Multiplication of Matrices - Pre-Calculus

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Question

Find the product.

$-3\begin{bmatrix} 4& 11& 0 \ 8 & -1 & 14\ 5 & 2 & -3 \end{bmatrix}$

Answer

When we multiply a scalar (regular number) by a matrix, all we need to do is mulitply it to every entry inside the matrix:

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Question

Find the product.

$8\begin{bmatrix} -2& 10& -4\ 6& 30& 1 \end{bmatrix}$

Answer

When we multiply a scalar (regular number) by a matrix, all we need to do is mulitply it through to every entry inside the matrix:

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Question

Find the product.

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

Answer

When we multiply a scalar (regular number) by a matrix, all we need to do is mulitply it through to every entry inside the matrix:

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Question

We consider the following matrix:

$A=\begin{bmatrix} 2^m & 2^m &\cdots & 2^m &2^m \2^m & 2^m &\cdots & 2^m &2^m \ \vdots &\vdots&\vdots& \vdots &\vdots\ 2^m & 2^m &\cdots & 2^m &2^m \ 2^m & 2^m &\cdots & 2^m &2^m \end{bmatrix}$

let $k= \frac{1}{2^{m-1}}$

what matrix do we get when we perform the following product:

Answer

We note k is simply a scalar. To do this multiplication all we need to do is to multply each entry of the matrix by k.

we see that when we multiply we have :

$k\times 2^m=\frac{2^m}{2^{m-1}}=2^{m-m+1}=2$

this gives the entry of the matrix kA.

Therefore the resulting matrix is :

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

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Question

We consider the matrix defined below.

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

Find the sum : $A+A+\cdots +A (n \quad times)$

Answer

Since we are adding the matrix to itself, we 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 form A to the entry from B which is the same as A. This means that to add A+A we simply add each entry of A to itself.

Since the entries from A are the same and given by 1 and the entries from B=A are the same and given by 1, we add these two to obtain:

1+1 and this means that each entry of A+A is 2. We continue in this fashion by additing the entries of A each one to itself n times to obtain that the entries of A+A+....A( n times ) are given by:

$\begin{bmatrix} n &n&\cdots &n \ n &n&\cdots &n\ n&\vdots&\vdots&n \ n &n&\cdots &n \end{bmatrix}$

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Question

Let be a positive integer and let be defined as below:

$A=\begin{bmatrix} n\ \ \vdots \ n \end{bmatrix}$

Find the product .

Answer

We note n is simply a scalar. To do this multiplication all we need to do is to multply each entry of the matrix by n.

We see that when we multiply we have : $n\times n=n^2$ .

This means that each entry of the resulting matrix is .

This gives the nA which is :

$\begin{bmatrix} n^2\ ^2\ \vdots \ n^2 \end{bmatrix}$

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Question

Evaluate: $-3\begin{bmatrix} 2& 3 & 4\ 4&-5 & 10 \end{bmatrix}$

Answer

This problem involves a scalar multiplication with a matrix. Simply distribute the negative three and multiply this value with every number in the 2 by 3 matrix. The rows and columns will not change.

$-3\begin{bmatrix} 2& 3 & 4\ 4&-5 & 10 \end{bmatrix}=\begin{bmatrix} -6&-9 &-12 \ -12& 15&-30 \end{bmatrix}$

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Question

Compute: $2\begin{bmatrix} -3& 4 \end{bmatrix}$

Answer

A scalar that multiplies a one by two matrix will result in a one by two matrix.

Multiply the scalar value with each value in the matrix.

$2\begin{bmatrix} -3& 4 \end{bmatrix} = \begin{bmatrix} -6&8 \end{bmatrix}$

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Question

Simplify:

$\begin{bmatrix} 13 & 4\ 71 &3 \end{bmatrix} + 4\begin{bmatrix} 5 & 3\ 11 &22 \end{bmatrix}$

Answer

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first:

$\begin{bmatrix} 13 & 4\ 71 &3 \end{bmatrix} + 4\begin{bmatrix} 5 & 3\ 11 &22 \end{bmatrix}= \begin{bmatrix} 13 & 4\ 71 &3 \end{bmatrix}+\begin{bmatrix} 20 & 12\ 44 &88 \end{bmatrix}$

The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

$\begin{bmatrix} 13 & 4\ 71 &3 \end{bmatrix} + \begin{bmatrix} 20 & 12\ 44 &88 \end{bmatrix} = \begin{bmatrix} 13 +20 & 4+12\ 71+44 &3 +88\end{bmatrix}$

$\begin{bmatrix} 13 +20 & 4+12\ 71+44 &3 +88\end{bmatrix}= \begin{bmatrix} 33 & 16\ 115 & 91\end{bmatrix}$

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Question

$5\begin{bmatrix} a \ b \end{bmatrix} = 14\begin{bmatrix} 20 \ 12 \end{bmatrix}$

What is ?

Answer

You can begin by treating this equation just like it was:

That is, you can divide both sides by :

$\begin{bmatrix} a \ b \end{bmatrix} = \frac{14}{5}\begin{bmatrix} 20 \ 12 \end{bmatrix}$

Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component:

$\begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} \frac{14}{5}20 \ \frac{14}{5}12 \end{bmatrix}$

Then, simplify:

$\begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} 56 \ \frac{168}{5} \end{bmatrix}$

Therefore, $a+b=56+\frac{168}{5}=\frac{280}{5}+\frac{168}{5}=\frac{448}{5}$

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Question

If $\frac{1}{2}$\bigl(\begin{smallmatrix} 2a\4b \end{smallmatrix} \bigr)$=$\bigl(\begin{smallmatrix} 53 \ 98 \end{smallmatrix} \bigr)$$ , what is ?

Answer

Begin by distributing the fraction through the matrix on the left side of the equation. This will simplify the contents, given that they are factors of :

$\frac{1}{2}\begin{bmatrix} 2a\4b \end{bmatrix}=\begin{bmatrix} a\2b \end{bmatrix}$

Now, this means that your equation looks like:
$\begin{bmatrix} a\2b \end{bmatrix}=\begin{bmatrix} 53 \ 98 \end{bmatrix}$

This simply means:

and

or

Therefore,

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Question

Find 3A given:

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

Answer

To multiply a scalar and a matrix, simly multiply each number in the matrix by the scalar. Thus,

$3A=3\begin{bmatrix} 2 &1 &-3 \2 &7 &-4 \end{bmatrix}=\begin{bmatrix} (3)(2) &(3)(1) &(3)(-3) \(3)(2) & (3)(7)&(3)(-4) \end{bmatrix}=\begin{bmatrix} 6 &3 &-9 \6 &21 &-12 \end{bmatrix}$

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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 .

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 .

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 .

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 .

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.

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:

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 $A^{4}}$ ?

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?

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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