So, I believe the answer here would tan y but could it possibly be sin y?

Mathematics

1 answer:

nadezda [96]3 years ago

4 0

sorry it took me so long to answer this!!

Yes, the answer would be tan y.

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Evgen [1.6K]

Answer:

2. $\left[\begin{array}{ccc}1&4\\0&3\\\end{array}\right]$

3. $\left[\begin{array}{ccc}-3&21&60\\-15&9&-45\\\end{array}\right]$

4. $\left[\begin{array}{ccc}6&-14\\-2&-6\\\end{array}\right]$

Step-by-step explanation:

2. This matrix is easy, as it just requires addition.

$\left[\begin{array}{ccc}1&4\\0&3\\\end{array}\right]$ + $\left[\begin{array}{ccc}0&0\\0&0\\\end{array}\right]$ = $\left[\begin{array}{ccc}1&4\\0&3\\\end{array}\right]$

3. This matrix requires for the matrices to be multiplied first, then added.

$\left[\begin{array}{ccc}9&-15&18\\-27&15&-9\\\end{array}\right]$ + $\left[\begin{array}{ccc}-12&36&42 \\12& -6 & -36 \\\end{array}\right]$ = $\left[\begin{array}{ccc}-3&21&60\\-15&9&-45\\\end{array}\right]$

4. Here we can add the last 2 matrices to find x.

$\left[\begin{array}{ccc}2&-8\\-4&2\\\end{array}\right]$ + $\left[\begin{array}{ccc}4&-6\\2&-8 \\\end{array}\right]$ = $\left[\begin{array}{ccc}6&-14\\-2&-6\\\end{array}\right]$