11 exercises on matrix multiplication

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Study with the 11 exercises on matrix multiplication, all with step-by-step resolution so you can solve your doubts and do well in exams and entrance exams.

question 1

Given the following matrices, check the option that indicates only possible products.

start style math size 18px bold A with bold 2 bold x bold 1 subscript end of subscript bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space B with bold 3 bold x bold 3 subscript end of subscript bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space C with bold 1 bold x bold 3 bold subscript space end of subscript bold bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space D with bold 3 bold x bold 2 subscript end of subscript end of style

a) C.A, B.A, A.D.
b) D.B, D.C, A.D.
c) AC, D.A, C.D.
d) B.A, A.B, D.C
e) A.D., D.C., C.A.

Correct answer: c) AC, D.A, C.D

A.C is possible because the number of columns in A (1) is equal to the number of rows in C (1).

D.A is possible, because the number of columns in D (2) is equal to the number of rows in A (2).

C.D is possible because the number of columns in C (3) is equal to the number of rows in D (3).

question 2

Make matrix product A. B.

A equal to open square brackets table row with 3 cell minus 2 end of cell 1 row with 1 5 cell with minus 1 end of cell end of table closes square brackets space space space space space space space space space space space B equal to open square brackets table row with 1 3 row with 0 cell with minus 5 end of cell row with 4 1 end of table close brackets

First we must check if it is possible to carry out the multiplication.

Since A is a 2x3 matrix and B a 3x2 matrix, it is possible to multiply, as the number of columns in A is equal to the number of rows in B.

We checked the dimensions of the matrix resulting from the multiplication.

Calling the result matrix of product A. B of matrix C, this will have two rows and two columns. Remember that the result matrix of the product "inherits" the number of rows from the first and the number of columns from the second.

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Therefore, matrix C will be of type 2x2. Building the generic matrix C, we have:

C = open square brackets table row with cell with c with 11 subscript end of cell cell with c with 12 subscript end of cell row with cell with c with 21 subscript end of cell cell with c with 22 subscript end of cell end of table close brackets

To calculate c11, we multiply the first line of A for the first column of B, adding the multiplied terms.

c11 = 3.1 + (-2).0 + 1.4 = 3 + 0 + 4 = 7

To calculate c12, we multiply the first line of A for the second column of B, adding the multiplied terms.

c12 = 3.3 + (-2).(-5) + 1.1 = 9 + 10 + 1 = 20

To calculate c21, we multiply the second line of A for the first column of B, adding the multiplied terms.

c21 = 1.1 + 5.0 + (-1).4 = 1 + 0 + (-4) = -3

To calculate c22, we multiply the second line of A for the second column of B, adding the multiplied terms.

c22 = 1.3 + 5.(-5) + (-1).1 = 3 + (-25) + (-1) = -23

Writing matrix C with its terms.

C = open brackets table row with 7 20 row with cell with minus 3 end of cell cell with minus 23 end of cell end of table close square brackets

question 3

Solve the matrix equation and determine the values ​​of x and y.

open square brackets table row with cell minus 1 end of cell 2 row with 4 cell minus 3 end of cell end of table closes square brackets. open brackets table row with x row with y end of table closes square brackets equal to open brackets table row with 3 row with cell with minus 4 end of cell end of table close brackets

We verified that it is possible to multiply the matrices before equality, as they are of type 2x2 and 2x1, that is, the number of columns in the first is equal to the number of rows in the second. The result is the 2x1 matrix on the right side of the equality.

We multiply row 1 of the first matrix by column 1 of the second matrix and equal to 3.

-1.x + 2.y = 3
-x + 2y = 3 (equation I)

We multiply row 2 of the first matrix by column 1 of the second matrix and equal to -4.

4.x + (-3).y = -4
4x - 3y = -4 (equation II)

We have two equations and two unknowns and we can solve a system to determine x and y.

Multiplying both sides of equation I by 4 and adding I + II, we have:

opens keys table attributes column alignment left end attributes row with cell with minus x plus 2 y equals 3 space left parenthesis and q u a tion space I right parenthesis end of cell row with cell with 4 x minus 3 y space equals minus 4 space left parenthesis e q u a tio n space I I right parenthesis end of cell end of table close open keys table attributes column alignment left end of attributes row with cell with 4. left parenthesis minus x plus 2 y right parenthesis equal to 4.3 space left parenthesis I right parenthesis end of cell row with cell with 4x minus 3 y space equal to minus 4 space left parenthesis I I right parenthesis end of cell end of table close stack attributes charalign center stackalign right end attributes row minus 4 x plus 8 y equal to 12 end row row plus 4 x minus 3 y equal to minus 4 end row horizontal line row 0 x plus 5 y equal to 8 end row end stack space space 5 y equal to 8 y equal to 8 about 5

Substituting y in equation I and solving for x, we have:

minus x plus 2 y equals 3 minus x plus 2.8 over 5 equals 3 minus x plus 16 over 5 equals 3 minus x equals 3 minus 16 over 5 minus x equals 15 over 5 minus 16 over 5 minus x. left parenthesis minus 1 right parenthesis equals minus 1 fifth. left parenthesis minus 1 right parenthesis x equals 1 fifth

So we have x equals 1 fifth space and y space equals 8 over 5

question 4

Given the following linear system, associate a matrix equation.

open braces table attributes column alignment left end attributes row with cell with a space more space b space more space 2 c space equal to space 3 end of cell row with cell with minus a space minus space b space plus space c space equal to space 4 end of cell row with cell with 5 a space plus space 2 b space minus space c space equal to space 6 end of cell end of table closes

There are three equations and three unknowns.

To associate a matrix equation to the system, we must write three matrices: the coefficients, the unknowns and the independent terms.

Coefficients matrix

open square brackets table row with 1 1 2 row with cell with minus 1 end of cell cell with minus 1 end of cell 1 row with 5 2 cell with minus 1 end of cell end of table close square brackets

Unknown matrix

open brackets table row with row with b row with c end of table close brackets

Matrix of independent terms

open brackets table row with 3 row with 4 row with 6 end of table close brackets

matrix equation

Matrix of coefficients. matrix of unknowns = matrix of independent terms

open square brackets table row with 1 1 2 row with cell with minus 1 end of cell cell with minus 1 end of cell 1 row with 5 2 cell with minus 1 end of cell end of table closes square brackets. open brackets table row with row with b row with c end of table close brackets equal to open brackets table row with 3 row with 4 row with 6 end of table close brackets

question 5

(UDESC 2019)

Given the matrices and knowing that A. B = C, so the value of x + y is equal to:

a) 1/10
b) 33
c) 47
d) 1/20
e) 11

Correct answer: c) 47

To determine the values ​​of x and y, we solve the matrix equation by obtaining a system. When solving the system, we get the values ​​of x and y.

THE. B equals C opens square brackets table row with cell with 2 x minus 1 end of cell cell with 5 y plus 2 end of cell row with cell with 3x minus 2 end of cell cell with 4 y plus 3 end of cell end of table close brackets. open square brackets table row with 4 row with cell minus 2 end of cell end of table closes square brackets equal to open square brackets table row with cell with 2 y minus 12 end of cell row with cell with 6 x plus 2 end of cell end of table close square brackets

Multiplying the matrices:

opens keys table attributes column alignment left end attributes row with cell with left parenthesis 2 x minus 1 right parenthesis space. space 4 space plus space left parenthesis 5 y plus 2 right parenthesis space. space left parenthesis minus 2 right parenthesis space equals space 2 y minus 12 space left parenthesis space e q u action space I right parenthesis end of cell row with cell with left parenthesis 3 x minus 2 right parenthesis space. space 4 space plus space left parenthesis 4 y plus 3 right parenthesis space. space left parenthesis minus 2 right parenthesis space equals space 6 x plus 2 space left parenthesis e q u a tion space I I right parenthesis end of cell end of table close opens keys table attributes column alignment left end attributes row with cell with 8 x minus 4 space plus space left parenthesis minus 10 y right parenthesis space minus 4 equals 2 y minus 12 space left parenthesis e q u a tion space I right parenthesis end of cell row to cell with 12 x minus 8 plus left parenthesis minus 8 y right parenthesis minus 6 equals 6 x plus 2 space left parenthesis e q u a tion space I I right parenthesis end of cell end of table close opens keys table attributes column alignment left end attributes row with cell with 8 x minus 12 y equals minus 12 plus 4 plus 4 space left parenthesis e q u a ç ã o space I right parenthesis end of cell row to cell with 6 x minus 8 y equals 2 plus 6 plus 8 space left parenthesis e q u a tion space I I right parenthesis end of cell end of table closes open keys table attributes column alignment left end of attributes row with cell 8 x minus 12 y equals minus 4 space parenthesis left and q u a tion space I right parenthesis end of cell row to cell with 6 x minus 8 y equal to 16 space left parenthesis and q u a tion space I I right parenthesis end of cell end of table closes

Isolating x in equation I

8 x space equal to space minus 4 plus 12 y x space equal to space numerator minus 4 over denominator 8 end of fraction plus numerator 12 y over denominator 8 end of fraction

Substituting x in equation II

6. open parentheses minus 4 over 8 plus numerator 12 y over denominator 8 end of fraction close parenthesis minus 8 y equals 16 minus 24 over 8 plus numerator 72 y over denominator 8 end of fraction minus 8 y equal to 16

matching the denominators

minus 24 over 8 plus numerator 72 y over denominator 8 end of fraction minus 8 y equals 16 minus 24 over 8 plus numerator 72 y over denominator 8 end of fraction minus numerator 64 y over denominator 8 end of fraction equal to 16 1 about 8. left parenthesis 72 y space minus space 24 space minus space 64 y right parenthesis equal to 16 72 y minus 64 y space minus space 24 equals 16 space. space 8 8 y equal to 128 plus 24 8 y equal to 152 y equal to 152 over 8 equal to 19

To determine x, we substitute y into equation II

6 x minus 8 y equal to 16 6 x minus 8.19 equal to 16 6 x minus 152 equal to 16 6 x equal to 16 plus 152 6 x equal to 168 x equal to 168 over 6 space equal to 28

Thus,

x + y = 19 + 18
x + y = 47

question 6

(FGV 2016) Given the matrix and knowing that the matrix is the inverse matrix of matrix A, we can conclude that the matrix X, which satisfies the matrix equation AX = B, has as the sum of its elements the number

a) 14
b) 13
c) 15
d) 12
e) 16

Correct answer: b) 13

Any matrix multiplied by its inverse is equal to the identity matrix In.

straight A. straight A to the power of minus 1 end of exponential equal to open square brackets table row with 1 0 row with 0 1 end of table close square brackets

Multiplying both sides of the equation AX = B by A to the power of minus 1 end of the exponential.

A to the power of minus 1 end of the exponential. THE. X equals A to the power of minus 1 end of the exponential. B I with n subscript. X equals A to the power of minus 1 end of the exponential. B I with n subscript. X equal to open square brackets table row with 2 cell with minus 1 end of cell row with 5 3 end of table closes square brackets. open square brackets table row with 3 row with cell minus 4 end of cell end of table closes square brackets

Making the product on the right side of the equation.

I with n subscribed. X equals open square brackets table row with cell with 2.3 space plus space left parenthesis minus 1 right parenthesis. left parenthesis minus 4 right parenthesis space space end of cell row with cell with 5.3 space plus space 3. left parenthesis minus 4 right parenthesis end of cell end of table closes square brackets I with n subscript. X equal to open square brackets table row with cell with 6 plus 4 end of cell row with cell with 15 minus 12 end of cell end of table closes I brackets with n subscript. X equals open square brackets table row with 10 row with 3 end of table close brackets

How the identity matrix is ​​the neutral element of the matrix product

X equals open square brackets table row with 10 row with 3 end of table close brackets

Thus, the sum of its elements is:

10 + 3 = 13

question 7

Given the matrix following matrix A, calculate its inverse matrix, if any.

A equal to open brackets table row with 3 7 row with 5 12 end of table close brackets

A is invertible, or invertible if there is a square matrix of the same order which, when multiplying or being multiplied by A, results in the identity matrix.

We intend to identify the existence, or not, of a matrix A to the power of minus 1 end of the exponential for what:

THE. A to the power of minus 1 end of the exponential equals A to the power of minus 1 end of the exponential. A equals I with n subscript

Since A is a square matrix of order 2, A to the power of minus 1 end of the exponential must also have order 2.

Let's write the inverse matrix with its values ​​as unknowns.

A to the power of minus 1 end of exponential equal to open square brackets table row with a b row with c d end of table close square brackets

Writing the matrix equation and solving the product.

THE. A to the power of minus 1 end of exponential equal to I with n subscript open square brackets table row with 3 7 row with 5 12 end of table close square brackets. open brackets table row with a b row with c d end of table closes square brackets equal to open brackets table row with 1 0 row with 0 1 end of table close square brackets open square brackets table row with cell with 3 a plus 7 c end of cell cell with 3 b plus 7 d end of cell row with cell with 5 a plus 12 c end of cell cell with 5 b plus 12 d end of cell end of table closes square brackets equal to open square brackets table row of 1 0 row of 0 1 end of table close brackets

Equating the equivalent terms on both sides of the equality.

3a + 7c = 1
5a + 12c = 0
3b + 7d = 0
5b + 12d = 1

We have a system with four equations and four unknowns. In this case, we can split the system into two. Each with two equations and two unknowns.

open keys table attributes column alignment left end attributes row with cell 3 a space plus 7 c space equal space a space 1 space end of cell row with cell with 5 a space plus space 12 c space equal to space 0 end of cell end of table close

solving the system
Isolating a in the first equation

3 a space equals space 1 space minus space 7 c space equals space numerator space 1 space minus space 7 c over denominator 3 end of fraction

Substituting a in the second equation.

5. open parentheses numerator 1 minus 7 c over denominator 3 end of fraction closes parentheses plus 12 c equal to 0 numerator 5 minus 35 c over denominator 3 end of fraction plus 12 c equal to 0 numerator 5 minus 35 c over denominator 3 end of fraction plus numerator 3.12 c over denominator 3 end of fraction equal to 0 5 minus 35 c plus 36 c equal to 0 bold italic c bold equals bold minus bold 5

Replacing c

a equal to numerator 1 minus 7. left parenthesis minus 5 right parenthesis over denominator 3 end of fraction a equal to numerator 1 plus 35 over denominator 3 end of fraction a equals 36 over 3 bold italic bold equals bold 12

and the system:

open keys table attributes column alignment left end attributes row with cell with 3 b space plus 7 d space equal space a space 0 space end of cell row with cell with 5 b space plus space 12 d space equals space 1 end of cell end of table close

Isolating b in the first equation

3 b equals minus 7 d b equals numerator minus 7 d over denominator 3 end of fraction

Substituting b in the second equation

5. open parentheses minus numerator 7 d over denominator 3 end of fraction closes parenthesis plus 12 d equals 1 numerator minus 35 d over denominator 3 end of fraction plus 12 d space equals space 1 numerator minus 35 d over denominator 3 end of fraction plus numerator 36 d over denominator 3 end of fraction equal to 1 minus 35 d plus 36 d equal to 1.3 bold italic d bold equal to bold 3

Substituting d to determine b.

b equals numerator minus 7.3 over denominator 3 end of fraction bold italic b bold equals bold minus bold 7

Replacing the determined values ​​in the inverse unknown matrix

A to the power of minus 1 end of exponential equal to open square brackets table row with a b row with c d end of table close square brackets equal to open square brackets table row with 12 cell minus 7 end of cell row with cell minus 5 end of cell 3 end of table close brackets

Checking whether the calculated matrix is, in fact, the inverse matrix of A.

For this, we must perform the multiplications.

THE. A to the power of minus 1 end of the exponential equal to I with n subscript space and space A to the power of minus 1 end of the exponential. A equals I with n subscript
P a r to space A. A to the power of minus 1 end of the exponential equal to I with n subscript
open square brackets table row with 3 7 row with 5 12 end of table closes square brackets. open square brackets table row with 12 cell minus 7 end of cell row with cell minus 5 end of cell 3 end of table close square brackets equal to open brackets table row with 1 0 row with 0 1 end of table close brackets open brackets table row with cell with 3.12 plus 7. left parenthesis minus 5 right parenthesis end of cell cell with 3. left parenthesis minus 7 right parenthesis plus 7.3 end of cell row to cell with 5.12 plus 12. left parenthesis minus 5 right parenthesis end of cell cell with 5. left parenthesis minus 7 right parenthesis plus 12.3 end of cell end of table closes square brackets equals open square brackets table row with 1 0 row with 0 1 end of table closes square brackets opens square brackets table row with cell with 36 minus 35 end of cell cell with minus 21 plus 21 end of cell row with cell with 60 minus 60 end of cell cell with minus 35 plus 36 end of cell end of table closes square brackets equal to open square brackets table row with 1 0 row with 0 1 end of table close square brackets open square brackets table row with 1 0 row with 0 1 end of table close brackets equal to open square brackets table row with 1 0 row with 0 1 end of table close brackets
P a r a space A to the power of minus 1 end of the exponential. A equal to I with n subscript opens square brackets table row with 12 cell with minus 7 end of cell row with cell with minus 5 end of cell 3 end of table closes square brackets. open brackets table row with 3 7 row with 5 12 end of table close brackets equal to open brackets table row with 1 0 row with 0 1 end of table close brackets open square brackets table row with cell with 12.3 plus left parenthesis minus 7 right parenthesis.5 end of cell cell with 12.7 plus left parenthesis minus 7 right parenthesis.12 end of cell row with cell with minus 5.3 plus 3.5 end of cell cell with minus 5.7 plus 3.12 end of cell end of table close square brackets equal to open square brackets table row with 1 0 row with 0 1 end of table close square brackets open square brackets table row with cell with 36 minus 35 end of cell cell with 84 minus 84 end of cell row with cell with minus 15 plus 15 end of cell cell with minus 35 plus 36 end of cell end of table closes square brackets equal to open square brackets table row with 1 0 row with 0 1 end of table close brackets open brackets table row with 1 0 row with 0 1 end of table close brackets equal to open brackets table row with 1 0 row with 0 1 end of table close brackets

Therefore, fractions are invertible.

question 8

(EsPCEx 2020) Be the matrices A equal to open square brackets table row with 1 cell with minus 1 end of cell 1 row with 2 1 cell with minus 3 end of cell row with 1 1 cell with minus 1 end of cell end of table closes square brackets comma B space equals open square brackets table row with x row with y row with z end of table closes square brackets space and space C equals space open square brackets table row 0 row with cell minus 12 end of cell row with cell minus 4 end of cell end of table close brackets. If AB=C, then x+y+z is equal to

a) -2.
b) -1.
c) 0.
d) 1.
e) 2.

Correct answer: e) 2.

To determine the unknowns x, y and z, we must perform the matrix equation. As a result, we will have a linear system of three equations and three unknowns. When solving the system, we determine x, y, and z.

THE. B equals C open square brackets table row with 1 cell with minus 1 end of cell 1 row with 2 1 cell with minus 3 end of cell row with 1 1 cell with minus 1 end of cell end of table closes brackets. open brackets table row with x row with y row with z end of table close brackets equal to open brackets table row with 0 row with cell with minus 12 end of cell row with cell with minus 4 end of cell end of table close square brackets open square brackets table row with cell with 1. x plus left parenthesis minus 1 right parenthesis. y plus 1. z end of cell row to cell with 2. x plus 1. y plus left parenthesis minus 3 right parenthesis. z end of cell row to cell with 1. x plus 1. y plus left parenthesis minus 1 right parenthesis. z end of cell end of table closes square brackets equal to open square brackets table row 0 row with cell minus 12 end of cell row with cell minus 4 end of cell end of table close square brackets open square brackets table row with cell with x minus y plus z end of cell row with cell with 2 x plus y minus 3 z end of cell row with cell with x plus y minus z end of cell end of table closes square brackets equal to open square brackets table row 0 row with cell minus 12 end of cell row with cell minus 4 end of cell end of table close brackets

By the equality of matrices, we have:

open braces table attributes column alignment left end attributes row with cell with x minus y plus z equal to 0 bold space left parenthesis bold italic and bold italic q bold italic u bold italic a bold italic ç bold italic ã bold italic o bold space bold italic I bold right parenthesis end of cell row with cell with 2 x plus y minus 3 z equals minus 12 space bold left parenthesis bold italic and bold italic q bold italic u bold italic a bold italic ç bold italic ã bold italic o bold space bold italic I bold italic I bold right parenthesis end of cell row with cell with x plus y minus z equals minus 4 space bold left parenthesis bold italic and bold italic q bold italic u bold italic a bold italic ç bold italic ã bold italic bold space bold italic I bold italic I bold italic I bold right parenthesis end of cell end of table closes

Adding equations I and III

stack attributes charalign center stackalign right end row attributes x minus y plus z equals nothing 0 end row row x plus y minus z equals minus 4 end row horizontal line row 2 x equals minus 4 end row end stack

So x = -4/2 = -2

Substituting x = -2 in equation I and isolating z.

minus 2 minus y plus z equals 0 z equals y plus 2

Substituting the values ​​of x and z in equation II.

2. left parenthesis minus 2 right parenthesis plus y minus 3. left parenthesis y plus 2 right parenthesis equals minus 12 minus 4 plus y minus 3 y minus 6 equals minus 12 minus 2 y equals a minus 12 plus 6 plus 4 minus 2 y equals minus 2 y equals numerator minus 2 over denominator minus 2 end of fraction y equals 1

Substituting the values ​​of x and y in equation I, we have:

minus 2 minus 1 plus z equals 0 minus 3 plus z equals 0 z equals 3

Thus, we have to:

x plus y plus z equals minus 2 plus 1 plus 3 equals minus 2 plus 4 equals 2

Therefore, the sum of the unknowns is equal to 2.

question 9

(PM-ES) About matrix multiplication, Fabiana wrote the following sentences in her notebook:

I space minus A space with 4 X 2 subscript end of subscript space. space B with 2 X 3 subscript end of subscript space equals space C with 4 X 3 subscript end of subscript space space I I space minus space A with 2 X 2 subscript end of subscript space. space B with 2 X 3 subscript end of subscript space equal to space C with 3 X 2 subscript end of subscript space space I I I space minus space A with 2 X 4 subscript end of subscript space. space B with 3 X 4 subscript end of subscript space equal to space C with 2 X 4 subscript end of subscript space space I V space minus space A with 1 X 2 subscript end of subscript space. B space with 2 X 1 subscript end of subscript space equal to C space with 1 x 1 subscript end of subscript

What Fabiana says is correct:

a) only in I.
b) only in II.
c) only in III.
d) only in I and III.
e) only in I and IV

Correct answer: e) only in I and IV

It is only possible to multiply matrices when the number of columns in the first is equal to the number of rows in the second.

Therefore, sentence III is already discarded.

The matrix C, will have the number of rows of A and the number of columns of B.

Thus, sentences I and IV are correct.

question 10

Given matrix A, determine A squared. A to the power of t.

A equal to open square brackets table row with 3 2 row with cell with minus 1 end of cell cell with minus 4 end of cell end of table close square brackets

Step 1: Determine A squared.

A squared equals A. A squared equal to open square brackets table row with 3 2 row with cell with minus 1 end of cell cell with minus 4 end of cell end of table closes square brackets. open square brackets table row with 3 2 row with cell with minus 1 end of cell cell with minus 4 end of cell end of table closes square brackets A equals open square brackets table row with cell with 3.3 plus 2. left parenthesis minus 1 right parenthesis end of cell cell with 3.2 plus 2. left parenthesis minus 4 right parenthesis end of cell row with cell minus 1.3 plus left parenthesis minus 4 right parenthesis. left parenthesis minus 1 right parenthesis cell end cell minus 1.2 plus left parenthesis minus 4 right parenthesis. left parenthesis minus 4 right parenthesis end of cell end of table closes square brackets A equals open square brackets table row with cell with 9 minus 2 end of cell cell with 6 minus 8 end of cell row with cell with minus 3 plus 4 end of cell cell with minus 2 plus 16 end of cell end of table closes square brackets A squared equals open square brackets table row with 7 cell with minus 2 end of cell row with 1 14 end of table close brackets

Step 2: Determine the transposed matrix A to the power of t.

We obtain the transposed matrix of A by orderly swapping the rows for the columns.

A to the power of t equal to open square brackets table row with 3 cell with minus 1 end of cell row with 2 cell with minus 4 end of cell end of table close square brackets

Step 3: Solve the matrix product A squared. A to the power of t.

open square brackets table row with 7 cell with minus 2 end of cell row with 1 14 end of table closes square brackets. open square brackets table row with 3 cell minus 1 end of cell row with 2 cell minus 4 end of cell end of table close square brackets equal to open square brackets table row with cell with 7.3 plus left parenthesis minus 2 right parenthesis.2 end of cell cell with 7. left parenthesis minus 1 right parenthesis plus left parenthesis minus 2 right parenthesis. left parenthesis minus 4 right parenthesis end of cell row with cell with 1.3 plus 14.2 end of cell cell with 1. left parenthesis minus 1 right parenthesis plus 14. left parenthesis minus 4 right parenthesis end of cell end of table closes square brackets open square brackets table row with cell with 21 minus 4 end of cell cell minus 7 plus 8 end of cell row with cell 3 plus 28 end of cell cell minus 1 minus 56 end of cell end of table closes square brackets open square brackets table row with 17 1 row with 31 cell minus 57 end of cell end of table close brackets

Therefore, the result of the matrix product is:

A squared. A to the power of t equal to open square brackets table row with 17 1 row with 31 cell minus 57 end of cell end of table closes squares

question 11

(UNICAMP 2018) The and B real numbers such that the matrix A equal to open brackets table row with 1 2 row with 0 1 end of table close brackets satisfies the equation A squared space equals space a A space plus space b I, on what I is the order 2 identity matrix. Therefore, the product ab it's the same as

a) −2.
b) −1.
c) 1.
d) 2.

Correct answer: a) -2.

Step 1: Determine A squared.

A squared equal to open square brackets table row with 1 2 row with 0 1 end of table closes square brackets. open brackets table row with 1 2 row with 0 1 end of table close brackets A squared equals open brackets table row with cell with 1.1 plus 2.0 end of cell cell with 1.2 plus 2.1 end of cell row with cell with 0.1 plus 1.0 end of cell cell with 0.2 plus 1.1 end of cell end of table closes square brackets A squared equals open square brackets table row with 1 4 row with 0 1 end of table close brackets

Step 2: Determine a. THE.

The. A equal to opens square brackets table row with cell with a.1 end of cell cell with a.2 end of cell row with cell with a.0 end of cell cell with a.1 end of cell end of table closes square brackets equal to open square brackets table row with cell with 2 end of cell row with 0 end of table close brackets

Step 3: Determine b. I, where I is the identity matrix.

B. I equals b. open brackets table row with 1 0 row with 0 1 end of table close brackets equal to open brackets table row with b 0 row with 0 b end of table close brackets

Step 4: Add aA + bI.

open square brackets table row with cell with 2 end of cell row with 0 end of table close square brackets more open brackets table row with b 0 row with 0 b end of table close square brackets equal to open square brackets table row with cell with a plus b end of cell cell with 2 end of cell row with 0 cell with a plus b end of cell end of table close brackets

Step 5: Match the corresponding terms inA squared space equals space a A space plus space b I.

A squared space equals space a A space plus space b I open square brackets table row with 1 4 row with 0 1 end of table close square brackets equal to open square brackets table row with cell with a plus b end of cell cell with 2 end of cell row with 0 cell with a plus b end of cell end of table closes square brackets open braces attributes of table column alignment left end of attributes row with cell with a plus b equal to 1 end of cell row with cell with 2 a equal to 4 end of cell end of table closes

Step 6: Solve the system by isolating a in equation I.

a equals 1 minus b

Substituting in equation II.

2. left parenthesis 1 minus b right parenthesis equals 4 2 minus 2 b equals 4 minus 2 b equals 4 minus 2 minus 2 b equals 2 b equals numerator 2 over denominator minus 2 end of fraction equal to minus 1

Replacing the value of b

a equals 1 minus left parenthesis minus 1 right parenthesis a equals 1 plus 1 equals 2

Step 7: perform the multiplication a.b.

The. b equals 2. left parenthesis minus 1 right parenthesis equals minus 2

learn more about Matrix Multiplication.

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