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Linear Algebra : Matrix-Matrix Product

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Example Questions

Example Question #11 : Matrices

Compute $A\cdot B$ , where

$A=\begin{bmatrix}8&4 \\ 2&4 \\ 6&1 \end{bmatrix}$ $b=\begin{bmatrix}4&2 \\ 4&9 \\ 2&1 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 16\\-28 \\ 11 \end{bmatrix}$

Not Possible

$\begin{bmatrix} 32&16 &8 \\ 8&36 &24 \end{bmatrix}$

$\begin{bmatrix}8&32 \\ 36&8 \\ 1&12 \end{bmatrix}$

$\begin{bmatrix} 32&8 \\ 8&36 \\ 12&1 \end{bmatrix}$

Correct answer:

Not Possible

Explanation:

In order to be able to multiply matrices, the number of columns of the 1st matrix must equal the number of rows in the second matrix. Here, the first matrix has dimensions of (3x2). This means it has three rows and two columns. The second matrix has dimensions of (3x2), also three rows and two columns. Since $\tiny \tiny \small 2\neq3$ , we cannot multiply these two matrices together

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Example Question #11 : Matrices

$C = \begin{bmatrix} 4& 0.7\\ 7& -0.4 \end{bmatrix}$ .

Evaluate $C^{T}C$ .

Possible Answers:

$\begin{bmatrix} 0.49 & 0 \\0 & 16\end{bmatrix}$

$\begin{bmatrix}0.65 & 0 \\ 0 & 65 \end{bmatrix}$

$\begin{bmatrix} 16 & 0 \\0 & 0.49\end{bmatrix}$

$\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$

$\begin{bmatrix}65 & 0 \\ 0 & 0.65 \end{bmatrix}$

Correct answer:

$\begin{bmatrix}65 & 0 \\ 0 & 0.65 \end{bmatrix}$

Explanation:

The transpose of a matrix switches the rows and the columns. Therefore, the first column of $C^{T}$ has the same entries, in order, as the first row of , and so forth. Since

$C = \begin{bmatrix} 4& 0.7\\ 7& -0.4 \end{bmatrix}$ ,

$C^{T} = \begin{bmatrix} 4& 7\\ 0.7& -0.4 \end{bmatrix}$ .

The entry in column , row of the product $C^{T}C$ is the product of row of $C^{T}$ and column of - the sum of the products of the numbers that appear in the corresponding positions of the row and the column. $C^{T}C$ can consequently be calculated as follows:

$C^{T}C$

$= \begin{bmatrix} 4& 7\\ 0.7& -0.4 \end{bmatrix} \begin{bmatrix} 4& 0.7\\ 7& -0.4 \end{bmatrix}$

$= \begin{bmatrix} 4 (4) + 7 (7) & 4 (0.7) + 7 (-0.4)\\ 0.7 (4) + (-0.4) (7) & 0.7 (0.7) + (-0.4) (-0.4) \end{bmatrix}$

$= \begin{bmatrix}16+49 & 2.8 + (-2.8) \\ 2.8 + (-2.8) & 0.49 + 0.16 \end{bmatrix}$

$= \begin{bmatrix}65 & 0 \\ 0 & 0.65 \end{bmatrix}$ ,

the correct choice.

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Example Question #11 : Matrix Matrix Product

$C = \begin{bmatrix} 3&0 & 6\\ -7&1 &2 \end{bmatrix}$

Evaluate $C^{T}C$ .

Possible Answers:

$\begin{bmatrix} 9 & 49\\ 0& 1\\ 36& 4 \end{bmatrix}$

$\begin{bmatrix} 45 & -9 \\ -9 & 54 \end{bmatrix}$

$C^{T}C$ is an undefined product.

$\begin{bmatrix} 58 &-7 & 4 \\ -7& 1 & 2 \\ 4 & 2& 40\end{bmatrix}$

$\begin{bmatrix} 9&0 & 36\\49&1 &4 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 58 &-7 & 4 \\ -7& 1 & 2 \\ 4 & 2& 40\end{bmatrix}$

Explanation:

The transpose of a matrix switches the rows and the columns. Therefore, the first column of $C^{T}$ has the same entries, in order, as the first row of , and so forth. Since

$C = \begin{bmatrix} 3&0 & 6\\ -7&1 &2 \end{bmatrix}$ ,

it follows that

$C ^{T} = \begin{bmatrix} 3& -7\\ 0 & 1\\ 6& 2 \end{bmatrix}$

$C^{T}C$

$= \begin{bmatrix} 3& -7\\ 0 & 1\\ 6& 2 \end{bmatrix} \begin{bmatrix} 3&0 & 6\\ -7&1 &2 \end{bmatrix}$

$= \begin{bmatrix} 3(3)+ (-7)(-7)&3(0)+ (-7)(1)& 3(6)+ (-7)(2) \\ 0(3)+1(-7)&0(0)+ 1(1) & 0(6)+1(2) \\ 6(3)+2(-7)&6(0)+ 2 (1) & 6(6)+ 2(2) \end{bmatrix}$

$= \begin{bmatrix} 9+49 &0 + (-7)& 18+ (-14 ) \\ 0 + (-7)&0+1 & 0+2 \\ 18 + (-14)&0+2& 36+4\end{bmatrix}$

$= \begin{bmatrix} 58 &-7 & 4 \\ -7& 1 & 2 \\ 4 & 2& 40\end{bmatrix}$

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Example Question #11 : Matrix Matrix Product

$C = \begin{bmatrix} 3&0 & 6\\ -7&1 &2 \end{bmatrix}$

Evaluate $C^{2}$ .

Possible Answers:

$\begin{bmatrix} 45 & -9 \\ -9 & 54 \end{bmatrix}$

$\begin{bmatrix} 9&0 & 36\\49&1 &4 \end{bmatrix}$

$\begin{bmatrix} 9 & 49\\ 0& 1\\ 36& 4 \end{bmatrix}$

$C^{2}$ is an undefined product.

$\begin{bmatrix} 58 &-7 & 4 \\ -7& 1 & 2 \\ 4 & 2& 40\end{bmatrix}$

Correct answer:

$C^{2}$ is an undefined product.

Explanation:

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. Therefore, for the square of a matrix to be defined, the number of rows and columns in that matrix must be the same; that is, it must be a square matrix. , having two rows and three columns, is not square, so $C^{2}$ cannot exist.

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Example Question #12 : Matrices

Markov

The Markov triplets are playing a game whose board comprises three spaces, marked "A", "B", and "C", as shown above. For each turn, one player rolls a fair six-sided die, then moves to the space indicated based on where the player is already and what the player rolls. For example, if the player is on space "B" and rolls a "2", he moves to space "A", since one of the numbers on the arrow going from "B" to 'A" is "2"; similarly, if he is on space "B" and rolls a "4", he will stay where he is.

It is agreed that each player will start on a different space. If Mickey Markov starts at space "B", what is the probability that Mickey will be on space "A" after two rolls?

(Hint: think about that name "Markov")

Possible Answers:

$\frac{11}{36}$

$\frac{1}{4}$

$\frac{13}{36}$

$\frac{1}{3}$

$\frac{5}{18}$

Correct answer:

$\frac{11}{36}$

Explanation:

We can set up a stochastic matrix, or Markov chain, using the probabilities that, given a beginning space, a player will wind up on each other space.

If a player starts on space "A", the probabilities are:

P(A|A ) = 0 (zero rolls out of six)

$P(B|A )=\frac{3}{6} = \frac{1}{2}$ (three rolls out of six)

$P(C|A )=\frac{3}{6} = \frac{1}{2}$ (three rolls out of six)

If a player starts on space "B", the probabilities are:

$P(A | B) = \frac{3}{6} = \frac{1}{2}$ (three rolls out of six)

$P(B | B )=\frac{1}{6}$ (one roll out of six)

$P(C | B)=\frac{2}{6} = \frac{1}{3}$ (two rolls out of six)

If a player starts on space "C", the probabilities are:

$P(A |C ) = \frac{4}{6} = \frac{2}{3}$ (three rolls out of six)

P(B |C )=0 (zero rolls out of six)

$P(C |C)=\frac{2}{6} = \frac{1}{3}$ (two rolls out of six)

Let the three states be spaces "A", "B", and "C", represented by rows/columns 1, 2, and 3, respectively. We can form a stochastic matrix from these probabilities, as follows:

$P = \begin{bmatrix} 0& \frac{1}{2} & \frac{1}{2} \\ \\ \frac{1}{2}& \frac{1}{6} & \frac{1}{3}\\ \\ \frac{2}{3}& 0 & \frac{1}{3} \end{bmatrix}$

This is the matrix of probabilities for one move. For the matrix of probabilities for two moves, square this matrix. The entry in column , row of the product $PP= P^{2}$ is the product of row of and column of - the sum of the products of the numbers that appear in the corresponding positions of the row and the column.

We are only concerned with the probability that a player on space "B" will end up back at space "A", so we will only multiply the second row of by the first column of , as follows:

$P (A) = \frac{1}{2} \cdot 0 + \frac{1}{6} \cdot \frac{1}{2}+ \frac{1}{3} \cdot \frac{2}{3}$

$= 0 + \frac{1}{12}+ \frac{2}{9}$

$= \frac{11}{36}$

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Example Question #11 : Matrices

Your friend Hector wants to multiply two matrices and as follows: $A\cdot B$ . Unfortunately, Hector knows nothing about matrix dimensions. Which of the following statements will help Hector figure out whether it is possible for him to multiply $A\cdot B$ ?

Possible Answers:

The number of columns in matrix must be equal to the number of columns in matrix .

The number of rows in matrix must be equal to the number of columns in matrix .

The number of columns in matrix must be equal to the number of rows in matrix .

and must both be square matrices, otherwise you cannot multiply them.

The number of rows in matrix must be equal to the number of rows in matrix .

Correct answer:

The number of columns in matrix must be equal to the number of rows in matrix .

Explanation:

Whenever we multiple two matrices together we must always check first that the number of columns in the first matrix is equal to the number of rows in the second matrix. For example, consider these two matrices

$\begin{pmatrix} 1 & 2 & 3\\ 4&5&6 \end{pmatrix} \cdot \begin{pmatrix} 10 & 9 & 8 & 7\\ 6&5&4&3\\ 2&1&0&-1 \end{pmatrix}$

The first matrix has 3 columns, and the second matrix has 3 rows. We can multiple these two matrices together in this order. However, if we switch the order around, we will not be able to multiply these two matrices.

$\begin{pmatrix} 10 & 9 & 8 & 7\\ 6&5&4&3\\ 2&1&0&-1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3\\ 4&5&6 \end{pmatrix} = ?$

Now the first matrix has 4 columns and the second matrix has 2 rows. We cannot multiply these two matrices in this order.

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Example Question #12 : Matrices

Find the product of these two matrices, if it exists.

$\begin{pmatrix} 2 & 0 & 1 & 6\\ 1 & 9 & 9 & 6 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2\\ 5 & 0 \\ -3& 2\\ 1& 3\end{pmatrix}$

Possible Answers:

$\begin{pmatrix} 25 & 38\\ 5 & 24\\ 23 & 17\\ 4 & 9 \end{pmatrix}$

$\begin{pmatrix} 5 & 24\\ 25 & 38 \end{pmatrix}$

The product does not exist because the dimensions do not match.

$\begin{pmatrix} 25 & 38\\ 5 & 24\end{pmatrix}$

$\begin{pmatrix} 5 & 24\\ 79 & 38 \end{pmatrix}$

Correct answer:

$\begin{pmatrix} 5 & 24\\ 25 & 38 \end{pmatrix}$

Explanation:

First we check that the dimensions match. The first matrix has 4 columns, and the second matrix has 4 rows. So the matrix product does exist. We find the product by taking the dot product of rows and columns.

$\begin{pmatrix} {\color{Red} 2} & {\color{blue} 0} & {\color{DarkGreen} 1} & {\color{Purple} 6}\\ 1 & 9 & 9 & 6 \end{pmatrix} \cdot \begin{pmatrix} {\color{Red} 1} & 2\\ {\color{blue} 5} & 0 \\ {\color{DarkGreen} -3}& 2\\ {\color{Purple} 1}& 3\end{pmatrix} = \begin{pmatrix} {\color{Red} 2\cdot 1} + {\color{Blue} 0\cdot 5} + {\color{DarkGreen} 1\cdot -3} + {\color{Purple} 6\cdot 1} & ?\\ ?&? \end{pmatrix} =\begin{pmatrix} 5 & ?\\ ? & ? \end{pmatrix}$

$\begin{pmatrix} {\color{Red} 2} & {\color{blue} 0} & {\color{DarkGreen} 1} & {\color{Purple} 6}\\ 1 & 9 & 9 & 6 \end{pmatrix} \cdot \begin{pmatrix} 1 & {\color{Red} 2} \\ 5 & {\color{blue} 0} \\ -3 & {\color{DarkGreen} 2}\\ 1 & {\color{Purple} 3}\end{pmatrix} = \begin{pmatrix} 5 & {\color{Red} 2\cdot 2} + {\color{Blue} 0\cdot 0} + {\color{DarkGreen} 1\cdot 2} + {\color{Purple} 6\cdot 3} \\ ?&? \end{pmatrix} =\begin{pmatrix} 5 & 24 \\ ? & ? \end{pmatrix}$

We fill in the rest of the entries in the product matrix in the same way.

$\begin{pmatrix} 2 & 0 & 1 & 6\\ 1 & 9 & 9 & 6 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2\\ 5 & 0 \\ -3& 2\\ 1& 3\end{pmatrix} = \begin{pmatrix} 5 & 24\\ 25 & 38 \end{pmatrix}$

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Example Question #12 : Matrix Matrix Product

Find the product of these two matrices, if it exists.

$\begin{pmatrix} 6 & 4 & -2\\ 7 & 2 & 0\\ 9 & -6 & 2 \end{pmatrix} \cdot \begin{pmatrix} 5 & 2 & 9\\ 3 & -5 & 1\\ \end{pmatrix}$

Possible Answers:

$\begin{pmatrix} 42 & -8 & 58\\ 41 & 4 & 65\\ 28 & 48 & 75 \end{pmatrix}$

$\begin{pmatrix} 125 & -30 & 8\\ -8 & -8 & -4\end{pmatrix}$

The product does not exist because the dimensions do not match.

$\begin{pmatrix} 41 & 4 \\ 27 & 48 \end{pmatrix}$

$\begin{pmatrix} 6 & 4 & -2\\ 5 & 2 & 9 \\ 12 & -11 & 3 \end{pmatrix}$

Correct answer:

The product does not exist because the dimensions do not match.

Explanation:

The product does not exist because there are 3 columns in the first matrix and 2 rows in the second matrix. The dimensions do not match.

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Example Question #11 : Matrix Matrix Product

Find the product of these two matrices, if it exists.

$\begin{pmatrix} 0 & 4 & -2\\ 6 & -5 & 3 \end{pmatrix} \cdot \begin{pmatrix} 4 & -1\\ -8 & 3\\ 0 & 5 \end{pmatrix}$

Possible Answers:

$\begin{pmatrix} 2 & -32\\ -6 & 64 \end{pmatrix}$

$\begin{pmatrix} -34 & -2\\ -16 & -6 \end{pmatrix}$

$\begin{pmatrix} -32 & 2\\ 64 & -6 \end{pmatrix}$

The product does not exist because the dimensions do not match.

$\begin{pmatrix} -1 & 7 & 3\\ 10 & -13 & 3 \end{pmatrix}$

Correct answer:

$\begin{pmatrix} -32 & 2\\ 64 & -6 \end{pmatrix}$

Explanation:

First we check the dimensions of the matrices. The first matrix has 3 columns and the second matrix has 3 rows. The product exists. We find the product by taking the dot product of rows and columns.

$\begin{pmatrix} {\color{Red} 0} & {\color{Red}4 } & {\color{Red} -2}\\ 6 & -5 & 3 \end{pmatrix} \cdot \begin{pmatrix} {\color{Red} 4} & -1\\ {\color{Red} -8} & 3\\ {\color{Red} 0} & 5 \end{pmatrix}= \begin{pmatrix}{\color{Red} -32} & \\ && \end{pmatrix}$ .

$\begin{pmatrix} {\color{Red} 0} & {\color{Red}4 } & {\color{Red} -2}\\ 6 & -5 & 3 \end{pmatrix} \cdot \begin{pmatrix} 4 & {\color{Red} -1} \\ -8 & {\color{Red} 3} \\ 0 & {\color{Red} 5} \end{pmatrix}= \begin{pmatrix} -32 & {\color{Red} 2} \\ && \end{pmatrix}$

We fill in the rest of the matrix entries in the same way.

$\begin{pmatrix} 0 & 4 & -2\\ 6 & -5 & 3 \end{pmatrix} \cdot \begin{pmatrix} 4 & -1\\ -8 & 3\\ 0 & 5 \end{pmatrix} = \begin{pmatrix} -32 & 2\\ 64 & -6 \end{pmatrix}$ .

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Example Question #15 : Matrices

Your friend Hector found your advice incredibly helpful and he is back for more. He would like to multiply two matrices and He would like to multiply $A\cdot B$ and $B \cdot A$ . Which statement below gives Hector all of the information he needs to pick appropriate matrices?

Possible Answers:

it is impossible to find two matrices such that $A\cdot B$ and $B \cdot A$ both exist.

The number of columns in matrix must equal the number of rows in matrix and the number of rows in matrix must equal the number of columns in matrix .

The number of columns in matrix must equal the number of rows in matrix .

For all matrices, $A\cdot B = B\cdot A$ .

The matrices and must both be square matrices.

Correct answer:

The number of columns in matrix must equal the number of rows in matrix and the number of rows in matrix must equal the number of columns in matrix .

Explanation:

Hector wishes to multiply $A\cdot B$ so we know that the number of columns in matrix must equal the number of rows in matrix . He also wishes to multiply the $B\cdot A$ so we know that the number of columns in matrix must equal the number of rows in matrix .

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Linear Algebra : Matrix-Matrix Product

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #11 : Matrices

Example Question #11 : Matrices

Example Question #11 : Matrix Matrix Product

Example Question #11 : Matrix Matrix Product

Example Question #12 : Matrices

Example Question #11 : Matrices

Example Question #12 : Matrices

Example Question #12 : Matrix Matrix Product

Example Question #11 : Matrix Matrix Product

Example Question #15 : Matrices

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