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Precalculus : Pre-Calculus

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Example Questions

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 #551 : Pre Calculus

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:

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

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

We can't multiply and .

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

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

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 #552 : Pre Calculus

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

Possible Answers:

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

Invalid

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

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

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

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:

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

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

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

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

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

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 #1 : Matrices

Simplify:

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

Possible Answers:

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

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

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

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

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

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 #2 : Matrices

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

What is a+b ?

Possible Answers:

$\frac{448}{5}$

$\frac{41}{5}$

$\frac{191}{4}$

$\frac{148}{5}$

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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Example Question #553 : Pre Calculus

Find 3A given:

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

Possible Answers:

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

Not possible

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #551 : Pre Calculus

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

Possible Answers:

$\frac{103}{2}$

$\frac{99}{2}$

102

Correct answer:

102

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:

a=53

and

2b=98 or b=49

Therefore, a+b=53+49=102

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Example Question #11 : Multiplication Of Matrices

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

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

Find .

Possible Answers:

$\begin{bmatrix} -26\\ 12\\ 43 \end{bmatrix}$

No Solution

$\begin{bmatrix} -34\\ 4\\ 25 \end{bmatrix}$

$\begin{bmatrix} -45\\ 21\\ 27 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -45\\ 21\\ 27 \end{bmatrix}$

Explanation:

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 3x3 and 3x1).

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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Example Question #12 : Multiplication Of Matrices

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

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

Find .

Possible Answers:

$\begin{bmatrix} -6 & -15\\ 66 & 70 \end{bmatrix}$

$\begin{bmatrix} 34 & 25 \\ 18 & 30 \end{bmatrix}$

No Solution

$\begin{bmatrix} -6 & 0 \\ 80 & 30 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -6 & -15\\ 66 & 70 \end{bmatrix}$

Explanation:

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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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

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

Example Question #551 : Pre Calculus

Example Question #552 : Pre Calculus

Example Question #1 : Matrices

Example Question #1 : Matrices

Example Question #2 : Matrices

Example Question #553 : Pre Calculus

Example Question #551 : Pre Calculus

Example Question #11 : Multiplication Of Matrices

Example Question #12 : Multiplication Of Matrices

All Precalculus Resources

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