Pre-Calculus - Find the Product of a Matrix and a Scalar

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

What is ?

$\frac{448}{5}$

$\frac{41}{5}$

$\frac{191}{4}$

$\frac{148}{5}$

Explanation

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

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

$\frac{103}{2}$

$\frac{99}{2}$

Explanation

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

Therefore,

Simplify:

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

$\begin{bmatrix} 33 & 16\ 115 & 91\end{bmatrix}$

$\begin{bmatrix} 49\ 206\end{bmatrix}$

$\begin{bmatrix} 49 & 206\end{bmatrix}$

$\begin{bmatrix} 20 & 12\ 44 &88 \end{bmatrix}$

$\begin{bmatrix} 31 & 93\ 3 &21 \end{bmatrix}$

Explanation

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.

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

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

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

$\begin{bmatrix} -6& -9& -12\ -12& 15& -30 \end{bmatrix}$

$\begin{bmatrix} -1& 0& 1\ -12& 15&-30 \end{bmatrix}$

$\begin{bmatrix} -18& 6& -42 \end{bmatrix}$

$\begin{bmatrix} -12&-6 \ 15& -9\ -30&-12 \end{bmatrix}$

$\begin{bmatrix} -12& -1\ 15& 0\ -30& 1 \end{bmatrix}$

Explanation

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.

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

Find the product.

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

$\begin{bmatrix} -12 & -33 & 0 \ -24 & 3 & -42 \ -15 & -6 & 9 \end{bmatrix}$

$\begin{bmatrix} -12 & -33& 0 \ -24 & -3 & -42 \ -15 & -6 & -9 \end{bmatrix}$

$\begin{bmatrix} 1 & 8 & 0 \ 5 & -4 & 11\ 2 & -1 & -6 \end{bmatrix}$

$\begin{bmatrix} -45\ -63\ -12 \end{bmatrix}$

Explanation

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

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:

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

We can't perform this multiplication.

The product depends on knowing the value of m.

The product depends on knowing the size of A

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

Explanation

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 :

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

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 : $A+A+\cdots +A (n \quad times)$

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

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

$\begin{bmatrix} 1 &1 &\cdots &1 \ 1 &1 &\cdots &1\ 1&\vdots&\vdots&1 \ n &n &\cdots &n \end{bmatrix}$

$\begin{bmatrix} n&n &\cdots &n \ 1 &1 &\cdots &1\ 1&\vdots&\vdots&1 \ 1 &1 &\cdots &1 \end{bmatrix}$

$\begin{bmatrix} 1 &1 &\cdots &1 \ n &n &\cdots &n\ 1&\vdots&\vdots&1 \ 1 &1 &\cdots &1 \end{bmatrix}$

Explanation

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:

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

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

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

Find the product .

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

We can't multiply and .

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

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

$\begin{bmatrix} n+1\ +1\ \vdots \ n+1 \end{bmatrix}$

Explanation

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 :

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

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

$\begin{bmatrix} -6& 8 \end{bmatrix}$

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

$\begin{bmatrix} -6\8 \end{bmatrix}$

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

Explanation

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.

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

Find 3A given:

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

$\begin{bmatrix} 6 &3 &-9 \6 &21 &-12 \end{bmatrix}$

$\begin{bmatrix} 5 & 4& 0\ 5 & 10 &-1 \end{bmatrix}$

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

Not possible

Explanation

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}$

Find the Product of a Matrix and a Scalar

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