# Subject: matrices Determine: begin{bmatrix}1 & -1 &2 end{bmatrix}left{ begin{bmatrix} 4 -1 0 end{bmatrix}begin{bmatrix}2 & 3 end{bmatrix} + begin{bmatrix}1 & 0 0 & 1 -1 &-1 end{bmatrix}right}

Subject: matrices
Determine:
$\left[\begin{array}{ccc}1& -1& 2\end{array}\right]\left\{\left[\begin{array}{c}4\\ -1\\ 0\end{array}\right]\left[\begin{array}{cc}2& 3\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\\ -1& -1\end{array}\right]\right\}$
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casincal
Step 1
Given matrix expression:
$\left[\begin{array}{ccc}1& -1& 2\end{array}\right]\left\{\left[\begin{array}{c}4\\ -1\\ 0\end{array}\right]\left[\begin{array}{cc}2& 3\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\\ -1& -1\end{array}\right]\right\}$
Step 2
Using BODMAS rule first solve bracket then multiplication and then addition:
Multiplying $\left[\begin{array}{c}4\\ -1\\ 0\end{array}\right]$ and $\left[\begin{array}{cc}2& 3\end{array}\right]$ multiplication is possible because columns of first matrix is equal to the rows of the second matrix, we get: $\left[\begin{array}{ccc}1& -1& 2\end{array}\right]\left\{\left[\begin{array}{cc}4×2& 4×3\\ -1×2& -1×3\\ 0×2& 0×3\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\\ -1& -1\end{array}\right]\right\}$
$\left[\begin{array}{ccc}1& -1& 2\end{array}\right]\left\{\left[\begin{array}{cc}8& 12\\ -2& -3\\ 0& 0\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\\ -1& -1\end{array}\right]\right\}$
$\left[\begin{array}{ccc}1& -1& 2\end{array}\right]\left\{\left[\begin{array}{cc}8+1& 12+0\\ -2+0& -3+1\\ 0-1& 0-1\end{array}\right]\right\}$
$\left[\begin{array}{ccc}1& -1& 2\end{array}\right]\left[\begin{array}{cc}9& 12\\ -2& -2\\ -1& -1\end{array}\right]$
Multiply the matrices,we get: $\left[\begin{array}{cc}1×9+\left(-1\right)×\left(-2\right)+2×\left(-1\right)& 1×12+\left(-1\right)×\left(-2\right)+2×\left(-1\right)\end{array}\right]$
$\left[\begin{array}{cc}9+2-2& 12+2-2\end{array}\right]$
$\left[\begin{array}{cc}9& 12\end{array}\right]$
Step 3
Therefore, the solution of the given matrix is $\left[\begin{array}{cc}9& 12\end{array}\right]$
Jeffrey Jordon