Matrices - Linear Algebra

Card 0 of 696

Question

Compute $A\cdot B$ , where

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

Answer

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 $\tiny \tiny \small 2 eq3$ , we cannot multiply these two matrices together

Compare your answer with the correct one above

Question

Compute $A\cdot B$ , where

$A = \begin{bmatrix} 3&4 \ 2&7 \ 6&5 \end{bmatrix}$

$B = \begin{bmatrix} 2&2 \ 5&7 \end{bmatrix}$

Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (3x2). The product matrix equals, $\begin{bmatrix} (3\cdot 2+4\cdot 5)&(3\cdot 2 +4\cdot 7) \ (2\cdot 2 +7\cdot 5)&(2\cdot 2 +7\cdot 7) \ (6\cdot 2 +5\cdot 5)& (6\cdot 2 +5\cdot 7) \end{bmatrix}$

Compare your answer with the correct one above

Question

Compute $A\cdot B$ , where

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

$B=\begin{bmatrix} 8&4 \ 3&5 \end{bmatrix}$

Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (2x2). Next, we notice that matrix A is the identity matrix. Any matrix multiplied by the identity matrix remains unchanged.

Compare your answer with the correct one above

Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 3&4 &0 \ 2&7 &1 \ 6&5 &7 \end{bmatrix}$

$B =\begin{bmatrix} 2&2 &8 \ 5&7 &0 \ 6&4 &3 \end{bmatrix}$

Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (3x3). The product matrix equals, $\begin{bmatrix} (3\cdot 2+4\cdot 5+0\cdot 6)&(3\cdot 2 +4\cdot 7+0\cdot 4)&(3\cdot 8 +4\cdot 0+0\cdot 3) \ (2\cdot 2 +7\cdot 5+1\cdot 6)&(2\cdot 2 +7\cdot 7+1\cdot 4)&(2\cdot 8 +7\cdot 0+1\cdot 3) \ (6\cdot 2 +5\cdot 5+7\cdot 6)& (6\cdot 2 +5\cdot 7+7\cdot 4)&(7\cdot 8 +5\cdot 0+7\cdot 3) \end{bmatrix}$

Compare your answer with the correct one above

Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 4&5 & 6 \end{bmatrix}$

$B=\begin{bmatrix} 0&1 & 4 \end{bmatrix}$

Answer

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 (1x3). This means it has one row and three columns. The second matrix has dimensions of (1x3), also one row and three columns. Since $\tiny \tiny \small 3 eq1$ , we cannot multiply these two matrices together

Compare your answer with the correct one above

Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 4&5 &2 \end{bmatrix}$

$B=\begin{bmatrix} 0\ 1 \ 3 \end{bmatrix}$

Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (1x1). $\begin{bmatrix} (4\cdot 0+5\cdot 1+3\cdot 2) \end{bmatrix}$

Compare your answer with the correct one above

Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 4& 5 \end{bmatrix}$

$B=\begin{bmatrix} 0& 4&2 \ 7&0 &1 \end{bmatrix}$

Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (1x3). The product matrix equals, $\begin{bmatrix} (4\cdot 0+5\cdot 7)&(4\cdot 4 +5\cdot 0)&(4\cdot 2 +5\cdot 1) \end{bmatrix}$

Compare your answer with the correct one above

Question

Compute $A\cdot B$ where,

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

$B=\begin{bmatrix} 7\ 5 \end{bmatrix}$

Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (2x1). The product matrix equals, $\begin{bmatrix} (0\cdot 7+1\cdot 5)\(1\cdot 7+0\cdot 5)\end{bmatrix}$

Compare your answer with the correct one above

Question

Compute $A\cdot B$ where,

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

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

Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (2x2). The product matrix equals, $\begin{bmatrix} (0\cdot 1+0\cdot 0)&(0\cdot 1 +0\cdot 0)\(1\cdot 1 +1\cdot 0)&(1\cdot 1 +1\cdot 0) \end{bmatrix}$

Compare your answer with the correct one above

Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 1\ 2\ 5 \end{bmatrix}$

$B=\begin{bmatrix} 1\ 5 \end{bmatrix}$

Answer

Since the number of columns in the first matrix does not equal the number of rows in the second matrix, you cannot multiply these two matrices.

Compare your answer with the correct one above

Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 12 \end{bmatrix}$

$B=\begin{bmatrix} 4&5 &1 &0 \end{bmatrix}$

Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (1x4). The product matrix equals, $\begin{bmatrix} 12\cdot 4&12\cdot 5&12\cdot 1 &12\cdot 0 \end{bmatrix}$

Compare your answer with the correct one above

Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 0&1 &4 \ 4&0 &7 \ 2& 5& 0 \end{bmatrix}$

$B=\begin{bmatrix} 1\ 2\ 0 \end{bmatrix}$

Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (3x1). The product matrix equals, $\begin{bmatrix} (0\cdot 1+1\cdot 2+4\cdot 0)\(4\cdot 1+0\cdot 2+7\cdot 0)\(2\cdot 1+5\cdot 2+0\cdot 0)\end{bmatrix}$

Compare your answer with the correct one above

Question

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

Evaluate $C^{T}C$ .

Answer

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.

Compare your answer with the correct one above

Question

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 $A\cdot B$ ?

Answer

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.

Compare your answer with the correct one above

Question

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

Evaluate $C^{T}C$ .

Answer

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

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

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. Therefore,

$C^{T}C$

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

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

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

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

Compare your answer with the correct one above

Question

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

Evaluate $C^{2}$ .

Answer

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.

Compare your answer with the correct one above

Question

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

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

Answer

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.

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

Compare your answer with the correct one above

Question

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

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

Answer

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.

Compare your answer with the correct one above

Question

Mary and Marty are playing a game whose board comprises two spaces, marked "X' and "Y", 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 "X" and rolls a "2", he moves to space "Y", since one of the numbers on the arrow going from "X" to 'Y" is "2"; similarly, if he is on space "X" and rolls a "6", he will stay where he is.

Give the stochastic matrix which represents the probabilities that, given that a player is on a particular space, he or she will be on each given space two turns later.

Answer

A stochastic matrix is a matrix of probabilities in which the entry $p_{m,n}$ is the probability that, given the fact that a given system is in a state , the system will be in state next.

We will let State 1 be that the player is on Space "X" and State 2 be that the player is on Space "Y". If the player is on Space "X", the probability that he will still be on that space after one turn is $\frac{1}{6}$ , since one roll out of six will allow him to stay there; similarly, the probability that he will be on Space "Y" is $\frac{5}{6}$ . If a player is on Space "B", the same probabilities are, respectively, $\frac{1}{2}$ and $\frac{1}{2}$ . The stochastic matrix representing these probabilities is

$P = \begin{bmatrix} \frac{1}{6}& \frac{1}{2}\ \ \frac{5}{6}& \frac{1}{2} \end{bmatrix}$ .

The stochastic matrix representing the probabilities of being on a particular space after two terms is the square of this matrix $P^{2}$ , which is

$P ^{2}= \begin{bmatrix} \frac{1}{6}& \frac{1}{2}\ \ \frac{5}{6}& \frac{1}{2} \end{bmatrix}\begin{bmatrix} \frac{1}{6}& \frac{1}{2}\ \ \frac{5}{6}& \frac{1}{2} \end{bmatrix}$

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. Therefore,

$P ^{2}= \begin{bmatrix} \frac{1}{6} \cdot \frac{1}{6}+ \frac{1}{2} \cdot \frac{5}{6} & \frac{1}{6} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \ \ \frac{5}{6} \cdot \frac{1}{6}+ \frac{1}{2}\cdot \frac{5}{6} &\frac{5}{6} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{36}+ \frac{5}{12} & \frac{1}{12} + \frac{1}{4} \ \ \frac{5}{36}+ \frac{5}{12} &\frac{5}{12} + \frac{1}{4} \end{bmatrix}$

$= \begin{bmatrix} \frac{4}{9} & \frac{1}{3} \ \ \frac{5}{9} &\frac{2}{3} \end{bmatrix}$ ,

the correct choice.

Compare your answer with the correct one above

Question

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

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

Answer

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.

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

Compare your answer with the correct one above

Tap the card to reveal the answer