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Linear Algebra : Matrices

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Example Questions

Example Question #1 : Matrices

Compute $A\cdot B$ where,

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

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

Possible Answers:

$\begin{bmatrix} 45&24 &10 \\ 45&57 &19 \\ 14&80 &70 \end{bmatrix}$

Not Possibe

$\begin{bmatrix} 26&24 &34 \\ 45&57 &19 \\ 69&75 &71 \end{bmatrix}$

$\begin{bmatrix} 79&75 &69 \\ 45&57 &19 \\ 26&34 &24 \end{bmatrix}$

$\begin{bmatrix} 26&34 &24 \\ 45&57 &19 \\ 79&75 &69 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 26&34 &24 \\ 45&57 &19 \\ 79&75 &69 \end{bmatrix}$

Explanation:

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, $\begin{bmatrix} (3\cdot 2+4\cdot 5+0\cdot 6)&(3\cdot 2 +4\cdot 7+0\cdot 4)&(3\cdot 8 +4\cdot 0+0\cdot 3) \\ (2\cdot 2 +7\cdot 5+1\cdot 6)&(2\cdot 2 +7\cdot 7+1\cdot 4)&(2\cdot 8 +7\cdot 0+1\cdot 3) \\ (6\cdot 2 +5\cdot 5+7\cdot 6)& (6\cdot 2 +5\cdot 7+7\cdot 4)&(7\cdot 8 +5\cdot 0+7\cdot 3) \end{bmatrix}$

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Example Question #2 : Matrices

Compute $A\cdot B$ where,

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

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

Possible Answers:

Not Possible

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

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

$\begin{bmatrix} 11 \end{bmatrix}$

$\begin{bmatrix} 4\\ 5 \\ 6 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 11 \end{bmatrix}$

Explanation:

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Example Question #3 : Matrices

Compute $A\cdot B$ where,

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

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

Possible Answers:

$\begin{bmatrix} 29 \end{bmatrix}$

$\begin{bmatrix} 0\\ 5 \\ 24 \end{bmatrix}$

$\begin{bmatrix} 0&5 & 24 \end{bmatrix}$

$\begin{bmatrix} 4& 8\\ 8&7 \end{bmatrix}$

Not Possible

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 (1x3). This means it has one row and three columns. The second matrix has dimensions of (1x3), also one row and three columns. Since $\tiny \tiny \small 3\neq1$ , we cannot multiply these two matrices together

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Example Question #1 : Matrices

Compute $A\cdot B$ , where

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

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

Possible Answers:

$\begin{bmatrix} 6&8 \\ 4&14 \\ 12&10 \end{bmatrix}$

$\begin{bmatrix} 34&26 \\ 53&39 \\ 47&37 \end{bmatrix}$

$\begin{bmatrix} -14&-22 \\ -31&-45 \\ -13&-23 \end{bmatrix}$

$\begin{bmatrix} 26&34 \\ 39&53 \\ 37&47 \end{bmatrix}$

Not Possible

Correct answer:

$\begin{bmatrix} 26&34 \\ 39&53 \\ 37&47 \end{bmatrix}$

Explanation:

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, $\begin{bmatrix} (3\cdot 2+4\cdot 5)&(3\cdot 2 +4\cdot 7) \\ (2\cdot 2 +7\cdot 5)&(2\cdot 2 +7\cdot 7) \\ (6\cdot 2 +5\cdot 5)& (6\cdot 2 +5\cdot 7) \end{bmatrix}$

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Example Question #691 : Linear Algebra

Compute $A\cdot B$ where,

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

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

Possible Answers:

$\begin{bmatrix} 35\\ 16 \\ 13 \end{bmatrix}$

Not Possible

$\begin{bmatrix} 0&35 \\ 16& 13 \end{bmatrix}$

$\begin{bmatrix} 13&16 & 35 \end{bmatrix}$

$\begin{bmatrix} 35&16 & 13 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 35&16 & 13 \end{bmatrix}$

Explanation:

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Example Question #692 : Linear Algebra

Compute $A\cdot B$ where,

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

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

Possible Answers:

$\begin{bmatrix} 7&5 \\ 5& 7 \end{bmatrix}$

Not Possible

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

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

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

Correct answer:

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

Explanation:

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Example Question #2 : Matrices

Compute $A\cdot B$ where,

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

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

Possible Answers:

$\begin{bmatrix} 2\\ 7 \\25 \end{bmatrix}$

$\begin{bmatrix} 2&25 \\ 7& 3 \end{bmatrix}$

$\begin{bmatrix} 1\\ 25 \end{bmatrix}$

$\begin{bmatrix} 2& 25 \end{bmatrix}$

Not Possible

Correct answer:

Not Possible

Explanation:

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.

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Example Question #2 : Matrices

Compute $A\cdot B$ where,

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

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

Possible Answers:

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

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

Not Possible

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

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

Correct answer:

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

Explanation:

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, $\begin{bmatrix} (0\cdot 1+0\cdot 0)&(0\cdot 1 +0\cdot 0)\\(1\cdot 1 +1\cdot 0)&(1\cdot 1 +1\cdot 0) \end{bmatrix}$

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Example Question #3 : Matrices

Compute $A\cdot B$ where,

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

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

Possible Answers:

Not Possible

$\begin{bmatrix} 48\\ 60\\ 12\\ 0 \end{bmatrix}$

$\begin{bmatrix} 48& 60&12 & 0 \end{bmatrix}$

$\begin{bmatrix} 0\\ 12\\ 60\\ 48 \end{bmatrix}$

$\begin{bmatrix} 16&17 & 13& 12 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 48& 60&12 & 0 \end{bmatrix}$

Explanation:

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Example Question #692 : Linear Algebra

Compute $A\cdot B$ where,

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

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

Possible Answers:

$\begin{bmatrix} 2& 4&12 \end{bmatrix}$

$\begin{bmatrix} 2\\ 4\\ 12 \end{bmatrix}$

$\begin{bmatrix} 0&1 &4 \\ 8&0 &14 \\ 0&0 &0 \end{bmatrix}$

$\begin{bmatrix} 25 \end{bmatrix}$

Not Possible

Correct answer:

$\begin{bmatrix} 2\\ 4\\ 12 \end{bmatrix}$

Explanation:

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, $\begin{bmatrix} (0\cdot 1+1\cdot 2+4\cdot 0)\\(4\cdot 1+0\cdot 2+7\cdot 0)\\(2\cdot 1+5\cdot 2+0\cdot 0)\end{bmatrix}$

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Linear Algebra : Matrices

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #1 : Matrices

Example Question #2 : Matrices

Example Question #3 : Matrices

Example Question #1 : Matrices

Example Question #691 : Linear Algebra

Example Question #692 : Linear Algebra

Example Question #2 : Matrices

Example Question #2 : Matrices

Example Question #3 : Matrices

Example Question #692 : Linear Algebra

All Linear Algebra Resources

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