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Precalculus : Matrices and Vectors

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Example Questions

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:

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

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

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

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

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

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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Example Question #11 : Find The Product Of Two Matrices

Let

$A=\begin{bmatrix} 1 & 1& 1&1 \\ -1& -1 & -1&-1 \\1 &1 &1 &1 \\ -1&-1 &-1 &-1 \end{bmatrix}$ and

$B=\begin{bmatrix} -1\\ 1\\ -1 \\1 \end{bmatrix}$

What is the matrix ?

Possible Answers:

Product cannot be found.

$A=\begin{bmatrix} 1 & 1& 1&1 \\ -1& 1 & -1&-1 \\1 &1 &0 &1 \\ -1&-1 &-1 &-1 \end{bmatrix}$

$A=\begin{bmatrix} 0& 1& 1&0 \\ -1& -1 & -1&-1 \\0 &1 &1 &0 \\ -1&-1 &-1 &-1 \end{bmatrix}$

$A=\begin{bmatrix} -1 & 1& 1&-1 \\ -1& -1 & -1&-1 \\1 &1 &1 &1 \\ -1&-1 &-1 &-1 \end{bmatrix}$

$A=\begin{bmatrix} 1 & 1& 1&1 \\ -1& -1 & -1&-1 \\1 &1 &1 &1 \\ 1&1 &1 &1 \end{bmatrix}$

Correct answer:

Product cannot be found.

Explanation:

We note first that A is 4x4 , B is 4x1.

To be able to do BA the number of columns of B must equal the number of rows

of A.

Since the number of columns of B is 1 and the number of rows of A is 4, we do not have equality and therefore we can't have the product BA.

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Example Question #24 : Multiplication Of Matrices

We consider the two matrices and given below, what is the simplest formula possible for A+B (assume that and have the same size).

$A=\begin{bmatrix} ln(2) & ln(2) &\cdots & ln(2)&ln(2) \\ln(2)& ln(2) &\cdots & ln(2) &ln(2)\\ \vdots &\vdots&\vdots& \vdots &\vdots\\ ln(2) & ln(2) &\cdots & ln(2) &ln(2) \\ ln(2) & ln(2) &\cdots & ln(2) &ln(2) \end{bmatrix},B=\begin{bmatrix} ln(3) & ln(3) &\cdots & ln(3)&ln(3) \\ln(3)& ln(3) &\cdots & ln(3) &ln(3)\\ \vdots &\vdots&\vdots& \vdots &\vdots\\ ln(3) & ln(3) &\cdots & ln(3) &ln(3) \\ ln(3) & ln(3) &\cdots & ln(3) &ln(3) \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} ln(5) & ln(5) &\cdots & ln(5)&ln(5) \\ln(5)& ln(5) &\cdots & ln(5) &ln(5)\\ \vdots &\vdots&\vdots& \vdots &\vdots\\ ln(5) & ln(5) &\cdots & ln(5) &ln(5) \\ ln(5) & ln(5) &\cdots & ln(5) &ln(5) \end{bmatrix}$

$\begin{bmatrix} ln(2) & ln(2) &\cdots & ln(2)&ln(2) \\ln(2)& ln(2) &\cdots & ln(2) &ln(2)\\ \vdots &\vdots&\vdots& \vdots &\vdots\\ ln(2) & ln(2) &\cdots & ln(2) &ln(2) \\ ln(2) & ln(2) &\cdots & ln(2) &ln(2) \end{bmatrix}$

$\begin{bmatrix} ln(6) & ln(6) &\cdots & ln(6)&ln(6) \\ln(6)& ln(6) &\cdots & ln(6) &ln(6)\\ \vdots &\vdots&\vdots& \vdots &\vdots\\ ln(6) & ln(6) &\cdots & ln(6) &ln(6) \\ ln(6) & ln(6) &\cdots & ln(6) &ln(6) \end{bmatrix}$

$\begin{bmatrix} ln(3)) & ln(3) &\cdots & ln(3)&ln(3) \\ln(3)& ln(3) &\cdots & ln(3) &ln(3)\\ \vdots &\vdots&\vdots& \vdots &\vdots\\ ln(3) & ln(3) &\cdots & ln(3) &ln(3) \\ ln(3) & ln(3) &\cdots & ln(3) &ln(3) \end{bmatrix}$

We can't find the sum of the two matrices.

Correct answer:

Explanation:

Since we are assuming that the two matrices 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 from A to the entry from B:

Since the entries from A are the same and given by ln(2) and the entries from B are the same and given by ln(3), we add these two to obtain :

ln(2)+ln(3) and by the properties of the logarithm we have ln(2)+ln(3)=ln(2x3)=ln(6).

Therefore our matrix is given by:

$\begin{bmatrix} ln(6)& ln(6) &\cdots & ln(6)&ln(6) \\ln(6)& ln(6) &\cdots & ln(6) &ln(6)\\ \vdots &\vdots&\vdots& \vdots &\vdots\\ ln(6) & ln(6) &\cdots & ln(6) &ln(6) \\ ln(6) & ln(6) &\cdots & ln(6) &ln(6) \end{bmatrix}$

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Precalculus : Matrices and Vectors

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

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

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

Example Question #11 : Find The Product Of Two Matrices

Example Question #24 : Multiplication Of Matrices

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