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Precalculus : Multiplication of Matrices

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Example Questions

Example Question #41 : Matrices And Vectors

Find 3A given:

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

Possible Answers:

Not possible

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

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

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

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 #2 : Matrices

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 #1 : Find The Product Of Two 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:

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

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

No Solution

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

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 #2 : Find The Product Of Two Matrices

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

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

Find .

Possible Answers:

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

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

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

No Solution

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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Example Question #1 : Find The Product Of Two Matrices

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

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

Find .

Possible Answers:

No Solution

$\begin{bmatrix} 45 \end{bmatrix}$

$\begin{bmatrix} 56\\ 15\\ 26 \end{bmatrix}$

$\begin{bmatrix} -9 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 45 \end{bmatrix}$

Explanation:

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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Example Question #1 : Find The Product Of Two Matrices

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

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

Find .

Possible Answers:

$\begin{bmatrix} -6\\ 59 \end{bmatrix}$

$\begin{bmatrix} 105 & 0 & -3\\ 42 & -10 & 9 \end{bmatrix}$

$\begin{bmatrix} 15 & 0 & -21 \\ 6 & -10 & 63 \end{bmatrix}$

No Solution

Correct answer:

No Solution

Explanation:

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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Example Question #5 : Find The Product Of Two Matrices

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.

Possible Answers:

x=1,0

x=1

x=0,1,3

$x=\frac{3}{2}$

There is no that satisfies the above equality.

Correct answer:

There is no that satisfies the above equality.

Explanation:

To have the above equality we need to have $x^{2}=x$ and x=3 .

x^2=x means that x=0 , or x=1 . 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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Example Question #51 : Matrices And Vectors

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:

Possible Answers:

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

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

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

$\begin{bmatrix} n & n+2\\ n+2&n \end{bmatrix}$

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

Correct answer:

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

Explanation:

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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Example Question #7 : Find The Product Of Two Matrices

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 $A^{4}}$ ?

Possible Answers:

The correct answer is itself.

$\begin{bmatrix} 4 &0 &0 &0 &0 \\ 4&0 &0 &0 &0 \\ 4&0 &0 &0 &0 \\ 4 &0 &0 &0 &0 \\ 4 &0 & 0&0 &0 \end{bmatrix}$

$\begin{bmatrix} 4 &4 &4 &4 &4 \\ 4&0 &0 &0 &0 \\4 &0 &0 &0 &0 \\ 4 &0 &0 &0 &0 \\ 4 &0 &0 &0 &0 \end{bmatrix}$

$\uplus \begin{bmatrix} 0 &0 &0 & 0 & 0\\ 0&0 &0 &0 &0 \\0 &0 &0 &0 &0 \\ 0&0 &0 &0 &0 \\ 4 &4 & 4 &4 &4 \end{bmatrix}$

$\begin{bmatrix} 4 &4 & 4& 4 & 4\\ 0 &0 &0 &0 &0 \\ 0&0 &0 &0 &0 \\ 0&0 &0 &0 &0 \\ 0 &0 &0 &0 &0 \end{bmatrix}$

Correct answer:

The correct answer is itself.

Explanation:

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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Example Question #8 : Find The Product Of Two Matrices

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

Possible Answers:

n=-1

n=1

n=2

n=4

n=3

Correct answer:

n=2

Explanation:

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

n^2-n-2=0

(n-2)(n+1)=0

n=2, n=-1

Since n is a natural number n=2 is the only possible solution.

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Precalculus : Multiplication of Matrices

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #41 : Matrices And Vectors

Example Question #2 : Matrices

Example Question #1 : Find The Product Of Two Matrices

Example Question #2 : Find The Product Of Two Matrices

Example Question #1 : Find The Product Of Two Matrices

Example Question #1 : Find The Product Of Two Matrices

Example Question #5 : Find The Product Of Two Matrices

Example Question #51 : Matrices And Vectors

Example Question #7 : Find The Product Of Two Matrices

Example Question #8 : Find The Product Of Two Matrices

All Precalculus Resources

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