Home
Class 12
MATHS
Let A+2B=[{:(2,4,0),(6,-3,3),(-5,3,5):}]...

Let `A+2B=[{:(2,4,0),(6,-3,3),(-5,3,5):}] and 2A-B=[{:(6,-2,4),(6,1,5),(6,3,4):}]` , then tr (A) - tr (B) is equal to (where , tr (A) =n trace of matrix x A i.e. . Sum of the principle diagonal elements of matrix A)

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the values of matrices A and B from the given equations and then compute the traces of these matrices. Finally, we will find the difference between the traces of A and B. ### Step 1: Write down the equations We have two equations: 1. \( A + 2B = \begin{pmatrix} 2 & 4 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 5 \end{pmatrix} \) (Equation 1) 2. \( 2A - B = \begin{pmatrix} 6 & -2 & 4 \\ 6 & 1 & 5 \\ 6 & 3 & 4 \end{pmatrix} \) (Equation 2) ### Step 2: Multiply Equation 2 by 2 Multiply the second equation by 2: \[ 4A - 2B = 2 \times \begin{pmatrix} 6 & -2 & 4 \\ 6 & 1 & 5 \\ 6 & 3 & 4 \end{pmatrix} = \begin{pmatrix} 12 & -4 & 8 \\ 12 & 2 & 10 \\ 12 & 6 & 8 \end{pmatrix} \] (Equation 3) ### Step 3: Add Equation 1 and Equation 3 Now, we add Equation 1 and Equation 3: \[ (A + 2B) + (4A - 2B) = \begin{pmatrix} 2 & 4 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 5 \end{pmatrix} + \begin{pmatrix} 12 & -4 & 8 \\ 12 & 2 & 10 \\ 12 & 6 & 8 \end{pmatrix} \] This simplifies to: \[ 5A = \begin{pmatrix} 2 + 12 & 4 - 4 & 0 + 8 \\ 6 + 12 & -3 + 2 & 3 + 10 \\ -5 + 12 & 3 + 6 & 5 + 8 \end{pmatrix} = \begin{pmatrix} 14 & 0 & 8 \\ 18 & -1 & 13 \\ 7 & 9 & 13 \end{pmatrix} \] ### Step 4: Solve for A Now, divide by 5 to find A: \[ A = \begin{pmatrix} \frac{14}{5} & 0 & \frac{8}{5} \\ \frac{18}{5} & -\frac{1}{5} & \frac{13}{5} \\ \frac{7}{5} & \frac{9}{5} & \frac{13}{5} \end{pmatrix} \] ### Step 5: Calculate the trace of A The trace of A, \( \text{tr}(A) \), is the sum of the diagonal elements: \[ \text{tr}(A) = \frac{14}{5} - \frac{1}{5} + \frac{13}{5} = \frac{14 - 1 + 13}{5} = \frac{26}{5} = 5.2 \] ### Step 6: Substitute A back into Equation 1 to find B From Equation 1: \[ A + 2B = \begin{pmatrix} 2 & 4 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 5 \end{pmatrix} \] Substituting A: \[ \begin{pmatrix} \frac{14}{5} & 0 & \frac{8}{5} \\ \frac{18}{5} & -\frac{1}{5} & \frac{13}{5} \\ \frac{7}{5} & \frac{9}{5} & \frac{13}{5} \end{pmatrix} + 2B = \begin{pmatrix} 2 & 4 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 5 \end{pmatrix} \] This gives: \[ 2B = \begin{pmatrix} 2 - \frac{14}{5} & 4 - 0 & 0 - \frac{8}{5} \\ 6 - \frac{18}{5} & -3 + \frac{1}{5} & 3 - \frac{13}{5} \\ -5 - \frac{7}{5} & 3 - \frac{9}{5} & 5 - \frac{13}{5} \end{pmatrix} \] Calculating each element: \[ 2B = \begin{pmatrix} \frac{10 - 14}{5} & 4 & -\frac{8}{5} \\ \frac{30 - 18}{5} & -\frac{15 - 1}{5} & \frac{15 - 13}{5} \\ -\frac{25 + 7}{5} & \frac{15 - 9}{5} & \frac{25 - 13}{5} \end{pmatrix} \] \[ = \begin{pmatrix} -\frac{4}{5} & 4 & -\frac{8}{5} \\ \frac{12}{5} & -\frac{14}{5} & \frac{2}{5} \\ -\frac{32}{5} & \frac{6}{5} & \frac{12}{5} \end{pmatrix} \] ### Step 7: Solve for B Now, divide by 2 to find B: \[ B = \begin{pmatrix} -\frac{2}{5} & 2 & -\frac{4}{5} \\ \frac{6}{5} & -\frac{7}{5} & \frac{1}{5} \\ -\frac{16}{5} & \frac{3}{5} & \frac{6}{5} \end{pmatrix} \] ### Step 8: Calculate the trace of B The trace of B, \( \text{tr}(B) \), is: \[ \text{tr}(B) = -\frac{2}{5} - \frac{7}{5} + \frac{6}{5} = -\frac{3}{5} \] ### Step 9: Calculate \( \text{tr}(A) - \text{tr}(B) \) Now, we find: \[ \text{tr}(A) - \text{tr}(B) = 5.2 - \left(-\frac{3}{5}\right) = 5.2 + 0.6 = 5.8 \] ### Final Answer Thus, \( \text{tr}(A) - \text{tr}(B) = 5.8 \). ---
Promotional Banner

Similar Questions

Explore conceptually related problems

Let A+2B=[(1,2,0),(6,-3,3),(-5,3,1)] and 2A-B=[(2,-1,5),(2,-1,6),(0,1,2)], then find tr(A)-tr(B).

If A=[{:(1,2),(4,1),(5,6):}] and B=[{:(1,2),(6,4),(7,3):}] , then varify that (i) (A-B)'=A'-B'

If A=[{:(2,-1,3),(4,5,-6):}] and B=[{:(1,2),(3,4),(5,-6):}] then

For xgt 0. " let A"=[(x+(1)/(x),0,0),(0,1//x,0),(0,0,12)], B=[((x)/(6(x^(2)+1)),0,0),(0,(x)/(4),0),(0,0,(1)/(36))] be two matrices and C=AB+(AB)^(2)+….+(AB)^(n). Then, Tr(lim_(nrarroo)C) is equal to (where Tr(A) is the trace of the matrix A i.e. the sum of the principle diagonal elements of A)

"If A"=[{:(,8,0),(,4,-2),(,3,6):}] and B=[{:(,2,-2),(,4,2),(,-5,1):}] then find the matrix X, such that 2A+3X=5B

If A=[(2, 1,-1),(3, 5,2),(1, 6, 1)] , then tr(Aadj(adjA)) is equal to (where, tr (P) denotes the trace of the matrix P i.e. the sum of all the diagonal elements of the matrix P and adj(P) denotes the adjoint of matrix P)

if A=[{:(1,2,3),(4,5,6):}]and B=[{:(-3,-2),(0,1),(-4,-5):}], then find AB and BA ,

if A=[{:(1,2,-5),(-3,4,6):}]and B[{:(-2,3,-4),(1,2,3):}]' then find 2A+B.

if A=[{:(1,6),(2,4),(-3,5):}]B=[{:(3,4),(1,-2),(2,-1):}], then find a matrix C such that 2A-B+c=0

Consider the matrix A=[(x, 2y,z),(2y,z,x),(z,x,2y)] and A A^(T)=9I. If Tr(A) gt0 and xyz=(1)/(6) , then the vlaue of x^(3)+8y^(3)+z^(3) is equal to (where, Tr(A), I and A^(T) denote the trace of matrix A i.e. the sum of all the principal diagonal elements, the identity matrix of the same order of matrix A and the transpose of matrix A respectively)