||B| = (-14)*26 = -364$ --- (ii)

Since, values of $|BA|$ and $|A||B|$ are equal in equation (i) and equation (ii), i.e. -364, the given relation $|BA| = |A||B|$ is verified.


5 a) If A = $\left ( \displaylines{ 3 & 2 \\ 7 & 5} \right )$ and B = $\left ( \displaylines{ -2 & 7 \\ -3 & 9} \right ) $, verify that (AB)$^{-1}$ = B$^{-1}$A$^{-1}$.

Solution:
Given,

A = $\left ( \displaylines{ 3 & 2 \\ 7 & 5} \right)$

B = $\left ( \displaylines{ -2 & 7 \\ -3 & 9} \right )$

To verify: (AB)$^{-1}$ = B$^{-1}$A$^{-1}$

We know, for matrix A, $A^{-1} = \dfrac{1}{|A|} (\text{adjoint of a A})$

LHS

$= (AB)^{-1}$

$= \left [ \left ( \displaylines{ 3 & 2 \\ 7 & 5} \right ) \left ( \displaylines{-2 & 7 \\ -3 &9} \right ) \right ] ^{-1}$

$= \left [ \displaylines{-6-6 & 21+18 \\ -14-15 & 49+45} \right ] ^{-1}$

$= \left [ \displaylines{-12 & 39 \\ -29 & 94} \right ]^{-1}$

$= \dfrac{1}{ \left | \displaylines { -12 & 39 \\ -29 & 94} \right | } * \left ( \displaylines{ 94 & -39 \\ 29 & -12} \right )$

$= \dfrac{1}{-12(94) - (-29)39} * \left ( \displaylines{94 & -39 \\ 29 & 012} \right)$

$= \dfrac{1}{3} * \left ( \displaylines{ 94 & -39 \\ 29 & -12} \right)$


RHS

$= B^{-1} A^{-1}$

$= \dfrac{1}{|B|} * (\text{adjoint of B}) * \dfrac{1}{|A|} * (\text{adjoint of A})$

$= \dfrac{1}{ \left | \displaylines{-2 & 7 \\ -3 & 9} \right |} * \left ( \displaylines{9 & -7 \\ 3 & 2} \right ) * \dfrac{1}{ \left | \displaylines{3 & 2 \\ 7 & 5} \right | } * \left ( \displaylines{ 5 & -2 \\ -7 & 3} \right )$

$= \dfrac{1}{(-2)9 - (7)(-3)} * \left ( \displaylines{9 & -7 \\ 3 & -22} \right) * \dfrac{1}{3(5) - 7(2)} * \left ( \displaylines{5 & -2 \\ -7 &3} \right)$

$= \dfrac{1}{-18 + 21} * \dfrac{1}{15 -14} * \left ( \displaylines{ 45 + 49 & -18-21 \\ 15 + 14 & -6-6 } \right ) $

$= \dfrac{1}{3} * \dfrac{1}{1} * \left ( \displaylines{94 & -39 \\ 20 & -12} \right )$

$= \dfrac{1}{3} \left ( \displaylines{94 & -30 \\ 20 & -12} \right)$


Since, LHS = RHS, given relation is verified.


6 a) If the inverse of the matrix $\left ( \displaylines{ x & 2x-9 \\ -y & 3} \right )$ is the matrix $\left ( \displaylines{3 & 5 \\ y & x} \right)$, find the values of x and y.

Solution:

Let matrix A be $\left ( \displaylines{ x & 2x-9 \\ -y & 3} \right )$.

Let matrix B be $\left ( \displaylines{3 & 5 \\ y & x} \right)$.

When two matrices are inverse of each other, then the following relation holds true.

$AB = BA= I$

Taking $AB = I$

$\implies \left ( \displayline{ x & 2x-9 \\ -y & 3} \right ) \left ( \displaylines{3 & 5 \\ y & x} \right) = \left( \displaylines{ 1 & 0 \\ 0 & 1} \right)$

$\implies \left ( \displaylines{3x + y(2x-9) & 5x + x(2x-9) \\ -3y + 3y & -5y + 3x} \right ) = \left ( \displaylines{ 1 & 0 \\ 0 & 1} \right ) $

From relation of equal matrices, we get,

$(i) 3x + y(2x-9) = 1$

$(ii) 5x + x(2x-9) = 0$

$(iii) -3y + 3y = 0$

$(iv) -5y + 3x = 1$

Solving equation (ii), we get,

$or, 5x + 2x^2 - 9x = 0$

$or, 2x^2 - 4x = 0$

$or, 2x^ 2= 4x$

$\therefore x = 2$

Putting value of x = 2 in equation (i), we get, $y = 1$

Hence, the required values of x and y are $(x,y) = (2,1)$


About Readmore Optional Mathematics Book 10

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

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

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