SATHEE: Exercise 03

03 ମ୍ୟାଟ୍ରିକ୍ସ

ବ୍ୟାୟାମ 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}]ଅଧ୍ୟୟନ ସାମଗ୍ରୀକୁ ଫେରନ୍ତୁ JEE ଅଧ୍ୟୟନ ସାମଗ୍ରୀ (ଗଣିତ) 01 ସମ୍ପର୍କ ଏବଂ କାର୍ଯ୍ୟ ବ୍ୟାୟାମ 01 ବ୍ୟାୟାମ 02 ବ୍ୟାୟାମ 03 02 ଓଲଟା ଟ୍ରାଇଗୋନେଟ୍ରିକ୍ କାର୍ଯ୍ୟଗୁଡ଼ିକ ବ୍ୟାୟାମ 01 ବ୍ୟାୟାମ 02 ବିଭିନ୍ନ ସମାଧାନ ବ୍ୟାୟାମ କରନ୍ତୁ 03 ମ୍ୟାଟ୍ରିକ୍ସ ବ୍ୟାୟାମ 01 ବ୍ୟାୟାମ 02 ବ୍ୟାୟାମ 03 ବ୍ୟାୟାମ 04 ବିବିଧ ସମାଧାନ 04 ନିର୍ଣ୍ଣୟକାରୀ ବ୍ୟାୟାମ 01 ବ୍ୟାୟାମ 02 ବ୍ୟାୟାମ 03 ବ୍ୟାୟାମ 04 ବ୍ୟାୟାମ 05 ବ୍ୟାୟାମ 06 ବିବିଧ ବ୍ୟାୟାମ 05 ନିରନ୍ତରତା ଏବଂ ଭିନ୍ନତା ବ୍ୟାୟାମ 01 ବ୍ୟାୟାମ 02 ବ୍ୟାୟାମ 03 ବ୍ୟାୟାମ 04 ବ୍ୟାୟାମ 05 ବ୍ୟାୟାମ 06 ବ୍ୟାୟାମ 07 ବ୍ୟାୟାମ 08 ବିବିଧ ବ୍ୟାୟାମ 06 ଡେରିଭେଟିକ୍ସର ପ୍ରୟୋଗ ବ୍ୟାୟାମ 01 ବ୍ୟାୟାମ 02 ବ୍ୟାୟାମ 03 ବ୍ୟାୟାମ 04 07 ଇଣ୍ଟିଗ୍ରାଲ୍ ବ୍ୟାୟାମ ବିବିଧ ବ୍ୟାୟାମ 08 ଇଣ୍ଟିଗ୍ରାଲ୍ସର ପ୍ରୟୋଗ ବ୍ୟାୟାମ ବିବିଧ ବ୍ୟାୟାମ 09 ଭେକ୍ଟର୍ ବ୍ୟାୟାମ 01 ବିବିଧ ସମାଧାନ 10 ତିନୋଟି ଡାଇମେନ୍ସନାଲ୍ ଜ୍ୟାମିତି ବ୍ୟାୟାମ 01 ବ୍ୟାୟାମ 02 ବ୍ୟାୟାମ 03 ବିବିଧ ସେନା 11 ରେଖା ପ୍ରୋଗ୍ରାମିଂ ବ୍ୟାୟାମ 01 ବ୍ୟାୟାମ 02 ରେଖା ପ୍ରୋଗ୍ରାମିଂ ବିଭିନ୍ନ ବ୍ୟାୟାମ 12 ସମ୍ଭାବନା ବ୍ୟାୟାମ 01 ବ୍ୟାୟାମ 02 ବ୍ୟାୟାମ 03 ବ୍ୟାୟାମ 04 ବ୍ୟାୟାମ 05 ସମ୍ଭାବ୍ୟ ବିବିଧ ବ୍ୟାୟାମ ଆମ ପାଖରେ ପହଞ୍ଚ |.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;}ଅଧିକ ବିଷୟଗୁଡିକ ମୋ ସହିତ ସମାଧାନ କର JEE ବ୍ରେକ୍ ୱେବିନାର୍ JEE-ପରୀକ୍ଷା NCERT ବୁକ୍ ଏବଂ ଉଦାହରଣ ସମ୍ବାଦ ଘୋଷଣା NCERT ଅଧ୍ୟାୟ ବ୍ୟାୟାମ ଭିଡିଓ ସମାଧାନ | ଟ୍ୟୁଟୋରିଆଲ୍ ଅଧିବେଶନ NCERT ପୁସ୍ତକଗୁଡ଼ିକ ବିଦ୍ୟାଳୟଗୁଡ଼ିକ ପାଇଁ ପିତାମାତାମାନଙ୍କ ପାଇଁ ମେଣ୍ଟରସିପ୍ ପାଆନ୍ତୁ ମୋ ସହିତ ଯୋଗାଯୋଗ କର prutor@iitk.ac.in ସାଥେ ମେଣ୍ଟରସିପ୍ JEE ପାଇଁ ସିଲାବସ୍ (ମୁଖ୍ୟ) ଭିଡିଓଗୁଡିକ ସାହାଯ୍ୟ କରନ୍ତୁ NCERT ସମାଧାନ NCERT ପଦାର୍ଥ ବିଜ୍ଞାନ 11 ସମାଧାନ NCERT ପଦାର୍ଥ ବିଜ୍ଞାନ 12 ସମାଧାନ NCERT ରସାୟନ 11 ସମାଧାନ NCERT ରସାୟନ 12 ସମାଧାନ NCERT ଗଣିତ 11 ସମାଧାନ NCERT ଗଣିତ 12 ସମାଧାନ ଆମର ସାଇଟ୍ 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 ଗୋପନୀୟତା ନୀତି Prutor@IITK ଦ୍ଵାରା ନିର୍ମିତfunction disableBtn(id){
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