### Table of Contents

Difficulty: Easy
##### $\left[ \begin{matrix} \sqrt{2} & 0 \\ 0 & \sqrt{2} \\ \end{matrix} \right]$ is called ………... matrix.
Difficulty: Easy
Verified By ClassNotes
Sponsored AdsHide Ads
Difficulty: Easy
Difficulty: Easy
##### Adjoint of $\left[ \begin{matrix} 1 & 2 \\ 0 & -1 \\ \end{matrix} \right]$ is
Difficulty: Easy
Sponsored AdsHide Ads
##### Product of $\left[ \begin{matrix} x & y \\ \end{matrix} \right]\left[ \begin{matrix} 2 \\ -1 \\ \end{matrix} \right]$ is
Difficulty: Easy
Verified By ClassNotes
Difficulty: Easy
##### If $X+~\left[ \begin{matrix} -1 & -2 \\ 0 & -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ then $X$ is equal to
Difficulty: Easy
Sponsored AdsHide Ads

Definition

Null

Definition
##### $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ is called _________ matrix.

Unit

Sponsored AdsHide Ads
Definition
##### Additive inverse of $\left[ \begin{matrix} 1 & -2 \\ 0 & -1 \\ \end{matrix} \right]$ is

$\left[ \begin{matrix} -1 & 2 \\ 0 & 1 \\ \end{matrix} \right]$

Definition
##### In matrix multiplication, in general, AB _____ BA.

$\ne$ (is not equal to)

Definition
##### Matrix A + B may be found if order of A and B is

Same

Sponsored AdsHide Ads
Definition
##### A matrix is called _________ matrix if number of rows and columns are equal.

Square

Question:

If $\left[ \begin{matrix} a+3 & 4 \\ 6 & b-1 \\ \end{matrix} \right]=\left[ \begin{matrix} -3 & 4 \\ 6 & 2 \\ \end{matrix} \right]$ , then find a and b.

Difficulty: Easy

Solution:

As, $\left[ \begin{matrix} a+3 & 4 \\ 6 & b-1 \\ \end{matrix} \right]=\left[ \begin{matrix} -3 & 4 \\ 6 & 2 \\ \end{matrix} \right]$

By comparing the corresponding elements, we get

$a+3=-3$

$a=-3-3$

$a=-6$

and

$b-1=2$

$b=2+1$

$b=3$

Hence

$a=-6$ and $b=3$

##### (iv) $\frac{2}{3}$(2A $-$ 3B)
Difficulty: Easy

Solution:

(i) 2A $+$ 3B

= $2\times \left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]+3\times \left[ \begin{matrix} 5 & -4 \\ -2 & -1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2~\times 2 & 2\times 3 \\ 2\times 1 & 2\times 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 3\times 5 & 3\times \left( -4 \right) \\ 3\times \left( -2 \right) & 3\times \left( -1 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & 6 \\ 2 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 15 & -12 \\ -6 & -3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4+15 & 6-12 \\ 2-6 & 0-3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 19 & -6 \\ -4 & -3 \\ \end{matrix} \right]$

So, 2A $+$ 3B = $\left[ \begin{matrix} 19 & -6 \\ -4 & -3 \\ \end{matrix} \right]$

(ii) $-$3A $+$ 2B

= $-3\times \left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]+2\times \left[ \begin{matrix} 5 & -4 \\ -2 & -1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -3~\times 2 & -3\times 3 \\ -3\times 1 & -3\times 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 2\times 5 & 2\times \left( -4 \right) \\ 2\times \left( -2 \right) & 2\times \left( -1 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -6 & -9 \\ -3 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 10 & -8 \\ -4 & -2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \left( -6 \right)+10 & \left( -9 \right)+\left( -8 \right) \\ -3+\left( -4 \right) & 0+(-2) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & -17 \\ -7 & -2 \\ \end{matrix} \right]$

So, $-$3A $+$ 2B  = $\left[ \begin{matrix} 4 & -17 \\ -7 & -2 \\ \end{matrix} \right]$

(iii) $-$3(A $+$ 2B)

= $-3\left( \left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }+\text{ }\!\!~\!\!\text{ }2\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 5 & -4 \\ -2 & -1 \\ \end{matrix} \right] \right)$

= $-3\left( \left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }+\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 2\times 5 & 2\times \left( -4 \right) \\ 2\times \left( -2 \right) & 2\times \left( -1 \right) \\ \end{matrix} \right] \right)$

= $-3\left( \left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }+\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 10 & -8 \\ -4 & -2 \\ \end{matrix} \right] \right)$

= $-3\left[ \begin{matrix} 2+10 & 3-8 \\ 1-4 & 0+-2 \\ \end{matrix} \right]$

= $-3\left[ \begin{matrix} 12 & -5 \\ -3 & -2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -3~\times 12 & -3~\times \left( -5 \right) \\ -3~\times \left( -3 \right) & -3~\times \left( -2 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -36 & 15 \\ 9 & 6 \\ \end{matrix} \right]$

So, $-$3(A $+$ 2B) = $\left[ \begin{matrix} -36 & 15 \\ 9 & 6 \\ \end{matrix} \right]$

(iv) $\frac{2}{3}$(2A $-$ 3B)

= $\frac{2}{3}\left( 2\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]-\text{ }\!\!~\!\!\text{ }3\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 5 & -4 \\ -2 & -1 \\ \end{matrix} \right] \right)$

= $\frac{2}{3}\left( \left[ \begin{matrix} 2~\times 2 & 2\times 3 \\ 2\times 1 & 2\times 0 \\ \end{matrix} \right]-\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 3\times 5 & 3\times \left( -4 \right) \\ 3\times \left( -2 \right) & 3\times \left( -1 \right) \\ \end{matrix} \right] \right)$

= $\frac{2}{3}\left( \left[ \begin{matrix} 4 & 6 \\ 2 & 0 \\ \end{matrix} \right]-\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 15 & -12 \\ -6 & -3 \\ \end{matrix} \right] \right)$

= $\frac{2}{3}\left( \left[ \begin{matrix} 4-15 & 6+12 \\ 2+6 & 0+3 \\ \end{matrix} \right] \right)$

= $\frac{2}{3}\times \left[ \begin{matrix} -11 & 18 \\ 8 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -\frac{22}{3} & \frac{36}{3} \\ \frac{16}{3} & \frac{6}{3} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -\frac{22}{3} & 12 \\ \frac{16}{3} & 2 \\ \end{matrix} \right]$

So, $\frac{2}{3}$(2A $-$ 3B) = $\left[ \begin{matrix} -\frac{22}{3} & 12 \\ \frac{16}{3} & 2 \\ \end{matrix} \right]$

Sponsored AdsHide Ads

Question:

Find the value of X, if $\left[ \begin{matrix} 2 & 1 \\ 3 & -3 \\ \end{matrix} \right] +$ X = $\left[ \begin{matrix} 4 & -2 \\ -1 & -2 \\ \end{matrix} \right]$.

Difficulty: Easy

Solution:

$\left[ \begin{matrix} 2 & 1 \\ 3 & -3 \\ \end{matrix} \right]~+$ X = $\left[ \begin{matrix} 4 & -2 \\ -1 & -2 \\ \end{matrix} \right]$

X = $\left[ \begin{matrix} 4 & -2 \\ -1 & -2 \\ \end{matrix} \right] - \left[ \begin{matrix} 2 & 1 \\ 3 & -3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4-2 & -2-1 \\ -1-3 & -2+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & -3 \\ -4 & 1 \\ \end{matrix} \right]$

So, X = $\left[ \begin{matrix} 2 & -3 \\ -4 & 1 \\ \end{matrix} \right]$

Question:

If A = $\left[ \begin{matrix} 0 & 1 \\ 2 & -3 \\ \end{matrix} \right]$, B = $\left[ \begin{matrix} -3 & 4 \\ 5 & -2 \\ \end{matrix} \right]$, then prove that

(i) AB $\ne$ BA

Difficulty: Easy

Solution:

(i) AB $\ne$ BA

AB = $\left[ \begin{matrix} 0 & 1 \\ 2 & -3 \\ \end{matrix} \right]\left[ \begin{matrix} -3 & 4 \\ 5 & -2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0\times \left( -3 \right)+1~\times 5 & 0\times 4+1~\times \left( -2 \right) \\ 2\times \left( -3 \right)+\left( -3 \right)~\times 5 & 2\times 4+\left( -3 \right)~\times \left( -2 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0+5 & 0-2 \\ -6-15 & 8+6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 5 & -2 \\ -21 & 14 \\ \end{matrix} \right] \quad\quad\quad\quad\quad ~...(1)$

BA = $\left[ \begin{matrix} -3 & 4 \\ 5 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 1 \\ 2 & -3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \left( -3 \right)\times 0+4~\times 2 & \left( -3 \right)\times 1+4~\times \left( -3 \right) \\ 5\times 0+\left( -2 \right)\times 2 & 5\times 1+\left( -2 \right)\times \left( -3 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0+8 & -3-12 \\ 0-4 & 5+6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 8 & -15 \\ -4 & 11 \\ \end{matrix} \right]\quad\quad\quad\quad\quad ~...(2)$

From $(1)$ and $(2)$, it is clear that AB $\ne$ BA

Question:

If $A = \left[ \begin{matrix} 3 & 2 \\ 1 & -1 \\ \end{matrix} \right]$, $B= \left[ \begin{matrix} 2 & 4 \\ -3 & -5 \\ \end{matrix} \right]$, then verify that

 (i)  $(AB)^{t} = B^{t}A^{t}$ (ii) $(AB)^{-1} = B^{-1}A^{-1}$

Difficulty: Easy

Solution:

(i)  $(AB)^{t} = B^{t}A^{t}$

Left Hand Side: $(AB)^{t}$

$AB$ = $\left[ \begin{matrix} 3 & 2 \\ 1 & -1 \\ \end{matrix} \right] \left[ \begin{matrix} 2 & 4 \\ -3 & -5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3\times 2+2\times \left( -3 \right) & 3\times 4+2\times \left( -5 \right) \\ 1\times 2+\left( -1 \right)\times \left( -3 \right) & 1\times 4+\left( -1 \right)\times \left( -5 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6-6 & 12-10 \\ 2+3 & 4+5 \\ \end{matrix} \right]$

$AB$= $\left[ \begin{matrix} 0 & 2 \\ 5 & 9 \\ \end{matrix} \right]$

${{\left( \text{AB} \right)}^{t}}$ = ${{\left[ \begin{matrix} 0 & 2 \\ 5 & 9 \\ \end{matrix} \right]}^{t}}$

${{\left( \text{AB} \right)}^{t}}$ = $\left[ \begin{matrix} 0 & 5 \\ 2 & 9 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

Right Hand Side: $B^{t}A^{t}$

${{\text{A}}^{t}}$ = ${{\left[ \begin{matrix} 3 & 2 \\ 1 & -1 \\ \end{matrix} \right]}^{t}}$ = $\left[ \begin{matrix} 3 & 1 \\ 2 & -1 \\ \end{matrix} \right]$

${{\text{B}}^{t}}$ = ${{\left[ \begin{matrix} 2 & 4 \\ -3 & -5 \\ \end{matrix} \right]}^{t}}$ = $\left[ \begin{matrix} 2 & -3 \\ 4 & -5 \\ \end{matrix} \right]$

${{\text{B}}^{t}}{{\text{A}}^{t}}$ = $\left[ \begin{matrix} 2 & -3 \\ 4 & -5 \\ \end{matrix} \right]$ $\left[ \begin{matrix} 3 & 1 \\ 2 & -1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2\times 3+\left( -3 \right)\times 2 & 2\times 1+\left( -3 \right)\times \left( -1 \right) \\ 4\times 3+\left( -5 \right)\times 2 & 4\times 1+\left( -5 \right)\times \left( -1 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6-6 & 2~+~3 \\ 12-10 & 4+5 \\ \end{matrix} \right]$

${{\text{B}}^{t}}{{\text{A}}^{t}}$ = $\left[ \begin{matrix} 0 & 5 \\ 2 & 9 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that $(AB)^{t} = B^{t}A^{t}$

(ii) $(AB)^{-1} = B^{-1}A^{-1}$

Left Hand Side: $(AB)^{-1}$

$AB$ = $\left[ \begin{matrix} 3 & 2 \\ 1 & -1 \\ \end{matrix} \right] \left[ \begin{matrix} 2 & 4 \\ -3 & -5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3\times 2+2\times \left( -3 \right) & 3\times 4+2\times \left( -5 \right) \\ 1\times 2+\left( -1 \right)\times \left( -3 \right) & 1\times 4+\left( -1 \right)\times \left( -5 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6-6 & 12-10 \\ 2+3 & 4+5 \\ \end{matrix} \right]$

$AB$= $\left[ \begin{matrix} 0 & 2 \\ 5 & 9 \\ \end{matrix} \right]$

$\left| \text{AB} \right|=0\times 9-2~\times 5=~0-10=~-10\ne 0$

${{\left( \text{AB} \right)}^{-1}}$= $\frac{1}{\left| \text{AB} \right|}Adj~\text{AB}$

= $\frac{1}{-10}\left[ \begin{matrix} 9 & -2 \\ -5 & 0 \\ \end{matrix} \right]$

${{\left( \text{AB} \right)}^{-1}}$ = $\left[ \begin{matrix} \frac{-9}{10} & \frac{1}{5} \\ \frac{1}{2} & 0 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

Right Hand Side: $B^{-1}A^{-1}$

$\text{A}=\left[ \begin{matrix} 3 & 2 \\ 1 & -1 \\ \end{matrix} \right]$

$\left| \text{A} \right|=3\times \left( -1 \right)-1~\times 2=~-3-2=~-5\ne 0$

${{\text{A}}^{-1}}=\frac{1}{\left| \text{A} \right|}Adj~\text{A}$

= $\frac{1}{-5}\left[ \begin{matrix} -1 & -2 \\ -1 & 3 \\ \end{matrix} \right]$

${{\text{A}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{5} & \frac{2}{5} \\ \frac{1}{5} & -\frac{3}{5} \\ \end{matrix} \right]$

$\text{B}=\left[ \begin{matrix} 2 & 4 \\ -3 & -5 \\ \end{matrix} \right]$

$\left| \text{B} \right|=2\times \left( -5 \right)-4~\times \left( -3 \right)=~-10+12=~2\ne 0$

${{\text{B}}^{-1}}=\frac{1}{\left| \text{B} \right|}Adj~\text{B}$

= $\frac{1}{2}\left[ \begin{matrix} -5 & -4 \\ 3 & 2 \\ \end{matrix} \right]$

${{\text{B}}^{-1}}$ = $\left[ \begin{matrix} -\frac{5}{2} & -2 \\ \frac{3}{2} & 1 \\ \end{matrix} \right]$

Now solving ${{\text{B}}^{-1}}{{\text{A}}^{-1}}$

= $\left[ \begin{matrix} -\frac{5}{2} & -2 \\ \frac{3}{2} & 1 \\ \end{matrix} \right]\times \left[ \begin{matrix} \frac{1}{5} & \frac{2}{5} \\ \frac{1}{5} & -\frac{3}{5} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{-5}{2}\times \frac{1}{5}+\left( -2 \right)\times \frac{1}{5} & \frac{-5}{2}\times \frac{2}{5}+\left( -2 \right)\times \frac{-3}{5} \\ \frac{3}{2}\times \frac{1}{5}+1\times \frac{1}{5} & \frac{3}{2}\times \frac{2}{5}+1\times \frac{-3}{5} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{-5}{10}+\frac{-2}{5} & \frac{-10}{10}+\frac{6}{5} \\ \frac{3}{10}+\frac{1}{5} & \frac{6}{10}-\frac{3}{5} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{-1}{2}-\frac{2}{5} & -1+\frac{6}{5} \\ \frac{3}{10}+\frac{1}{5} & \frac{3}{5}-\frac{3}{5} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{-5-4}{10} & \frac{-5+6}{5} \\ \frac{3+2}{10} & \frac{3-3}{5} \\ \end{matrix} \right]$

${{\text{B}}^{-1}}{{\text{A}}^{-1}}$ = $\left[ \begin{matrix} \frac{-9}{10} & \frac{1}{5} \\ \frac{1}{2} & 0 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that $(AB)^{-1} = B^{-1}A^{-1}$

Sponsored AdsHide Ads

## Top Your Class

Subscribe to the premium package and ace your exams using premium features

Go Premium with just USD 4.99/month!