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Precalculus : Find the Product of Two Matrices

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Example Questions

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:

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

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

No Solution

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

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 #1 : Find The Product Of Two Matrices

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:

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

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

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

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

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

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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Example Question #6 : Find The Product Of Two Matrices

We consider the matrices and below. We suppose that and are of the same size

$A=\begin{bmatrix} 1 &1 &\cdots &1 \\ 1 &1 &\cdots &1 \\\vdots &\vdots &\vdots &\vdots \\ 1 &1 &\cdots &1 \end{bmatrix},B=\begin{bmatrix} 2 &2 &\cdots &2 \\ 2 &2 &\cdots &2 \\\vdots &\vdots &\vdots &\vdots \\ 2&2 &\cdots &2 \end{bmatrix}$

What is the product ?

Possible Answers:

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

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

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

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

$\begin{bmatrix} 2n+1 &2n+1 &\cdots &2n+1\\ 2n+1 &2n+1 &\cdots &2n+1 \\\vdots &\vdots &\vdots &\vdots \\ 2n+1 &2n+1 &2n+1 &2n+1 \end{bmatrix}$

Correct answer:

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

Explanation:

Note that every entry of the product matrix is the sum of 2+2+...2 ( times) .

$AB=\begin{bmatrix} 1 &1 &\cdots &1 \\ 1 &1 &\cdots &1 \\\vdots &\vdots &\vdots &\vdots \\ 1 &1 &\cdots &1 \end{bmatrix} \times \begin{bmatrix} 2 &2 &\cdots &2 \\ 2 &2 &\cdots &2 \\\vdots &\vdots &\vdots &\vdots \\ 2&2 &\cdots &2 \end{bmatrix}$

$= \begin{bmatrix} 1(2)+1(2)+... 2 & 1(2)+1(2)+... 2 \\ 1(2)+1(2)+... 2 & 1(2)+1(2)+... 2\\\vdots& \vdots\end{bmatrix}=\begin{bmatrix} 2n & 2n &2n ... \\2n & 2n &2n ... \\\vdots &\vdots\end{bmatrix}$

This gives as every entry of the product of the two matrices.

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Example Question #10 : Find The Product Of Two Matrices

We will consider the two matrices

$A=\begin{bmatrix} 1&1&\cdots &1\\ 0 &0&\cdots &0\\\vdots &\vdots &\vdots &\vdots \\ 0 &0 &0 &0 \end{bmatrix},B=\begin{bmatrix} 1&0 &\cdots &0\\ 1 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 1&0 &0 &0 \end{bmatrix}$

We suppose that and have the same size

What is ?

Possible Answers:

$B=\begin{bmatrix} 1&0 &\cdots &0\\ 1 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ n&0 &0 &0 \end{bmatrix}$

$\begin{bmatrix} n&0 &\cdots &0\\ 0 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 0&0 &0 &0 \end{bmatrix}$

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

$B=\begin{bmatrix} n&0 &\cdots &0\\ n-1 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 1&0 &0 &0 \end{bmatrix}$

$\begin{bmatrix} 1&0 &\cdots &0\\ 1 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 1&0 &0 &0 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} n&0 &\cdots &0\\ 0 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 0&0 &0 &0 \end{bmatrix}$

Explanation:

$AB=\begin{bmatrix} 1&1&\cdots &1\\ 0 &0&\cdots &0\\\vdots &\vdots &\vdots &\vdots \\ 0 &0 &0 &0 \end{bmatrix} \times\begin{bmatrix} 1&0 &\cdots &0\\ 1 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 1&0 &0 &0 \end{bmatrix}$

$=\begin{bmatrix}1(1) +1(1)+...1(1) & 1(0)+0(0)+..0(0)\\ 0(1)+0(0)+...0(0) \\\vdots& \vdots \end{bmatrix}=\begin{bmatrix} 1(n) &0 &...\\0&0&...\\\vdots&\vdots\end{bmatrix}$

Note that when we multiply the first row by the first colum we get: 1+1...1 ( times), this gives the value of .

All other rows are zeros, and therefore we have zeros in the other entries.

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Precalculus : Find the Product of Two Matrices

Study concepts, example questions & explanations for Precalculus

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