Question

http://prntscr.com/nuhz4u

Answer 1

@Narad

Answer 2

\[\left[\begin{matrix}2 & 4\\ 1 & -6\end{matrix}\right]+\left(\begin{matrix}1 \\ 0\end{matrix}\right)\]
Is this a matrix addition?

Answer 3

Or a matrix multiplication?

Answer 4

Addition

Answer 5

Matrices must be of the same size for matrix addition

Answer 6

so multiplication?

Answer 7

Yes,
\[\left[\begin{matrix}2 & 4\\ 1 & -6\end{matrix}\right]*\left(\begin{matrix}1 \\ 0\end{matrix}\right)\]
=\[\left[\begin{matrix}2*1 & 4*0 \\ 1*1 & -6*0\end{matrix}\right] = \left[\begin{matrix}2 & 0 \\ 1 & 0\end{matrix}\right]\]

Answer 8

Okay

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Answer 9

The operation is possible because the first matrix is 2*2 and the second is 2*1

Answer 10

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Answer 11

nº7
This is not possible since the first matrix is 1*3 and the second is 2*1

Answer 12

Okay

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Answer 13

\[\left[\begin{matrix}1 & -6\\ 2& 0\end{matrix}\right]*\left[\begin{matrix}1 & 4 \\ 9 & 8\end{matrix}\right]\]

\[=\left[\begin{matrix}1*1+9*-6& 1*4+8*-6 \\ 2*1+0*8 & 2*4+0*8\end{matrix}\right]\]

\[=\left[\begin{matrix}-53 & -44 \\ 2& 8\end{matrix}\right]\]

Answer 14

Nº 8 is possible since the 2 matrices are of the same size

Answer 15

Okay

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Answer 16

nª9 this is possible sence the matrices are 3*1 ans 1*3
\[(1 7 3)+\left(\begin{matrix}2 \\ -1\\5\end{matrix}\right)\]
\[=(1*2+7*-1+3*5)\]
\[=(10)\]

Answer 17

Okay

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Answer 18

The matrix is
\[A=\left[\begin{matrix}5 & 6 \\ 1 & 9\end{matrix}\right]\]

The determinant is
\[detA =\left|\begin{matrix}5 & 6 \\ 1 & 9\end{matrix}\right|\]
\[=5*9-1*6\]
\[=45-6=39\]

Answer 19

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Answer 20

C?

Answer 21

Yes this is the correct matrix

Answer 22

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Answer 23

The matrix is
\[B=\left[\begin{matrix}2 & 7 \\ 9 & -10\end{matrix}\right]\]
The new matrix is
\[-2B=-2*\left[\begin{matrix}2 & 7 \\ 9 & 10\end{matrix}\right]\]
\[=\left[\begin{matrix}-4 & -14 \\ -18 & -20 \end{matrix}\right]\]
The answer is option C

Answer 24

Correct -10 to 10 in the first line

Answer 25

http://prntscr.com/nuk0n7

Answer 26

@Narad

Answer 27

The matrix is
\[B=\left[\begin{matrix}2 & 7 \\ 9& 10\end{matrix}\right]\]

Therefore,
\[2B=2*\left[\begin{matrix}2 & 7 \\ 9& 10\end{matrix}\right]\]
\[=\left[\begin{matrix}4 & 14 \\ 18 & 20\end{matrix}\right]\]
The answer is option B

Answer 28

Okay

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Answer 29

The matrices are
\[A=\left[\begin{matrix}1 & 4 \\ 5 & 2\end{matrix}\right]\] and
\[B=\left[\begin{matrix}2 & 7 \\ 9 & 10\end{matrix}\right]\]

Therefore,
\[B-A\]
\[=\left[\begin{matrix}2-1 & 7-4 \\ 9-5 & 10-2\end{matrix}\right]\]
\[=\left[\begin{matrix}1 & 3 \\ 4 & 8\end{matrix}\right]\]
The answer is option C

Answer 30

Okay

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Answer 31

If a matrix A has dimensions m * n and matrix B has dimensions n*p, then the product AB has dimensions m*p
The answer is TRUE

Answer 32

Got it

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Answer 33

Matrix A is
\[A=\left[\begin{matrix}5 & 3\\ 0 & 1\end{matrix}\right]\]
and matrx B is
\[B=\left[\begin{matrix}4 & 2 &-1\\ 0 & 1& 3\\ \end{matrix}\right]\]
The product is
\[AB=\left[\begin{matrix}20 & 13&4\\ 0 & 1&3\end{matrix}\right]\]
The answer is option B

Answer 34

okay

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Answer 35

@Narad

Answer 36

The matrices are
\[A=\left[\begin{matrix}2 & 3 \\ 4 & 5\end{matrix}\right]\]
and \[B=\left[\begin{matrix}2 & 4 \\ 5 & 8\end{matrix}\right]\]
The product is
\[AB=\left[\begin{matrix}19 & 32 \\ 33 & 56\end{matrix}\right]\]
The answer is option A

Answer 37

Okay

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Answer 38

The first matrix is 3*4 and the second matrix is 2*3
It is not possible to multiply the matrices
The solution does not exist and the answer is option D

Answer 39

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Answer 40

Question nº9
The matrix is
\[A=\left[\begin{matrix}6 & 7 \\ 8& 0\end{matrix}\right]\]
The determinant is
\[detA =\left|\begin{matrix}6 & 8 \\ 7& 0\end{matrix}\right|\]
\[=6*0-8*7=0-56=-56\]
The answer is option B

Answer 41

Question nº10
The matrix is
\[A=\left[\begin{matrix}1 & 1 \\ 2 & 2\end{matrix}\right]\]
The determinant is
\[detA=\left|\begin{matrix}1 & 1 \\ 2 & 2\end{matrix}\right|=2-2=0\]
As the determinant =0, the matrix is singular and the inverse does not exist.
The answer is no.

Answer 42

Makes sense

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Answer 43

Question nº11
The matrix is
\[A=\left[\begin{matrix}3 & 1\\ -4 & 1\end{matrix}\right]\]
The determinant is
\[detA=\left|\begin{matrix}3 & 1 \\ -4 & 1\end{matrix}\right|=3-(-4)=7\]
The answer is option B

Answer 44

Question nº12
The matrix is
\[A=\left[\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right]\]
The determinant is
\[detA=\left|\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right|= 3-2=1\]
The matrix has an inverse since \[detA \neq0\]
The answer is option D