Respuesta :
Answer:
[tex]\large\boxed{B=A+2I}[/tex]
Step-by-step explanation:
It's possible if dimensions of a matrix A and matrix B are n × n
[tex]AB=A^2+2A\qquad\text{multiply both sides on the left by}\ A^{-1}\\\\A^{-1}AB=A^{-1}A^2+A^{-1}(2A)\qquad\text{we know}\ A^{-1}A=I\\\\IB=A^{-1}A\cdot A+2A^{-1}A\\\\IB=IA+2I\qquad\text{we know}\ IA=A\\\\B=A+2I[/tex]
Matrices is an array of numbers, usually 2 dimensional, but can be single dimensional too.
A matrix B such that [tex]AB = A^2 + 2A[/tex] is given as
[tex]B = A + 2I = \left[\begin{array}{cc}4&0\\0&4\end{array}\right][/tex]
(Assuming A is left invertible and [tex]A = \left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex])
When can we cancel out matrix multiplied on both sides of an equation?
Suppose that there is an equation
[tex]AB = AC[/tex]
We cannot always say that [tex]B = C[/tex]
If we assume that A is left invertible, then only we can surely say that we have got [tex]B = C[/tex]
Similarly, for [tex]BA = CA[/tex] to imply [tex]B = C[/tex], we need A to be right invertible.
Assuming that we have A as a left invertible matrix, say
[tex]A = \left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex]
and [tex]L_A[/tex] be its left inverse, then [tex]L_A A = I[/tex] ([tex]I[/tex] is identity matrix)
Then,
[tex]AB = A^2 + 2A\\A(B) = A(A + 2I_2)\\\\\Multiplying L_{A}\text{ on left side of both terms,}\\\\L_{A} AB = L_{A}A(A + 2I_2)\\B = A + 2I_2\\\\B = \left[\begin{array}{cc}2&0\\0&2\end{array}\right] + \left[\begin{array}{cc}2&0\\0&2\end{array}\right] = \left[\begin{array}{cc}4&0\\0&4\end{array}\right] = 4I_2\\\\B = 4I_2[/tex]
Thus, i
A matrix B such that [tex]AB = A^2 + 2A[/tex] is given as
[tex]B = \left[\begin{array}{cc}4&0\\0&4\end{array}\right][/tex]
(Assuming A is left invertible and [tex]A = \left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex])
Learn more about invertible matrices here:
https://brainly.com/question/17027442
