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Precalculus : Find the Product of a Matrix and a Scalar

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Example Questions

Example Question #541 : Pre Calculus

Find the product.

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

Possible Answers:

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

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

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

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

Correct answer:

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

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:

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Example Question #1 : Multiplication Of Matrices

Find the product.

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

Possible Answers:

$\begin{bmatrix} -16& 80& -32\\ 48& 240& 8 \end{bmatrix}$

$\begin{bmatrix} 32\\ 320\\ -24 \end{bmatrix}$

$\begin{bmatrix} 32\\ 296 \end{bmatrix}$

$\begin{bmatrix} 16& 80& 32\\ 48& 24& 8 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -16& 80& -32\\ 48& 240& 8 \end{bmatrix}$

Explanation:

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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Example Question #2 : Multiplication Of Matrices

Find the product.

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

Possible Answers:

$\begin{bmatrix} 10& 25\\ 5& 15 \end{bmatrix}$

$\begin{bmatrix} 10& 5\\ 25& 15 \end{bmatrix}$

$\begin{bmatrix} 15\\ 40 \end{bmatrix}$

$\begin{bmatrix} 35\\ 20 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 10& 25\\ 5& 15 \end{bmatrix}$

Explanation:

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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Example Question #4 : Find The Product Of A Matrix And A Scalar

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:

Possible Answers:

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

The product depends on knowing the value of m.

The product depends on knowing the size of A

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

We can't perform this multiplication.

Correct answer:

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

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 :

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

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Example Question #5 : Find The Product Of A Matrix And A Scalar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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Example Question #31 : Matrices And Vectors

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

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

Find the product .

Possible Answers:

We can't multiply and .

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

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

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

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

Correct answer:

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

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 n^2 .

This gives the nA which is :

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

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Example Question #41 : Matrices And Vectors

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

Possible Answers:

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

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

Invalid

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

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

Correct answer:

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

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.

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

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Example Question #1 : Matrices

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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.

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

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Example Question #3 : Matrices

Simplify:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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.

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

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

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Example Question #1061 : Algebra

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

What is a+b ?

Possible Answers:

$\frac{191}{4}$

$\frac{448}{5}$

$\frac{148}{5}$

$\frac{41}{5}$

Correct answer:

$\frac{448}{5}$

Explanation:

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

5x=14y

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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Precalculus : Find the Product of a Matrix and a Scalar

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #541 : Pre Calculus

Example Question #1 : Multiplication Of Matrices

Example Question #2 : Multiplication Of Matrices

Example Question #4 : Find The Product Of A Matrix And A Scalar

Example Question #5 : Find The Product Of A Matrix And A Scalar

Example Question #31 : Matrices And Vectors

Example Question #41 : Matrices And Vectors

Example Question #1 : Matrices

Example Question #3 : Matrices

Example Question #1061 : Algebra

All Precalculus Resources

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