 # If A=begin{bmatrix}-2 & 1&-4 -2 & 4&-1 1 &-1 &-4 end{bmatrix} text{ and } B=begin{bmatrix}-2 & 4&2 -4 & -1&1 4 &1 &1 end{bmatrix} then AB=? BA=? True or false : AB=BA for any two square matrices A and B of the same size. Ava-May Nelson 2020-11-10 Answered
If
then AB=?
BA=?
True or false : AB=BA for any two square matrices A and B of the same size.
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Step 1
The given matrices are,

Step 2
To find the value of AB, BA and telling whether AB=BA is true or not for any two square matrices of the same size.
Step 3
Taking the product of matrices, and evaluating AB,
$AB=\left[\begin{array}{ccc}-2& 1& -4\\ -2& 4& -1\\ 1& -1& -4\end{array}\right]\left[\begin{array}{ccc}-2& 4& 2\\ -4& -1& 1\\ 4& 1& 1\end{array}\right]$
$=\left[\begin{array}{ccc}\left(-2\cdot -2\right)+\left(1\cdot -4\right)+\left(-4\cdot 4\right)& \left(-2\cdot 4\right)+\left(1\cdot -1\right)+\left(-4\cdot 1\right)& \left(-2\cdot 2\right)+\left(1\cdot 1\right)+\left(-4\cdot 1\right)\\ \left(-2\cdot -2\right)+\left(4\cdot -4\right)+\left(-1\cdot 4\right)& \left(-2\cdot 4\right)+\left(4\cdot -1\right)+\left(-1\cdot 1\right)& \left(-2\cdot 2\right)+\left(4\cdot 1\right)+\left(-1\cdot 1\right)\\ \left(1\cdot -2\right)+\left(-1\cdot -4\right)+\left(-4\cdot 4\right)& \left(1\cdot 4\right)+\left(-1\cdot -1\right)+\left(-4\cdot 1\right)& \left(1\cdot 2\right)+\left(-1\cdot 1\right)+\left(-4\cdot 1\right)\end{array}\right]$
$=\left[\begin{array}{ccc}4-4+-16& -8-1-4& -4+1-4\\ 4-16-4& -8-4-1& -4+4-1\\ -2+4+-16& 4+1-4& 2-1-4\end{array}\right]$
$=\left[\begin{array}{ccc}-16& -13& -7\\ -16& -13& -1\\ -14& 1& -3\end{array}\right]\dots \left(1\right)$
Step 4
Similarly, evaluating BA
$BA=\left[\begin{array}{ccc}-2& 4& 2\\ -4& -1& 1\\ 4& 1& 1\end{array}\right]\left[\begin{array}{ccc}-2& 1& -4\\ -2& 4& -1\\ 1& -1& -4\end{array}\right]$
$=\left[\begin{array}{ccc}\left(-2\cdot -2\right)+\left(4\cdot -2\right)+\left(2\cdot 1\right)& \left(-2\cdot 1\right)+\left(4\cdot 4\right)+\left(2\cdot -1\right)& \left(-2\cdot -4\right)+\left(4\cdot -1\right)+\left(2\cdot -4\right)\\ \left(-4\cdot -2\right)+\left(-1\cdot -2\right)+\left(1\cdot 1\right)& \left(-4\cdot 1\right)+\left(-1\cdot 4\right)+\left(1\cdot -1\right)& \left(-4\cdot -4\right)+\left(-1\cdot -1\right)+\left(1\cdot -4\right)\\ \left(4\cdot -2\right)+\left(1\cdot -2\right)+\left(1\cdot 1\right)& \left(4\cdot 1\right)+\left(1\cdot 4\right)+\left(1\cdot -1\right)& \left(4\cdot -4\right)+\left(1\cdot -1\right)+\left(1\cdot -4\right)\end{array}\right]$ Jeffrey Jordon