Given matrix A and matrix B. Find (if possible) the matrices: (a) AB (b) BA.

$A=\left[\begin{array}{cc}3& -2\\ 1& 5\end{array}\right],B=\left[\begin{array}{cc}0& 0\\ 5& -6\end{array}\right]$

lwfrgin
2021-03-02
Answered

Given matrix A and matrix B. Find (if possible) the matrices: (a) AB (b) BA.

$A=\left[\begin{array}{cc}3& -2\\ 1& 5\end{array}\right],B=\left[\begin{array}{cc}0& 0\\ 5& -6\end{array}\right]$

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Malena

Answered 2021-03-03
Author has **83** answers

Step 1

Given:$A={\left[\begin{array}{cc}3& -2\\ 1& 5\end{array}\right]}_{2\times 2},B={\left[\begin{array}{cc}0& 0\\ 5& -6\end{array}\right]}_{2\times 2}$

Step 2$AB=\left[\begin{array}{cc}3& -2\\ 1& 5\end{array}\right]\left[\begin{array}{cc}0& 0\\ 5& -6\end{array}\right]$

$=\left[\begin{array}{cc}3\times 0-2\times 5& 3\times 0+2\times 6\\ 1\times 0+5\times 5& 1\times 0-6\times 5\end{array}\right]$

$=\left[\begin{array}{cc}0-10& 0+12\\ 0+25& 0-30\end{array}\right]=\left[\begin{array}{cc}-10& 12\\ 25& -30\end{array}\right]$

$BA=\left[\begin{array}{cc}0& 0\\ 5& -6\end{array}\right]\left[\begin{array}{cc}3& -2\\ 1& 5\end{array}\right]=\left[\begin{array}{cc}0\times 3+0\times 1& 0\times (-2)+0\times 5\\ 5\times 3-6\times 1& 5\times (-1)-6\times 5\end{array}\right]$

$=\left[\begin{array}{cc}0+1& 0\\ 15-6& -10-30\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 9& -40\end{array}\right]$

Given:

Step 2

Jeffrey Jordon

Answered 2022-01-27
Author has **2262** answers

Answer is given below (on video)

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2. halve row 3,

3. add row 3 to row 1,

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5. subtract row 2 from each of the other rows,

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7. delete column 1 (column dimension is reduced by 1).

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(b) Write it again as a product of ABC (same B) of three matrices.

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If we rewrite the expression with new angles,

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$\mathrm{sin}\frac{11\pi}{12}\mathrm{sin}\frac{29\pi}{12}-\mathrm{cos}\frac{13\pi}{12}\mathrm{cos}\frac{41\pi}{12}$

If we rewrite the expression with new angles,

$\frac{29\pi}{12}\to \frac{5\pi}{12}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{13\pi}{12}\to \frac{11\pi}{12}$

we don't change the value of the expression. But, if we now use sine of sum of angles, we get $\frac{1}{2}$ instead of 0.

Why does this happen? Do angles need to be in the same quadrant for the formula to work?

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