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University of California - Los Angeles

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Problem 11

In Exercises 11-18, if possible, find (a) $A+B$, (b) $A-B$, (c ) $3A$, and (d) $3A-2B$.

$A = \left[ \begin{array}{r} 1 & -1 \\ 2 & -1 \end{array} \right]$, $B = \left[ \begin{array}{r} 2 & -1 \\ -1 & 8 \end{array} \right]$

Answer

a) $\left( \begin{array}{cc}{3} & {-2} \\ {1} & {7}\end{array}\right)$

b) $\left( \begin{array}{cc}{-1} & {0} \\ {3} & {-9}\end{array}\right)$

c) $\left( \begin{array}{cc}{3} & {-3} \\ {6} & {-3}\end{array}\right)$

d) $\left( \begin{array}{cc}{-1} & {-1} \\ {8} & {-19}\end{array}\right)$

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## Discussion

## Video Transcript

The question here gives us two different major, sees A and B and asks us to solve a series of questions. So the first of sub set of this question in part a states that it wants us to add The two major sees together. So we know that, um, through some of the properties of major season, has the same number of corresponding entries weaken, basically, just add them together. So in this particular place, a plus B is equivalent of doing one plus two by taking the first of each and then doing the same with Second, um, second row, first column, negative one plus negative one ah, two plus negative one. And, of course, negative one plus eight. So from here, we can state that it is equal to three one negative two and seven. So that would be the major extents obtained by doing a plus B for, um part beat. It states that we want to subtract it, and by that we know we can do the same thing here. So one minus two. Um, negative one minus negative. 12 minus negative one and negative one minus eight. So from here, we get negative 10 three and negative nine as such. So that would be the answer to this matrix here. I'm see states that it wants us to do a scaler times the matrix of A from here through the property of scaler multiplication. We know that we have to generally just do that if we put it as this. Rather, we can just multiply each single corresponding entry within the Matrix with the numerical value of three. And that would give us the scaler off such. So they'll be just, um three negative 36 and negative three as such. So the last sub part of this question us us to do three a minus to be minds to be so essentially if we use the first section here on 36 negative three negative three. And we subtract that with two times the value of all the corresponding entries in be I give 18 which is the same thing is of course, if we write that down like so, um, we would get minus four negative too negative, too. And 16. So if we basically just subtract both of these together, we would get the value of negative one um, negative one. Of course. Three plus two. I'm six plus two is of course, no one rather than be eight and then negative 19. So that would be the value of this matrix here. So those would be all four answers to the four subsets in this question.

## Recommended Questions

In Exercises 11-18, if possible, find (a) $A+B$, (b) $A-B$, (c ) $3A$, and (d) $3A-2B$.

$A = \left[ \begin{array}{r} 1 & 2 \\ 2 & 1 \end{array} \right]$, $B = \left[ \begin{array}{r} -3 & -2 \\ 4 & 2 \end{array} \right]$

In Exercises 11-18, if possible, find (a) $A+B$, (b) $A-B$, (c ) $3A$, and (d) $3A-2B$.

$A = \left[ \begin{array}{r} 3 \\ 2 \\ -1 \end{array} \right]$, $B = \left[ \begin{array}{r} -4 & 6 & 2 \end{array} \right]$

In Exercises 11-18, if possible, find (a) $A+B$, (b) $A-B$, (c ) $3A$, and (d) $3A-2B$.

$A = \left[ \begin{array}{r} 8 & -1 \\ 2 & 3 \\ -4 & 5 \end{array} \right]$, $B = \left[ \begin{array}{r} 1 & 6 \\ -1 & -5 \\ 1 & 10 \end{array} \right]$

In Exercises 11-18, if possible, find (a) $A+B$, (b) $A-B$, (c ) $3A$, and (d) $3A-2B$.

$A = \left[ \begin{array}{r} 6 & 0 & 3 \\ -1 & -4 & 0 \end{array} \right]$, $B = \left[ \begin{array}{r} 8 & -1 \\ 4 & -3 \end{array} \right]$

In Exercises 11-18, if possible, find (a) $A+B$, (b) $A-B$, (c ) $3A$, and (d) $3A-2B$.

$A = \left[ \begin{array}{r} 1 & -1 & 3 \\ 0 & 6 & 9 \end{array} \right]$, $B = \left[ \begin{array}{r} -2 & 0 & -5 \\ -3 & 4 & -7 \end{array} \right]$

In Exercises 11-18, if possible, find (a) $A+B$, (b) $A-B$, (c ) $3A$, and (d) $3A-2B$.

$A = \left[ \begin{array}{r} -1 & 4 & 0 \\ 3 & -2 & 2 \\ 5 & 4 & -1 \\ 0 & 8 & -6 \\ -4 & -1 & 0 \end{array} \right]$, $B = \left[ \begin{array}{r} -3 & 5 & 1 \\ 2 & -4 & -7 \\ 10 & -9 & -1 \\ 3 & 2 & -4 \\ 0 & 1 & -2 \end{array} \right]$

In Exercises $11-16,$ find (a) $A+B,(b) A-B,(e) 3 A,$ and $(d) 2 A-3 B$

$$

A=\left[\begin{array}{cccc}{-1} & {-2} & {0} & {3}\end{array}\right], \text { and } B=\left[\begin{array}{cccc}{1} & {2} & {-2} & {0}\end{array}\right]

$$

In Exercises $11-16,$ find (a) $A+B,(b) A-B,(e) 3 A,$ and $(d) 2 A-3 B$

$$

A=\left[\begin{array}{rr}{-3} & {1} \\ {0} & {-1} \\ {2} & {1}\end{array}\right], \quad B=\left[\begin{array}{rr}{4} & {0} \\ {-2} & {1} \\ {-3} & {-1}\end{array}\right]

$$

In Exercises $11-16,$ find (a) $A+B,(b) A-B,(e) 3 A,$ and $(d) 2 A-3 B$

$$

A=\left[\begin{array}{rr}{2} & {3} \\ {-1} & {5}\end{array}\right], \quad B=\left[\begin{array}{rr}{1} & {-3} \\ {-2} & {-4}\end{array}\right]

$$

In Exercises $11-16,$ find (a) $A+B,(b) A-B,(e) 3 A,$ and $(d) 2 A-3 B$

$$

A=\left[\begin{array}{rrr}{-1} & {0} & {2} \\ {4} & {1} & {-1} \\ {2} & {0} & {1}\end{array}\right], \quad B=\left[\begin{array}{rrr}{2} & {1} & {0} \\ {-1} & {0} & {2} \\ {4} & {-3} & {-1}\end{array}\right]

$$