If A and B are n times n square matrices such | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the equation: $(2 A+B)^2+(A-3 B)^2=5 A^2-2 A B+10 B^2$. Expanding the squares,we get: $(4 A^2+2 A B+2 B A+B^2)+(A^2-3 A B-3 B A+9 B^2)=5 A^2

Vedclass · Accepted Answer

$A^2 B^2$

Vedclass · Answer

(D) Given the equation: $(2 A+B)^2+(A-3 B)^2=5 A^2-2 A B+10 B^2$. Expanding the squares,we get: $(4 A^2+2 A B+2 B A+B^2)+(A^2-3 A B-3 B A+9 B^2)=5 A^2-2 A B+10 B^2$. Combining like terms: $(4 A^2+A^2)+(B^2+9 B^2)+(2 A B-3 A B)+(2 B A-3 B A)=5 A^2-2 A B+10 B^2$. This simplifies to: $5 A^2+10 B^2-A B-B A=5 A^2-2 A B+10 B^2$. Subtracting $5 A^2+10 B^2$ from both sides,we get: $-A B-B A=-2 A B$. Adding $A B$ to both sides,we get: $-B A=-A B$,which implies $A B=B A$. Since $A$ and $B$ commute,$A B A B=A(B A) B=A(A B) B=(A A)(B B)=A^2 B^2$.

If $A$ and $B$ are $n \times n$ square matrices such that $(2 A+B)^2+(A-3 B)^2=5 A^2-2 A B+10 B^2$,then $A B A B=$

Similar Questions

In an upper triangular matrix of order $n \times n$,the minimum number of zeros is:

If $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$,then $(aI + bA)^n$ is (where $I$ is the identity matrix of order $2$)

If $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that $A^{2} = I$,then

If $A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$ and $A^2$ is the identity matrix,then $x =$

If $A = [1\, 2\, 3]$ and $B = \begin{bmatrix} -5 & 4 & 0 \\ 0 & 2 & -1 \\ 1 & -3 & 2 \end{bmatrix}$,then $AB = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module