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Precalculus : Pre-Calculus

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Example Questions

Example Question #561 : Pre Calculus

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

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

Find .

Possible Answers:

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

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

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

No Solution

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 #562 : Pre Calculus

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

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

Find .

Possible Answers:

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

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

No Solution

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

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 #563 : Pre Calculus

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:

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

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

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

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

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

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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Example Question #21 : Multiplication Of Matrices

We consider the matrices and that we assume of the same size .

$A=\begin{bmatrix} 34&0 &\cdots &0\\ 34 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 34&0 &0 &0 \end{bmatrix},B=\begin{bmatrix} 0&2 &\cdots &0\\ 0 &2&\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 0&2 &0 &0 \end{bmatrix}$

Find the product .

Possible Answers:

$AB=\begin{bmatrix} 0&68 &\cdots &0\\ 0 &68 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 0&0 &0 &0 \end{bmatrix}$

$AB=\begin{bmatrix} 68&0 &\cdots &0\\ 68 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 68&0 &0 &0 \end{bmatrix}$

$AB=\begin{bmatrix} 0&0 &\cdots &0\\ 0 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 68&68 &\cdots&68 \end{bmatrix}$

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

$AB=\begin{bmatrix} 0&68 &\cdots &0\\ 0&68&\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 0&68&0 &0 \end{bmatrix}$

Correct answer:

$AB=\begin{bmatrix} 0&68 &\cdots &0\\ 0&68&\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 0&68&0 &0 \end{bmatrix}$

Explanation:

Note that multiplying every row of by the first column of gives .

Mutiplying every row of by the second column of gives .

Now the remaining columns are columns of zeros, and therefore this product gives zero in every row-column product.

Knowing these three aspects we get the resulting matrix.

$AB=\begin{bmatrix} 0&68 &\cdots &0\\ 0&68&\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 0&68&0 &0 \end{bmatrix}$

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Example Question #22 : Multiplication Of Matrices

We consider the two matrices and defined below:

$A=\begin{bmatrix} 1\\2 \\\3 \\4 \end{bmatrix}$ , $B=\begin{bmatrix} 1 &0 &1 \end{bmatrix}$

What is the matrix ?

Possible Answers:

$\begin{bmatrix} 1 & 1 &1 \\ 2& 0 & 2\\ 3 & 0 & 3\\4 & 0 &4 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 &1 \\ 2& 0 & 2\\ 3 & 0 & 3\\4 & 0 &4 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 &1 \\ 2& 0 & 2\\ 3 & 0 & 3\\0 & 0 &0 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 &0 \\ 2& 0 & 0\\ 3 & 0 & 0\\4 & 0 &0\end{bmatrix}$

We can't find the product

Correct answer:

$\begin{bmatrix} 1 & 0 &1 \\ 2& 0 & 2\\ 3 & 0 & 3\\4 & 0 &4 \end{bmatrix}$

Explanation:

The first matrix is (4x1) and the second matrix is (1x3). We can perform the matrix multiplication in this case. The resulting matrix is (4x3).

The first entry in the formed matrix is on the first row and the first column.

It is coming from the product of the first row of A and the first column of B.

This gives $AB(1,1)=1\cdot1=1$ .We continue in this fashion.

The entry (4,3) is coming from the 4th row of A and the 3rd column of B.

This gives $AB(4,3)=4\cdot1=4$ . To obtain the whole matrix we need to remember that any entry on AB say(i,j) is coming from the product of the rom i from A and the column j of B.

After doing all these calculations we obtain:

$\begin{bmatrix} 1 & 0 &1 \\ 2& 0 & 2\\ 3 & 0 & 3\\4 & 0 &4 \end{bmatrix}$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #561 : Pre Calculus

Example Question #562 : Pre Calculus

Example Question #5 : Find The Product Of Two Matrices

Example Question #563 : Pre Calculus

Example Question #7 : Find The Product Of Two Matrices

Example Question #8 : Find The Product Of Two Matrices

Example Question #6 : Find The Product Of Two Matrices

Example Question #10 : Find The Product Of Two Matrices

Example Question #21 : Multiplication Of Matrices

Example Question #22 : Multiplication Of Matrices

All Precalculus Resources

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