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Chapter 6: Problem 64

Find two matrices \(A\) and \(B\) such that \(A B=B A\)

Short Answer

Expert verified
So, two matrices \(A\) and \(B\) that satisfy \(A B=B A\) are \(A = \[ \[1, 0\], \[0, 1\] \] \) and \(B = \[ \[2, 0\], \[0, 2\] \]\).

Step by step solution

01

Choose Matrix A

Let's choose a simple diagonal matrix for \(A\). For instance, \(A = \[ \[1, 0\], \[0, 1\] \] \). This is the 2x2 Identity matrix.
02

Choose Matrix B

Likewise, for \(B\), let's also choose a diagonal matrix. For instance, \(B = \[ \[2, 0\], \[0, 2\] \] \). This is a scalar matrix, meaning all its entries are the same.
03

Verify the Commutative Property

Now, let's multiply \(A\) and \(B\) and \(B\) and \(A\) to see if they are indeed equal. \(A \times B = \[ \[1, 0\], \[0, 1\] \] \times \[ \[2, 0\], \[0, 2\] \] = \[ \[2, 0\], \[0, 2\] \] \) and \(B \times A = \[ \[2, 0\], \[0, 2\] \] \times \[ \[1, 0\], \[0, 1\] \] = \[ \[2, 0\], \[0, 2\] \] \). So, both are equal, which satisfies the problem statement.

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Most popular questions from this chapter

In Exercises \(27-36,\) find (if possible): \(\begin{array}{llll}\text { a. } A B & \text { and } & \text { b. } B A\end{array}\) $$ A=\left[\begin{array}{rrr} 1 & -1 & 4 \\ 4 & -1 & 3 \\ 2 & 0 & -2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 1 & 1 & 0 \\ 1 & 2 & 4 \\ 1 & -1 & 3 \end{array}\right] $$

Use Cramer's rule to solve each system or to determine that the system is inconsistent or contains dependent equations. $$ \begin{aligned}&2 x=7+3 y\\\&4 x-6 y=3\end{aligned} $$

Consider the system $$ \begin{array}{l}a_{1} x+b_{1} y=c_{1} \\\a_{2} x+b_{2} y=c_{2}\end{array} $$ Use Cramer's rule to prove that if the first equation of the system is replaced by the sum of the two equations, the resulting system has the same solution as the original system.

In Exercises \(27-36,\) find (if possible): \(\begin{array}{llll}\text { a. } A B & \text { and } & \text { b. } B A\end{array}\) $$ A=\left[\begin{array}{rrr} 1 & -1 & 1 \\ 5 & 0 & -2 \\ 3 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 1 & 1 & 0 \\ 1 & -4 & 5 \\ 3 & -1 & 2 \end{array}\right] $$

Use Cramer's rule to solve each system. $$ \begin{aligned}&3 x+2 z=4\\\&5 x-y=-4\\\&4 y+3 z=22\end{aligned} $$
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