# Given matrix A and matrix B. Find (if possible) the matrices: (a) AB (b) BA. A=begin{bmatrix}1 & 2 &3&4end{bmatrix} , B=begin{bmatrix}1 2 3 4 end{bmatrix}

Given matrix A and matrix B. Find (if possible) the matrices: (a) AB (b) BA. $A=\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right],B=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Dora
Step 1
the given matrices are:
$A=\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right],B=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$
we have to find (if possible) the matrices:
(a)AB
(b)BA
Step 2
as we know the product of the two matrices is possible if the former matrix in the product has no. of columns equal to the number of rows of the latter matrix.
for example if we have two matrices A and B having orders respectively then for product AB to be defined the number of columns of matrix A should be equal to the number of rows of matrix B that implies n=p and for product BA to be defined the number of columns of matrix B should be equal to the number of rows of matrix A that implies q=m therefore as the given matrices are:

matrix A has order $1×4$, that implies matrix A has 1 row and 4 columns.
matrix B has order $4×1$, that implies matrix B has 4 rows and 1 column.
Step 3
now as no. of columns of matrix A is equal to number of rows of matrix B.
therefore the product AB is defined.
therefore,$AB=\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$
$=\left[\begin{array}{cccc}1×1& 2×2& 3×3& 4×4\end{array}\right]$
$=\left[\begin{array}{cccc}1& 4& 9& 16\end{array}\right]$
$\left[\begin{array}{c}30\end{array}\right]$
therefore, $AB=\left[\begin{array}{c}30\end{array}\right]$
Step 4
now as number of columns of matrix B is equal to number of rows of matrix A. therefore the product BA is defined.
therefore, $BA=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right]$
$=\left[\begin{array}{cccc}1×1& 1×2& 1×3& 1×4\\ 2×1& 2×2& 2×3& 2×4\\ 3×1& 3×2& 3×3& 3×4\\ 4×1& 4×2& 4×3& 4×4\end{array}\right]$
$=\left[\begin{array}{cccc}1& 2& 3& 4\\ 2& 4& 6& 8\\ 3& 6& 9& 12\\ 4& 8& 12& 16\end{array}\right]$
therefore, $BA=\left[\begin{array}{cccc}1& 2& 3& 4\\ 2& 4& 6& 8\\ 3& 6& 9& 12\\ 4& 8& 12& 16\end{array}\right]$
Jeffrey Jordon