SATHEE: Exercise 03

03 आव्यूह

अभ्यास 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}]अध्ययन सामग्री पर वापस जेईई अध्ययन सामग्री (गणित) 01 संबंध एवं फलन अभ्यास 01 अभ्यास 02 अभ्यास 03 02 व्युत्क्रम त्रिकोणमितीय फलन अभ्यास 01 अभ्यास 02 अभ्यास विविध समाधान 03 आव्यूह अभ्यास 01 अभ्यास 02 अभ्यास 03 अभ्यास 04 विविध समाधान 04 सारणिक अभ्यास 01 अभ्यास 02 अभ्यास 03 अभ्यास 04 अभ्यास 05 अभ्यास 06 विविध अभ्यास 05 सांत्यता और अवकलनीयता अभ्यास 01 अभ्यास 02 अभ्यास 03 अभ्यास 04 अभ्यास 05 अभ्यास 06 अभ्यास 07 अभ्यास 08 विविध अभ्यास 06 अवकलज का अनुप्रयोग अभ्यास 01 अभ्यास 02 अभ्यास 03 अभ्यास 04 07 समाकलन अभ्यास करें विविध अभ्यास 08 समाकलन का अनुप्रयोग अभ्यास करें विविध अभ्यास 09 वैक्टर अभ्यास 01 विविध समाधान 10 त्रिविमीय ज्यामिति का परिचय अभ्यास 01 अभ्यास 02 अभ्यास 03 विविध अभ्यास 11 रैखिक प्रोग्रामिंग अभ्यास 01 अभ्यास 02 रैखिक प्रोग्रामिंग विविध अभ्यास 12 प्रायिकता अभ्यास 01 अभ्यास 02 अभ्यास 03 अभ्यास 04 अभ्यास 05 प्रायिकता विविध अभ्यास हमारे पास पहुंचें.footer-bottom-wrapper{background-color: #f9fafc; padding: 10px 0px; margin-left: -15px; margin-right: -15px; width: calc(100% + 30px);}
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.ncert-link li:last-child{margin-right: 0px; padding-right: 0px; border-right: 0px;}अधिक विषय मेरे साथ हल करें JEE का तोड़ Webinar JEE-टेस्ट एनसीईआरटी पुस्तक और उदाहरण समाचार घोषणाएं एनसीईआरटी अध्याय अभ्यास वीडियो समाधान ट्यूटोरियल सत्र एनसीईआरटी पुस्तकें स्कूलों के लिए अभिभावकों के लिए मेंटरशिप प्राप्त करें मुझसे संपर्क करें prutor@iitk.ac.in SATHEE मेंटरशिप JEE (Main) के लिए पाठ्यक्रम सहायता वीडियो एनसीईआरटी समाधान NCERT भौतिक विज्ञान 11 समाधान NCERT भौतिक विज्ञान 12 समाधान NCERT रसायन विज्ञान 11 समाधान NCERT रसायन विज्ञान 12 समाधान NCERT गणित 11 समाधान NCERT गणित 12 समाधान हमारी साइटें 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 गोपनीयता नीति Prutor@IITK द्वारा संचालितfunction disableBtn(id){
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