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Precalculus : Matrices and Vectors

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Example Questions

Example Question #25 : Multiplication Of Matrices

We consider the two matrice given below, find :

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

Possible Answers:

$\begin{bmatrix} 1 &2 &1 \\ 1&2 &2 \\ 1&2 &3 \end{bmatrix}$

$\begin{bmatrix} 1 &2 &3 \\ 1&2 &2 \\ 1&2 &3 \end{bmatrix}$

$\begin{bmatrix} 1 &1 &1 \\ 1&2 &2 \\ 1&1 &3 \end{bmatrix}$

$\begin{bmatrix} 1 &2 &3 \\ 1&2 &3 \\ 1&2 &3 \end{bmatrix}$

$\begin{bmatrix}1 &1 &0\\ 1&2&0\\1&2&3\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1 &1 &0\\ 1&2&0\\1&2&3\end{bmatrix}$

Explanation:

The number of columns of A is equal to the number of rows of B. Therefore we can perform this operation.

Any entry of the matrix product is the result of the sum of the product of the elements of the row of A with the colum of of B. To obtain the first entry of the matrix product, we use the the first row of A and the first column of B, multiplying componentwise and adding. Doing this operation for each entry,

we obtain our matrix:

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

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

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Example Question #26 : Multiplication Of Matrices

We recall the complex number satisfies : i^2=-1 .

We define the matrix as follows:

$A=\begin{bmatrix} i &i \\ i&i \end{bmatrix}$

Find the matrix .

Possible Answers:

$\begin{bmatrix} i &i \\ i & i\end{bmatrix}$

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

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

We can't find this multiplication.

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

Correct answer:

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

Explanation:

We can treat i as a scalar. To do this multiplication all we need to do is to multply each entry of the matrix by i.

We see that when we multiply each entry of the matrix by i, we obtain i^2 and we know that i^2=-1 .

This means that all the entries are equal to same scalar . Now placing all these scalar in the matrix entries we obtain:

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

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Example Question #27 : Multiplication Of Matrices

Multiply:

$\small \begin{bmatrix} 2 &1 &0 \\-3 &2 &1 \end{bmatrix} * \begin{bmatrix} 4 &1 \\ -1& 0\\ 3& 3\end{bmatrix}$

Possible Answers:

$\small \begin{bmatrix} 8 &2 \\ -7& 0\end{bmatrix}$

$\small \begin{bmatrix} 7 &2 \\ -11& 0\end{bmatrix}$

$\small \begin{bmatrix} 9 &2 \\ -11& 0 \end{bmatrix}$

$\small \begin{bmatrix} 2& 7\\ 0& -11\end{bmatrix}$

$\small \begin{bmatrix} 7& 2\\ 13& 6\end{bmatrix}$

Correct answer:

$\small \begin{bmatrix} 7 &2 \\ -11& 0\end{bmatrix}$

Explanation:

To find the product of 2 matrices, first line up the first row of the left matrix with the first column of the right matrix. Multiply the first, second, and third entries and then add them together.

Next, line up the second row of the left matrix with the second column of the right matrix. Then the first row and the second column, and finally the second row and the first column:

$\small \small \begin{bmatrix} 2 &1 &0 \\-3 &2 &1 \end{bmatrix} * \begin{bmatrix} 4 &1 \\ -1& 0\\ 3& 3\end{bmatrix} = \begin{bmatrix} 2(4)+1(-1)+0(3) & 2(1)+1(0)+0(3)\\ -3(4)+2(-1)+1(3)& -3(1)+2(0)+1(3)\end{bmatrix}$

$\small =\begin{bmatrix} 7 & 2\\ -11& 0 \end{bmatrix}$

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Example Question #11 : Find The Product Of Two Matrices

Multiply $\small \begin{bmatrix} 2 &-1 &4 \\ 1& 0& 2\\3 &1 &-3 \end{bmatrix}*\begin{bmatrix} 10\\-4 \\1 \end{bmatrix}$

Possible Answers:

$\small \begin{bmatrix} 20\\ 12\\ 37\end{bmatrix}$

$\small \begin{bmatrix} 28\\ 12\\ 23\end{bmatrix}$

$\small \begin{bmatrix} 20 &12 & 37 \end{bmatrix}$

$\small \begin{bmatrix} 28 & 12& 23\end{bmatrix}$

$\small \begin{bmatrix} 12\\ 12 \\ 23 \end{bmatrix}$

Correct answer:

$\small \begin{bmatrix} 28\\ 12\\ 23\end{bmatrix}$

Explanation:

To find the product, line up the rows of the left matrix individually with the one column in the right matrix:

$\small \begin{bmatrix} 2(10)+-1(-4)+4(1)\\1(10)+0(-4)+2(1) \\ 3(10)+1(-4)+-3(1)\end{bmatrix}$

$\small =\begin{bmatrix} 28\\ 12\\ 23\end{bmatrix}$

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Example Question #65 : Matrices And Vectors

Which of the following matrices can be multiplied?

Possible Answers:

$\begin{bmatrix} 5 \end{bmatrix}$ $\times\begin{bmatrix} 1& 3 \\ 4&5 \end{bmatrix}$

$\begin{bmatrix} 1&2 \\ 2&1 \\ 4& 2 \end{bmatrix}$ $\times \begin{bmatrix} 3&7 \\ 1&0 \\ 6& 2 \end{bmatrix}$

$\begin{bmatrix} 5 \\ 7 \end{bmatrix}$ $\times \begin{bmatrix} 5 \\ 6 \end{bmatrix}$

$\begin{bmatrix} 4 & 2\\ 6 & 2 \\ 5& 6 \end{bmatrix}$ $\times \begin{bmatrix} 4& 5 \end{bmatrix}$

$\begin{bmatrix} 5 \\ 6 \end{bmatrix}$ $\times \begin{bmatrix} 7 & 8 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 5 \\ 6 \end{bmatrix}$ $\times \begin{bmatrix} 7 & 8 \end{bmatrix}$

Explanation:

The size of every matrix can be written in the form rows x cols. The following matrix is of the size 2 x 1 because it has 2 rows and 1 column.

For two matrices to be able to be multiplied, their sizes must line up that the number of columns in the first matrix is equal to the number of rows of the first matrix. For example:

$2 \times {\color{Green} 1} | {\color{Green} 1}\times2$

So these two matrices can be multiplied. However, if the case were such that:

$\times \begin{bmatrix} 4\\ 3\\ 2 \end{bmatrix}$

$2\times{\color{Red} 1}|{\color{Red} 3}\times1$

Here, the # of columns in the first matrix does not line up with the # of rows in the second matrix, so the two matrices cannot be multiplied.

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Example Question #31 : Multiplication Of Matrices

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

Matrices and are shown above. Find the matrix product .

Possible Answers:

$AB=\begin{bmatrix} 12&8 \\20 &17 \end{bmatrix}$

$AB=\begin{bmatrix} -19& -21 \\-12 & -14 \end{bmatrix}$

$AB=\begin{bmatrix} 15&12 \\-21 &-17 \end{bmatrix}$

$AB=\begin{bmatrix} -22&30 \\-6 &18 \end{bmatrix}$

$AB=\begin{bmatrix} 16&-24 \\ -10 & 23 \end{bmatrix}$

Correct answer:

$AB=\begin{bmatrix} 12&8 \\20 &17 \end{bmatrix}$

Explanation:

First, note that the order of the matrix multiplication is important $(AB\neq BA)$ . Multiplication of two matrices is possible only if the number of columns of the first matrix (A) is equal to the number of rows of the second matrix (B) . Both and are $2\times2$ matrices (2 rows and 2 columns, respectively). Thus, is possible since the number of columns of (2) equals to the number of rows of (2). Furthermore, the size of is equal to the number of rows of and the number of columns of $(2\times2)$ .

To avoid confusion, I will use the notation , , and to denote the constituents of matrices , , and , respectively. For example, $a_{1}_{2}$ refers to the constituent in that is in row 1 and column 2. The general version of the three matrices are shown below:

$A=\begin{bmatrix} a_{1}_{1}&a_{1}_{2} \\a_{2}_{1} &a_{2}_{2} \end{bmatrix}$ $B=\begin{bmatrix} b_{1}_{1}&b_{1}_{2} \\b_{2}_{1} &b_{2}_{2} \end{bmatrix}$ $AB=\begin{bmatrix} c_{1}_{1}&c_{1}_{2} \\c_{2}_{1} &c_{2}_{2} \end{bmatrix}$

Using the rules of multiplying two matrices, the definition of is shown below:

$c_{1}_{1}=a_{1}_{1}\cdot b_{1}_{1}+a_{1}_{2}\cdot b_{2}_{1}=(1)(0)+(3)(4)=12$

$c_{2}_{1}=a_{2}_{1}\cdot b_{1}_{1}+a_{2}_{2}\cdot b_{2}_{1}=(-2)(0)+(5)(4)=20$

$c_{1}_{2}=a_{1}_{1}\cdot b_{1}_{2}+a_{1}_{2}\cdot b_{2}_{2}=(1)(-1)+(3)(3)=8$

$c_{2}_{2}=a_{2}_{1}\cdot b_{1}_{2}+a_{2}_{2}\cdot b_{2}_{2}=(-2)(-1)+(5)(3)=17$

Thus,

$AB=\begin{bmatrix} 12&8 \\20 &17 \end{bmatrix}$

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Example Question #572 : Pre Calculus

Calculate AB when

$A=\begin{bmatrix}-1 & 3\\0 & 3\\2 &1 \end{bmatrix}$ and $B=\begin{bmatrix} -2& 4\\ 3&1 \end{bmatrix}$ .

Possible Answers:

$AB=\begin{bmatrix} 9&-5 \\ -3&0 \\ -6&4 \end{bmatrix}$

$AB=\begin{bmatrix} 11& -1\\ 9& 3\\ -1&9 \end{bmatrix}$

$AB=\begin{bmatrix} 3&-2 \\ 5&0 \end{bmatrix}$

$AB=\begin{bmatrix} 6& -7\\ 5& 12\\ -1&3 \end{bmatrix}$

$AB=\begin{bmatrix} 4&3 \\ -2&4 \end{bmatrix}$

Correct answer:

$AB=\begin{bmatrix} 11& -1\\ 9& 3\\ -1&9 \end{bmatrix}$

Explanation:

In order to perform matrix multiplication, the number of columns in the first matrix has to be the same as the number of rows in the second column.

$A:3x2\:\, B:2x2$

$AB=\begin{bmatrix}-1 & 3\\0 & 3\\2 &1 \end{bmatrix}\cdot \begin{bmatrix} -2&4 \\ 3&1 \end{bmatrix}$

From here, we multiply each term in the first matrix's row by the first column in the second matrix. Continue in this fashion to get the product matrix.

$AB=\begin{bmatrix} (-1)(-2)+(3)(3)& (-1)(4)+(3)(1)\\ (0)(-2)+(3)(3)&(0)(4)+(3)(1) \\ (2)(-2)+(1)(3)&(2)(4)+(1)(1) \end{bmatrix}$

$AB=\begin{bmatrix} 11& -1\\ 9& 3\\ -1&9 \end{bmatrix}$

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Example Question #68 : Matrices And Vectors

Evaluate: $\begin{bmatrix} 2&7 \\ 6&-5 \end{bmatrix} \begin{bmatrix} 3& 4\\ 2&1 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 25&30 \\ -3&-4\end{bmatrix}$

$\begin{bmatrix} 20&15 \\ 8&19 \end{bmatrix}$

$\begin{bmatrix}34&11 \\ -2&7 \end{bmatrix}$

$\begin{bmatrix}11&34 \\ 7&-2 \end{bmatrix}$

$\begin{bmatrix} 6&28 \\ 12&-5 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 20&15 \\ 8&19 \end{bmatrix}$

Explanation:

Write the rule for multiplying a two by two matrix. The result will be a two by two matrix.

$\begin{bmatrix} a& b\\ c&d\end{bmatrix} \begin{bmatrix} w&x \\ y&z \end{bmatrix} = \begin{bmatrix} aw+by&ax+bz \\ cw+dy&cx+dz \end{bmatrix}$

Follow this rule for the given problem.

$\begin{bmatrix} 2&7 \\ 6&-5 \end{bmatrix} \begin{bmatrix} 3& 4\\ 2&1 \end{bmatrix}= \begin{bmatrix} (2)(3)+(7)(2)&(2)(4)+(7)(1) \\ (6)(3)+(-5)(2)&(6)(4)+(-5)(1) \end{bmatrix}$

Simplify, and the answer is: $\begin{bmatrix} 20&15 \\ 8&19 \end{bmatrix}$

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Example Question #31 : Multiplication Of Matrices

Multiply the matrices: $\begin{bmatrix} 1& 3 \end{bmatrix}\begin{bmatrix} 4&9 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 5& 12 \end{bmatrix}$

$\textup{Cannot be solved.}$

$\begin{bmatrix} 4& 27 \end{bmatrix}$

108

Correct answer:

$\textup{Cannot be solved.}$

Explanation:

In order to multiply these matrices we will need to consider the rows and columns for each matrix.

Both matrices have a dimension of $1\times 2$ .

The rule for multiplying matrices is where the number of columns of the first matrix must match the number of rows of the second matrix.

If the dimensions of the first matrix are $a\times b$ , and the dimensions of the second matrix are $b\times c$ , then we will get a dimension of $a \times c$ matrix as a result. If the value of are not matched, we cannot evaluate the product of the matrices.

The correct answer is: $\textup{Cannot be solved.}$

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Example Question #61 : Matrices And Vectors

Find the product of A and B.

$A=\begin{bmatrix} 3 &-1 &4 \end{bmatrix}$

$B=\begin{bmatrix} 2\\ 1\\-3 \end{bmatrix}$

Possible Answers:

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

$\begin{bmatrix} 6 & -2& -1\\4 &3 &-6 \\2 &5 &6 \end{bmatrix}$

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

$\begin{bmatrix} 6 &-1 &-12 \end{bmatrix}$

Correct answer:

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

Explanation:

Since you are multiplying a $1\times 3$ to a $3\times 1$ , the answer is going to be a $1\times 1$ .

To solve, simply multiply each corresponding element and add together.

Thus, your answer is

[-7] .

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Precalculus : Matrices and Vectors

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #25 : Multiplication Of Matrices

Example Question #26 : Multiplication Of Matrices

Example Question #27 : Multiplication Of Matrices

Example Question #11 : Find The Product Of Two Matrices

Example Question #65 : Matrices And Vectors

Example Question #31 : Multiplication Of Matrices

Example Question #572 : Pre Calculus

Example Question #68 : Matrices And Vectors

Example Question #31 : Multiplication Of Matrices

Example Question #61 : Matrices And Vectors

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