# compute the indicated matrices (if possible).

compute the indicated matrices (if possible). $D+BC$
Let $A=\left[\begin{array}{cc}3& 0\\ -1& 5\end{array}\right],B=\left[\begin{array}{ccc}4& -2& 1\\ 0& 2& 3\end{array}\right],C=\left[\begin{array}{cc}1& 2\\ 3& 4\\ 5& 6\end{array}\right],D=\left[\begin{array}{cc}0& -3\\ -2& 1\end{array}\right],E=\left[\begin{array}{cc}4& 2\end{array}\right],F=\left[\begin{array}{c}-1\\ 2\end{array}\right]$

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d2saint0

Step 1
The product of two matrices can be calculated if the number of columns of first matrix equals to number of rows of second matrix. Then the order of the resultant matrix is equal to number of rows of first matrix number of columns of second matrix. Generally the order of matrix is represented as $m×n$, where m is the number of rows of the matrix and n is the number of columns of the matrix. The addition of matrices is calculated only if the order of matrices are same.
Step 2
The given matrices are $D=\left[\begin{array}{cc}0& -3\\ -2& 1\end{array}\right],B=\left[\begin{array}{ccc}4& -2& 1\\ 0& 2& 3\end{array}\right],C=\left[\begin{array}{cc}1& 2\\ 3& 4\\ 5& 6\end{array}\right]$ .The order of the matrices D is $2×2$ , B is $2×3$ and C is $3×2$. As the number of columns of matrix B and number of rows of matrix C both equal to 3, then the matrix multiplication BC can be computed and the order of the resultant matrix BC is $2×2$, which is equal to order of matrix D. So, addition of matrices D and BC is computed. Calculate $D+BC$ as follows,
$D=\left[\begin{array}{cc}0& -3\\ -2& 1\end{array}\right],B=\left[\begin{array}{ccc}4& -2& 1\\ 0& 2& 3\end{array}\right],C=\left[\begin{array}{cc}1& 2\\ 3& 4\\ 5& 6\end{array}\right]$
$D+BC=\left[\begin{array}{cc}0& -3\\ -2& 1\end{array}\right]+\left[\begin{array}{ccc}4& -2& 1\\ 0& 2& 3\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\\ 5& 6\end{array}\right]$

$=\left[\begin{array}{cc}0+3& -3+6\\ -2+21& 1+26\end{array}\right]$

Hence, the value of matrix $D+BC$ is equal to $D+BC=$ $\left[\begin{array}{cc}3& 3\\ 19& 27\end{array}\right]$

Jeffrey Jordon