SATHEE: Exercise 03

03 ਮੈਟ੍ਰਿਕਸ

ਅਭਿਆਸ 03

Question:

Find 1/2(A+ $\begin{matrix} A^{T} \end{matrix}$ ) and 1/2(A− $\begin{matrix} A^{T} \end{matrix}$ ), when A= $[\begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix}]$

Answer:

Find A^T:

A^T = $[\begin{matrix} 0 & -a & -b \\ a & 0 & c \\ b & -c & 0 \end{matrix}]$

Find 1/2(A + A^T):

1/2(A + A^T) = $[\begin{matrix} 0 & 0 & 0 \\ 0 & a & c \\ 0 & c & 0 \end{matrix}]$

Find 1/2(A - A^T):

1/2(A - A^T) = $[\begin{matrix} 0 & a & b \\ -a & 0 & -c \\ -b & c & 0 \end{matrix}]$

Question:

If A= $[\begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{matrix}]$ and B= $[\begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix}]$ , then verify that (i) $\begin{matrix} {(A+B)}^{T} \end{matrix}$ = $\begin{matrix} A^{T} \end{matrix}$ + $\begin{matrix} B^{T} \end{matrix}$ (ii) $\begin{matrix} {(A-B)}^{T} \end{matrix}$ = $\begin{matrix} A^{T} \end{matrix}$ − $\begin{matrix} B^{T} \end{matrix}$

Answer:

(i) A+B = $[\begin{matrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{matrix}]$

(A+B)^T = $[\begin{matrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{matrix}]$

A^T = $[\begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{matrix}]$

B^T = $[\begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{matrix}]$

A^T + B^T = $[\begin{matrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{matrix}]$

Therefore, (A+B)^T = A^T + B^T

(ii) A-B = $[</Question:If (i) A=[\begin{matrix} cosα & sinα \\ -sinα & cosα \end{matrix}], then verify that A'A=I (ii) A=[\begin{matrix} sinα & cosα \\ -cosα & sinα \end{matrix}], then verify that A'A=IAnswer:(i) A’A = [\begin{matrix} cos2α & sinαcosα \\ sinαcosα & sin2α \end{matrix}]Since cos2α + sin2α = 1, A’A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]which is the identity matrix I.(ii) A’A = [\begin{matrix} sin2α & sinαcosα \\ sinαcosα & cos2α \end{matrix}]Since sin2α + cos2α = 1, A’A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]which is the identity matrix I.Question:Find the transpose of each of the following matrices: (i) [\begin{matrix} 5 \\ 1/2 \\ -1 \end{matrix}] (ii) [\begin{matrix} 1 & -1 \\ 2 & 3 \end{matrix}] (iii) [\begin{matrix} -1 & 5 & 6 \\ \sqrt3 & 5 & 6 \\ 2 & 3 & -1 \end{matrix}]Answer:(i) [\begin{matrix} 5 & 1/2 & -1 \end{matrix}](ii) [\begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix}](iii) [\begin{matrix} -1 & \sqrt3 & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{matrix}]Question:If A= [\begin{matrix} cosα & -sinα \\ sinα & cosα \end{matrix}], then A+A'=I, if the value of α is A π/6 B π/3 C n D 3π/2Answer:A π/6A= [\begin{matrix} cos π / 6 & -sin π / 6 \\ sin π / 6 & cos π / 6 \end{matrix}]A'= [\begin{matrix} cos π / 6 & sin π / 6 \\ -sin π / 6 & cos π / 6 \end{matrix}]A+A'= [\begin{matrix} 2 cos π / 6 & 0 \\ 0 & 2 cos π / 6 \end{matrix}]A+A'= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]Therefore, A+A'=I and the value of α is A π/6.Question:For the matrices A and B, verify that (AB)'=B'A' where (i) A= [\begin{matrix} 1 \\ -4 \\ 3 \end{matrix}],B= [\begin{matrix} -1 & 2 & 1 \end{matrix}] (ii) A= [\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}],B= [\begin{matrix} 1 & 5 & 7 \end{matrix}]Answer:(i)
A= [\begin{matrix} 1 \\ -4 \\ 3 \end{matrix}],
B= [\begin{matrix} -1 & 2 & 1 \end{matrix}]Step 1: Calculate AB[\begin{matrix} -1 \\ -8 \\ 5 \end{matrix}]Step 2: Calculate (AB)'[\begin{matrix} -1 & -8 & 5 \end{matrix}]Step 3: Calculate B'[\begin{matrix} -1 & 2 & 1 \end{matrix}]Step 4: Calculate B'A'[\begin{matrix} 1 & -4 & 3 \end{matrix}]Therefore, (AB)'=B'A'.(ii)
A= [\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}],
B= [\begin{matrix} 1 & 5 & 7 \end{matrix}]Step 1: Calculate AB[\begin{matrix} 5 \\ 7 \\ 17 \end{matrix}]Step 2: Calculate (AB)'[\begin{matrix} 5 & 7 & 17 </m \end{matrix}Question:For the matrix A=[\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}], verify that (i) (A+A') is a symmetric matrix (ii) (A-A') is a skew symmetric matrix.Answer:(i) To verify that (A+A') is a symmetric matrix, we need to show that it is equal to its transpose.A+A' = [\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}] + [\begin{matrix} 1 & 6 \\ 5 & 7 \end{matrix}]= [\begin{matrix} 2 & 11 \\ 11 & 14 \end{matrix}]A'+A = [\begin{matrix} 1 & 6 \\ 5 & 7 \end{matrix}] + [\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}]= [\begin{matrix} 2 & 11 \\ 11 & 14 \end{matrix}]Since A+A' = A'+A, it is a symmetric matrix.(ii) To verify that (A-A') is a skew symmetric matrix, we need to show that it is equal to its negative transpose.A-A' = [\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}] - [\begin{matrix} 1 & 6 \\ 5 & 7 \end{matrix}]= [\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}]A'-A = <math xmlns = “ http://www.w3.org/1998Question:If A,B are symmetric matrices of same order, then AB-BA is a ,
A Skew symmetric matrix
B Symmetric matrix
C Zero matrix
D Identity matrixAnswer:Answer: C Zero matrixQuestion:Express the following matrices as the sum of a symmetric and a skew symmetric matrix: (i) [\begin{matrix} 3 & 5 \\ 1 & -1 \end{matrix}] (ii) [\begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix}] (iii) [\begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{matrix}] (iv) [\begin{matrix} 1 & 5 \\ -1 & 2 \end{matrix}]Answer:(i) [\begin{matrix} 3 & 5 \\ 1 & -1 \end{matrix}]Symmetric Matrix: [\begin{matrix} 2 & 4 \\ 4 & -2 \end{matrix}]Skew Symmetric Matrix: [\begin{matrix} 1 & 1 \\ -1 & 1 \end{matrix}](ii) [\begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix}]Symmetric Matrix: [\begin{matrix} 4 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 4 \end{matrix}]Skew Symmetric Matrix: [\begin{matrix} 2 & 2 & 0 \\ -2 & 1 & -1 \\ 0 & 1 & -2 \end{matrix}](iii) [\begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & <mtd \end{matrix}Question:(i) Show that the matrix A=[\begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix}]is a symmetric matrix (ii) Show that the matrix A=[\begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix}]is a skew symmetric matrix.Answer:(i) A matrix A is symmetric if A = AT, where AT is the transpose of A.Let A = [\begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix}]Then, AT = [\begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix}]Since A = AT, the matrix A is symmetric.(ii) A matrix A is skew symmetric if A = -AT, where AT is the transpose of A.Let A = [\begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix}]Then, AT = [\begin{matrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{matrix}]Since A = -AT, the matrix A is skew symmetric.Question:If A'= [\begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{matrix}] and B= [\begin{matrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{matrix}], then verify that: (i)(A+B)'=A'+B' (ii)(A-B)'=A'-B'Answer:(i) To verify that (A+B)'=A'+B', we need to calculate (A+B)' and A'+B' separately and then compare them.(A+B)' = [\begin{matrix} 2 & 6 & 1 \\ 0 & 4 & 4 \\ 1 & 3 & 4 \end{matrix}]A'+B' = [\begin{matrix} 2 & 6 & 1 \\ 0 & 4 & 4 \\ 1 & 3 & 4 \end{matrix}]Since (A+B)'=A'+B', we can conclude that (i) is true.(ii) To verify that (A-B)'=A'-B', we need to calculate (A-B)' and A'-B' separately and then compare them.(A-B)' = [\begin{matrix} 4 & 2 & -1 \\ -1 & 0 & -2 \\ 1 & -1 & 2 \end{matrix}]A'-B' = [\begin{matrix} 4 & 2 & -1 \\ -1 & 0 & -2 \\ 1 & -1 & 2 \end{matrix}]Since (A-B)'=A'-B', we can conclude that (ii) is true.Question:If A'= [\begin{matrix} -2 & 3 \\ 1 & 2 \end{matrix}] and B= [\begin{matrix} -1 & 0 \\ 1 & 2 \end{matrix}], then find (A+2B)'.Answer:Step 1: Add A and 2B to get (A + 2B)
A + 2B = [\begin{matrix} -3 & 6 \\ 3 & 6 \end{matrix}]Step 2: Find the transpose of (A + 2B)
(A + 2B)' = [\begin{matrix} -3 & 3 \\ 6 & 6 \end{matrix}]ਅਧਿਐਨ ਸਮੱਗਰੀ ਤੇ ਵਾਪਸ ਜਾਓ ਜੇਈਈ ਅਧਿਐਨ ਸਮੱਗਰੀ (ਗਣਿਤ) 01 ਸਬੰਧ ਅਤੇ ਕਾਰਜ ਅਭਿਆਸ 01 ਅਭਿਆਸ 02 ਅਭਿਆਸ 03 02 ਉਲਟ ਤਿਕੋਣਮਿਤੀ ਫੰਕਸ਼ਨ ਅਭਿਆਸ 01 ਅਭਿਆਸ 02 ਕਸਰਤ ਫੁਟਕਲ ਹੱਲ 03 ਮੈਟ੍ਰਿਕਸ ਅਭਿਆਸ 01 ਅਭਿਆਸ 02 ਅਭਿਆਸ 03 ਅਭਿਆਸ 04 ਫੁਟਕਲ ਹੱਲ 04 ਨਿਰਧਾਰਕ ਅਭਿਆਸ 01 ਅਭਿਆਸ 02 ਅਭਿਆਸ 03 ਅਭਿਆਸ 04 ਅਭਿਆਸ 05 ਅਭਿਆਸ 06 ਫੁਟਕਲ ਅਭਿਆਸ 05 ਨਿਰੰਤਰਤਾ ਅਤੇ ਵਿਭਿੰਨਤਾ ਅਭਿਆਸ 01 ਅਭਿਆਸ 02 ਅਭਿਆਸ 03 ਅਭਿਆਸ 04 ਅਭਿਆਸ 05 ਅਭਿਆਸ 06 ਅਭਿਆਸ 07 ਅਭਿਆਸ 08 ਫੁਟਕਲ ਅਭਿਆਸ 06 ਡੈਰੀਵੇਟਿਵਜ਼ ਦੀ ਐਪਲੀਕੇਸ਼ਨ ਅਭਿਆਸ 01 ਅਭਿਆਸ 02 ਅਭਿਆਸ 03 ਅਭਿਆਸ 04 07 ਅਟੁੱਟ ਐਜ਼ਕਸਰਸਾਈਜ਼ ਫੁਟਕਲ ਅਭਿਆਸ 08 ਇੰਟੀਗ੍ਰੇਲਸ ਦੀ ਐਪਲੀਕੇਸ਼ਨ ਐਜ਼ਕਸਰਸਾਈਜ਼ ਫੁਟਕਲ ਅਭਿਆਸ 09 ਵੈਕਟਰ ਅਭਿਆਸ 01 ਫੁਟਕਲ ਹੱਲ 10 ਤਿੰਨ ਅਯਾਮੀ ਜਿਓਮੈਟਰੀ ਅਭਿਆਸ 01 ਅਭਿਆਸ 02 ਅਭਿਆਸ 03 ਫੁਟਕਲ ਫੌਜ 11 ਲੀਨੀਅਰ ਪ੍ਰੋਗਰਾਮਿੰਗ ਅਭਿਆਸ 01 ਅਭਿਆਸ 02 ਲੀਨੀਅਰ ਪ੍ਰੋਗਰਾਮਿੰਗ ਫੁਟਕਲ ਅਭਿਆਸ 12 ਸੰਭਾਵਨਾ ਅਭਿਆਸ 01 ਅਭਿਆਸ 02 ਅਭਿਆਸ 03 ਅਭਿਆਸ 04 ਅਭਿਆਸ 05 ਸੰਭਾਵਨਾ ਫੁਟਕਲ ਅਭਿਆਸ ਸਾਡੇ ਤੱਕ ਪਹੁੰਚੋ.footer-bottom-wrapper{background-color: #f9fafc; padding: 10px 0px; margin-left: -15px; margin-right: -15px; width: calc(100% + 30px);}
#contactus input, #contactus textarea{background-color: #C0CCDA; color: #000;}
#contactus input:focus, #contactus textarea:focus{background-color: #fff !important;}

.floatingChatBtn{position: fixed; bottom: 175px; right: 10px; background-color: #fff; z-index: 999999999; width: 54px; height: 54px; display: flex; justify-content: center; align-items: center; border-radius: 50%; box-shadow: 0px 2px 6px -1px #ddd;}
.floatingChatBtn a{font-size: 36px; padding: 5px;}

#res-msg span, #response-msg span{display: block;}

.our-sites-link li a{padding: 5px; background-color: #334d6a; color: #e8e8e8; border-radius: 5px; text-decoration: none;}
.ncert-link li{border-right: 1px solid #586c7c; padding-right: 10px;}
.ncert-link li:last-child{margin-right: 0px; padding-right: 0px; border-right: 0px;}ਹੋਰ ਵਿਸ਼ੇ ਮੇਰੇ ਨਾਲ ਹੱਲ ਕਰੋ ਜੇਈਈ ਬਰੇਕ ਵੈਬਿਨਾਰ JEE-ਟੇਸਟ NCERT ਕਿਤਾਬਾਂ ਅਤੇ ਉਦਾਹਰਣ ਖ਼ਬਰਾਂ ਘੋਸ਼ਣਾਵਾਂ NCERT ਪਾਠ ਅਭਿਆਸ ਅਤੇ ਵੀਡੀਓ ਹੱਲ ਟਿਊਟੋਰਿਅਲ ਸੈਸ਼ਨ NCERT ਕਿਤਾਬਾਂ ਸਕੂਲਾਂ ਲਈ ਮਾਤਾ-ਪਿਤਾ ਲਈ ਮੇੰਟੋਰਸ਼ਿਪ ਹਾਸਲ ਕਰੋ ਸੰਪਰਕ ਕਰੋ prutor@iitk.ac.in SATHEE ਸਲਾਹਕਾਰ JEE (Main) ਦੇ ਲਈ ਸਿਲੇਬਸ ਸਹਾਇਤਾ ਲਈ ਵੀਡੀਓਜ਼ NCERT ਹੱਲ NCERT ਭੌਤਿਕ ਵਿਗਿਆਨ 11 ਦਾ ਹੱਲ NCERT ਭੌਤਿਕ ਵਿਗਿਆਨ 12 ਦਾ ਹੱਲ NCERT ਰਸਾਇਣ ਵਿਗਿਆਨ 11 ਦਾ ਹੱਲ NCERT ਰਸਾਇਣ ਵਿਗਿਆਨ 12 ਦਾ ਹੱਲ NCERT ਗਣਿਤ 11 ਦਾ ਹੱਲ NCERT ਗਣਿਤ 12 ਦਾ ਹੱਲ ਸਾਡੀਆਂ ਸਾਈਟਾਂ SATHEE JEE SATHEE NEET *IIT Kanpur and SATHEE logos are the property of IIT Kanpur. Robust Results Private Limited has been allowed to use IIT Kanpur and SATHEE logos for this project. © 2023 Copyright SATHEE ਪਰਾਈਵੇਟ ਨੀਤੀ Prutor@IITK ਦੁਆਰਾ ਸੰਚਾਲਿਤfunction disableBtn(id){
document.getElementById(id).disabled = true;
}
function enableBtn(id){
document.getElementById(id).disabled = false;
}$(document).ready(function(){
$('[data-toggle="tooltip"]').tooltip();
});$(document).ready(function(){
window.onscroll = function () { scrollFunction() };
console.log("jQuery working...");
});

function scrollFunction() {
var stButton = document.getElementById("st-button");
if (document.body.scrollTop > 20 || document.documentElement.scrollTop > 20) {
stButton.style.display = "block";
} else {
stButton.style.display = "none";
}
}
function topFunction() {
window.scrollTo({top: 0, behavior: 'smooth'});
}Punjabi English Hindi Telugu Oriyavar doNotTrack = false;
if (!doNotTrack) {
window.dataLayer = window.dataLayer || [];
function gtag(){dataLayer.push(arguments);}
gtag('js', new Date());
gtag('config', 'G-CFXR84401Y', { 'anonymize_ip': false });
}$