Find the Product of Two Matrices - Pre-Calculus

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Question

$A=\begin{bmatrix} -16 & -4 & 3 \ 2 & 6 & -7 \ 5 & -1 & 21 \end{bmatrix}$

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

Find .

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Answer

The dimensions of A and B are as follows: A= 3x3, B= 3x1

When we mulitply two matrices, we need to keep in mind their dimensions (in this case 3x_3 and 3_x1).

The two inner numbers need to be the same. Otherwise, we cannot multiply them. The product's dimensions will be the two outer numbers: 3x1.

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Question

$A=\begin{bmatrix} -3 & 0 \ 8 & 5 \end{bmatrix}$

$B=\begin{bmatrix} 2 & 5 \ 10 & 6 \end{bmatrix}$

Find .

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Answer

The dimensions of both A and B are 2x2. Therefore, the matrix that results from their product will have the same dimensions.

$\begin{bmatrix} a & b \ c & d \end{bmatrix} \cdot \begin{bmatrix} e & f \ g & h \end{bmatrix}=\begin{bmatrix} ae+bg & af+bh \ ce+dg & cf+dh \end{bmatrix}$

Thus plugging in our values for this particular problem we get the following:

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Question

$A=\begin{bmatrix} 14 & -5 & -13 \end{bmatrix}$

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

Find .

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Answer

The dimensions of A and B are as follows: A=1x3, B= 3x1.

Because the two inner numbers are the same, we can find the product.

The two outer numbers will tell us the dimensions of the product: 1x1.

$\begin{bmatrix} a & b & c \end{bmatrix} \cdot \begin{bmatrix} d \ e\ f\end{bmatrix}= \begin{bmatrix} ad + be +cf \end{bmatrix}$

Therefore, plugging in our values for this problem we get the following:

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Question

$A=\begin{bmatrix} 3\ -10\ 21 \end{bmatrix}$

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

Find .

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Answer

The dimensions of A and B are as follows: A= 3x1, B= 2x3

In order to be able to multiply matrices, the inner numbers need to be the same. In this case, they are 1 and 2. As such, we cannot find their product.

The answer is No Solution.

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Question

We consider the matrix equality:

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

Find the that makes the matrix equality possible.

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Answer

To have the above equality we need to have $x^{2}=x$ and .

means that , or . Trying all different values of , we see that no can satisfy both matrices.

Therefore there is no that satisfies the above equality.

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Question

Let be the matrix defined by:

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

The value of $A^{n}$ ( the nth power of ) is:

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Answer

We will use an induction proof to show this result.

We first note the above result holds for n=1. This means $A^{1}=\begin{bmatrix} 1 &1 \ 1 &1 \end{bmatrix}$

We suppose that $A^n=\begin{bmatrix} 2^{n-1}& 2^{n-1} \ 2^{n-1}& 2^{n-1} \end{bmatrix}$ and we need to show that: $A^{n+1}=\begin{bmatrix} 2^{n} &2^{n} \2^{n}&2^{n} \end{bmatrix}$

By definition $A^{n+1}=A^{n}A$ . By inductive hypothesis, we have:

$A^{n}=\begin{bmatrix} 2^{n-1}&2^{n-1}\ 2^{n-1}&2^{n-1}\end{bmatrix}$

Therefore, $A^{n+1}=\begin{bmatrix} 2^{n-1}&2^{n-1} \ 2^{n-1} &2^{n} \end{bmatrix}\begin{bmatrix} 1 & 1\ 1&1 \end{bmatrix}$

$A^{n+1}=\begin{bmatrix} 2^{n} &2^{n} \ 2^{n} &2^{n} \end{bmatrix}$

This shows that the result is true for n+1. By the principle of mathematical induction we have the result.

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Question

We will consider the 5x5 matrix defined by:

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

what is the value of $A^{4}}$ ?

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Answer

Note that:

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

Since $A^{4}=A^{2}A^{2}=AA=A^{2}=A$ .

This means that $A^{4}=A$

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Question

Let have the dimensions of a $m\times n^{2}}$ matrix and a $(n+2)\times k$ matrix. When is possible?

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Answer

We know that to be able to have the product of the 2 matrices, the size of the column of A must equal the size of the row of B. This gives :

$n^{2}=n+2$ .

Solving for n, we find

Since n is a natural number is the only possible solution.

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Question

We consider the matrices and below. We suppose that and are of the same size

$A=\begin{bmatrix} 1 &1 &\cdots &1 \ 1 &1 &\cdots &1 \\vdots &\vdots &\vdots &\vdots \ 1 &1 &\cdots &1 \end{bmatrix},B=\begin{bmatrix} 2 &2 &\cdots &2 \ 2 &2 &\cdots &2 \\vdots &\vdots &\vdots &\vdots \ 2&2 &\cdots &2 \end{bmatrix}$

What is the product ?

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Answer

Note that every entry of the product matrix is the sum of ( times) .

$AB=\begin{bmatrix} 1 &1 &\cdots &1 \ 1 &1 &\cdots &1 \\vdots &\vdots &\vdots &\vdots \ 1 &1 &\cdots &1 \end{bmatrix} \times \begin{bmatrix} 2 &2 &\cdots &2 \ 2 &2 &\cdots &2 \\vdots &\vdots &\vdots &\vdots \ 2&2 &\cdots &2 \end{bmatrix}$

$= \begin{bmatrix} 1(2)+1(2)+... 2 & 1(2)+1(2)+... 2 \ 1(2)+1(2)+... 2 & 1(2)+1(2)+... 2\\vdots& \vdots\end{bmatrix}=\begin{bmatrix} 2n & 2n &2n ... \2n & 2n &2n ... \\vdots &\vdots\end{bmatrix}$

This gives as every entry of the product of the two matrices.

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Question

We will consider the two matrices

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

We suppose that and have the same size

What is ?

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Answer

$AB=\begin{bmatrix} 1&1&\cdots &1\ 0 &0&\cdots &0\\vdots &\vdots &\vdots &\vdots \ 0 &0 &0 &0 \end{bmatrix} \times\begin{bmatrix} 1&0 &\cdots &0\ 1 &0 &\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 1&0 &0 &0 \end{bmatrix}$

$=\begin{bmatrix}1(1) +1(1)+...1(1) & 1(0)+0(0)+..0(0)\ 0(1)+0(0)+...0(0) \\vdots& \vdots \end{bmatrix}=\begin{bmatrix} 1(n) &0 &...\0&0&...\\vdots&\vdots\end{bmatrix}$

Note that when we multiply the first row by the first colum we get: ( times), this gives the value of .

All other rows are zeros, and therefore we have zeros in the other entries.

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Question

We consider the matrices and that we assume of the same size .

$A=\begin{bmatrix} 34&0 &\cdots &0\ 34 &0 &\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 34&0 &0 &0 \end{bmatrix},B=\begin{bmatrix} 0&2 &\cdots &0\ 0 &2&\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 0&2 &0 &0 \end{bmatrix}$

Find the product .

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Answer

Note that multiplying every row of by the first column of gives .

Mutiplying every row of by the second column of gives .

Now the remaining columns are columns of zeros, and therefore this product gives zero in every row-column product.

Knowing these three aspects we get the resulting matrix.

$AB=\begin{bmatrix} 0&68 &\cdots &0\ 0&68&\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 0&68&0 &0 \end{bmatrix}$

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Question

We consider the two matrices and defined below:

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

What is the matrix ?

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Answer

The first matrix is (4x1) and the second matrix is (1x3). We can perform the matrix multiplication in this case. The resulting matrix is (4x3).

The first entry in the formed matrix is on the first row and the first column.

It is coming from the product of the first row of A and the first column of B.

This gives $AB(1,1)=1\cdot1=1$ .We continue in this fashion.

The entry (4,3) is coming from the 4th row of A and the 3rd column of B.

This gives $AB(4,3)=4\cdot1=4$ . To obtain the whole matrix we need to remember that any entry on AB say(i,j) is coming from the product of the rom i from A and the column j of B.

After doing all these calculations we obtain:

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

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Question

Let

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

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

What is the matrix ?

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Answer

We note first that A is 4x4 , B is 4x1.

To be able to do BA the number of columns of B must equal the number of rows

of A.

Since the number of columns of B is 1 and the number of rows of A is 4, we do not have equality and therefore we can't have the product BA.

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Question

We consider the two matrices and given below, what is the simplest formula possible for (assume that and have the same size).

$A=\begin{bmatrix} ln(2) & ln(2) &\cdots & ln(2)&ln(2) \ln(2)& ln(2) &\cdots & ln(2) &ln(2)\ \vdots &\vdots&\vdots& \vdots &\vdots\ ln(2) & ln(2) &\cdots & ln(2) &ln(2) \ ln(2) & ln(2) &\cdots & ln(2) &ln(2) \end{bmatrix},B=\begin{bmatrix} ln(3) & ln(3) &\cdots & ln(3)&ln(3) \ln(3)& ln(3) &\cdots & ln(3) &ln(3)\ \vdots &\vdots&\vdots& \vdots &\vdots\ ln(3) & ln(3) &\cdots & ln(3) &ln(3) \ ln(3) & ln(3) &\cdots & ln(3) &ln(3) \end{bmatrix}$

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Answer

Since we are assuming that the two matrices have the same size, we can perform the matrices addition.

We know that when adding matrices, we add them componenwise. Let (i,j) be any entry of the addition matrix. We add the entry from A to the entry from B:

Since the entries from A are the same and given by ln(2) and the entries from B are the same and given by ln(3), we add these two to obtain :

ln(2)+ln(3) and by the properties of the logarithm we have ln(2)+ln(3)=ln(2x3)=ln(6).

Therefore our matrix is given by:

$\begin{bmatrix} ln(6)& ln(6) &\cdots & ln(6)&ln(6) \ln(6)& ln(6) &\cdots & ln(6) &ln(6)\ \vdots &\vdots&\vdots& \vdots &\vdots\ ln(6) & ln(6) &\cdots & ln(6) &ln(6) \ ln(6) & ln(6) &\cdots & ln(6) &ln(6) \end{bmatrix}$

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Question

We consider the two matrice given below, find :

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

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Answer

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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Question

We recall the complex number satisfies : .

We define the matrix as follows:

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

Find the matrix .

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Answer

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 and we know that .

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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Question

Multiply:

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

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Answer

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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Question

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

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Answer

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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Question

Which of the following matrices can be multiplied?

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Answer

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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Question

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

Matrices and are shown above. Find the matrix product .

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Answer

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 is equal to the number of rows of the second matrix . 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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