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University of California - Los Angeles

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Problem 13

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

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

Answer

a) $\left| \begin{array}{cc}{9} & {5} \\ {3} & {-2} \\ {-3} & {15}\end{array}\right|$

b) $\left| \begin{array}{cc}{7} & {-7} \\ {1} & {8} \\ {-5} & {-5}\end{array}\right|$

c) $\left| \begin{array}{cc}{24} & {-3} \\ {6} & {9} \\ {-12} & {15}\end{array}\right|$

d) $\left| \begin{array}{cc}{22} & {-15} \\ {4} & {19} \\ {-14} & {-5}\end{array}\right|$

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## Discussion

## Video Transcript

The question here gives us two. Major sees A and B it wants us to answer, um, a Siri's of sub questions related to, um, using some of the properties with major cities. So the first subsection, um, ask us to do a plus B through we know that through on the property off matrix addition. Nonetheless, we know if it has the same number of a raise, that is, if the dimensions are saying we could just simply add each corresponding Rohan column to each other. So from this property, then we can say that a plus B is the same thing as, of course, one plus negative three two plus negative, too. Two plus four and one plus two. So from here, we know that each individual value would give us negative too. 06 and three. And they'll be the answer to this, um, first sub question here, Um, for the 2nd 1 were given, um, you want to find nonetheless a minus b, so we can do the same thing here If we find the first row first. Call him one minus negative. Three to minus negative too. To minus four and one minus two. We would be given. Ah, 44 negative too. And they have one. And that'll be that subsection there. Um, for the 3rd 1 we want to find 38 We know that through the scaler property of matrices if we basically wanna find the scaler of such major extreme multiply each corresponding entry with the Valley of Three in this particular case. So if we do 12 to 1 like such, we would be given the scaler of this particular matrix would therefore be 3663 and I'll be the equivalent value. So part D asks us to find 38 minus to be so from using this valley that we've gone from part See, we can just plug that in and we want to find the value minus two times the value of B 34 Native tune too. So from here, we can state that the matrix would be that subtracted by of course, um, negative six negative four eat and four. So from here we can used the subtraction, the subtraction property of matrix is and we just simply subtracted so that final matrix value would be three plus six. So that is 96 plus four is 10 6 minus eight is native to and three minus four is negative one. So that would be the final matrix here.

## Recommended Questions

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

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

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

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

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) $3A-2B$.

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

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

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

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

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

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

$$

A=\left[\begin{array}{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}{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}{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}{rrrr}{5} & {-2} & {3} & {1} \\ {-1} & {0} & {2} & {2}\end{array}\right], \quad B=\left[\begin{array}{rrrr}{-2} & {3} & {1} & {0} \\ {4} & {0} & {-1} & {-2}\end{array}\right]

$$