Midterm 1&2 eventually Final

Let A = [a₁₁ a₁₂] [a₂₁ a₂₂] with det(A)≠0, then A is nonsingular and

A^-1 = 1/det(A) [a₁₁ a₁₂] [a₂₁ a₂₂]

2x₁ + x₂ = -1 x₁ + x₂ = 3 inverse is:

A^-1 = [1. -1] [-1. 2] Matrix equation: [2. 1] [x₁] = [-1] [1. 1] [x₂] = [3] Solution: [x₁]= A^-1 vector b = [1. -1] [-1] [x₂] [-1. 2] [3] [x₁] = [-4] [x₂] = [7]

A= [1 5] [-1 2] =

A^T= [1 -1] [5. 2] det (A) = 2 + 5 = 7 det ( A^T) = 2 +5 =7

Cofactor Expansion

Aᵢⱼ= (-1)^i+j*det(Mᵢⱼ)

Let A= -1. 0. -2 2. 1. 0 0. 3. 4 a) find the (1,2) and (2,2) cofactors of A.

A₁₂ = (-1)¹⁺²[2. 0] = -1(2*4-0*0) = -8 [0. 4] A₂₂ = (-1)²⁺²[-1 -2] = 1(-4-0*1-2) = -4 [0. 4] 2.1 Notes

For the following, compute (i) det (A), (ii) adj A and (iii) A^-1(d) A=[2 0 0] [ 0 3 0] [ 0 0 1]

Finally we will calculate the inverse. Step 1 - Find the determinantA^−1=1/det(A)*adj(A)= 2 [3 0] + 0 [0 1]= 2 * 3 + 0 + 0 6 Step 2 - Since the determinant is non-zero, the inverse exists. Set up a 3x6 matrix where the left half is the original matrix and the right half is its identity matrix. Step 3 - Find the reduced row echelon form. R2*1/3 1/2 0 0 0 1/3 0 0 0 1

Please find one elementary matrix of the following:A=([2,5,4],[3,4,1],[4,-1,0]),B=([4,-1,0],[3,4,1],[2,5,4])

If you R₁<->R₃, B= E×AE= [0. 0. 1] [0. 1. 0] [1. 0. 0]

1 3 0|0 0 0 1|0 0 0 0|1, inconsistent or consistent?

Inconsistent - no solutions

Determine whether the following sets form subspaces of R³ : c) (x₁, x₂, x₃)^T|x₃=x₁+x₂}

Let x={(x₁,x₂, x₃)^T|x₃=x₁+x₂} (i) 0∈S since 0=0=0 S is nonempty (ii) if x,y∈S, x₃=x₁+x₂, y₃=,y₁+y₂, x₃+y₃)^T and x₃+y₃=x₁+x₂+y₁+y₂= x₁+y₁+x (iii) if x∈S and 𝛂∈R x₃=x₁+x₂, then 𝛂x= (𝛂x₁, 𝛂x₂, 𝛂x₃)∈ S since 𝛂x₃ = 𝛂(x₁+x₂) = 𝛂x₁+𝛂x₂ So S is a subspace of R³

a. b. c d. e. f g. h. i Find the determinant formula of a 3x3 matrix?

Mnemonic device: 1. Eskimos find Ham to store in Igloos. 2. Drumsticks, Feathers, Gills were found in an Igloo. 3. Don't Ever Glue your House...! |A|= a(ei-fh)-b(di-fg)+c(dh-eg)

Let A= -1. 0. -2 2. 1. 0 0. 3. 4 a) find the (1,2) and (2,2) minor matrices of A.

M₁₂ = [2. 0] [0. 4] M₂₂ = [-1 -2] [0. 4] Notes 2.1

is [0,0] R1 [1,5] R2 [0,0] in echelon form?

No, [0,0] R1 [1,5] R2 [0,0] is not in row echelon form

Is 5x₁-4x₂+3/x₃=2 a linear equation...?

No, because of 3/x₃

Is x₄-lnx₂+4x₃=0 a linear equation...?

No, because of ln(x₂)

Is this matrix in reduced echelon form? [1 0 4 2] [0 1 -1 0] [0 0 0 1]

No, it fails not all zeros on the bottom row.

Please find one elementary matrix of the following: A=([4,-1,1],[0,2,3],[5,1,0]),B=([4,-1,1],[10,4,3],[5,1,0])

Step 1: By comparing these matrices, we can see that Matrix B is a result of adding 2 times the 3rd row of matrix A to the 2nd row of matrix A. Step 2: Next we apply the row operation (2Row 3 + Row 2) to its identity matrix. I = [1 0 0] [1 0 0] [0 1 0] --> E---> [0 1 2] [0 0 1] [0 0 1]

give two elementary matrices E₁,E₂ such that B=E₂E₁A

Step 1: Determine what multiplied by A = B Step 2: By comparing these matrices, we can see that matrix B is a result of multiplying A by 2. For the second row operation, multiply R₁ by -1. Step 3. Next, apply the row operations to the identity matrix.I= [1. 0] -->E₁=[2. 0] [0. 1] --> [0. 2]-->E₂= [-2. 0] [0. 2]

2. 4. 7 2. 1 3 -3. 1. -1 Find the values of det(M₂₁), det(M₂₂), and det(M₂₃) det(M₂₁)

Step 1: det(M₂₁) - delete col 1 & row 2 4. 7 1. -1=(4)(-1)-7(1)=-11 det(M₂₂) 2. 7-3. -1 =-1(2)-7(-3) =19 det(M₂₃) 2 4-3 12(1)-(4)(-3)=14

2. 4. 7 2. 1 3 -3. 1. -1 Find the cofactors: A₂₁, A₂₂, & A₂₃

Step 1:If det(M₂₁)=-11, det(M₂₂)=19, and det(M₂₃)=14 Step 2: to find the cofactor, Aᵢⱼ=(-1)^(i+j)MᵢⱼTherefore, A₂₁= (-1)²⁺¹M₂₁ = -1(-11)=11A₂₂=(-1)²⁺²M₂₂=1*19=19A₂₃=(-1)²⁺³M₂₃=-1(14)=-14

Augmented matrix

a coefficient matrix with an extra column containing the constant terms, [A|b]

Coefficient Matrix

a matrix that contains only the coefficients of a system of equations

Cramer's Rule

a method that uses determinants to solve a system of linear equations

the absolute value of the determinant can be interpreted as:

an area or volume

Eigenvalues

are the values on its main diagonal. they are the roots of any square matrix by which eigen vectors are further scaled.

The matrix I in [A|I]

is used to "record" the operations of an elementary matrix

A is triangular if A has zeros below the diagonal

or A has zeros above the diagonal

1 0 0|4 0 1 0|-2 0 0 1|5 0 0 0|0 0 0 0|0, what is the number of leading ones and unknowns?

r = # of leading ones = 3 n = # of unknowns = 3 x1 = 4 , x₂= -2 , x₃= 5

Gauss-Jordan Elimination

solving a system of linear equations by transforming an augmented matrix so that it is in reduced row-echelon form

The "ith" row of A is....

the jth column of A^T A= (Transpose Operation Input) (aᵢⱼ)(m×n) matrix = A^T (Transpose Operation Output)= aᵢⱼ(n×m) matrix

If A is invertible, then the solution of A vector x =

vector b; vector x = A^-1 vector b

Determine whether the following vectors are linearly independent in P₃: b) 3, x, x², x-2

x-2=x +(-⅔)3. This shows that 3,x,x²,x-2 are not linearly independent but linearly dependent.

Determine whether the following vectors are linearly independent in P₃: a) x², 1, x²-1

x²-1 = (1)x²+(-1). This shows that x²-1 is a linesr combination of x² and 1. The vectors are not linearly independent but linearly dependent.

Is this matrix in reduced echelon form? [1 -1 0 1 -1] [0 0 1 2 -2] [0 0 0 0 0]

yes, it follows the definition ie, last row of zeros, the first leading entry is one, the non zero entry to the right of the leading one

C[a,b]=

{f: f is a continuous function from [a,b] to R} Addition (f+g)(x) = f(x)+g(x) Scalar Multiplication: (rf)(x) = rf(x) Zero for Addition: z(x)=0 for all x in [a,b]

We can find determinant A by transforming A to an echelon form...

(Not necessarily reduced echelon form) and translating the used elementary row operations to changes in the determinants.

A matrix is in echelon form if:

(i) First, non 0 entry (leading entry) of each row is 1 (ii) The leading one of each row is to the right of the leading 1 of the previous row. (iii) Zero rows are all in the bottom

Determine whether the following sets form subspaces of R³ : b) (x₁, x₂, x₃)^T|x₁+x₃=1}

(i) O∈S since 0=0=0 S is nonempty (ii) if x,y ∈S, x₁=x₂=x₃, y₁=y₂=y₃ then x+y ∈S since x+y = (x₁=x₂=x₃, y₁=y₂=y₃ then x+y∈S since x+y = (x+y, x₂+y₂, x₃+y₃)^T and x₁+y₁=x₂+y₂ = x₃+y₃

The matrix is in Reduced Row Echelon Form if

(i) The Matrix is in echelon form. (ii) The first non 0 entry of the row is the only non 0 entry in its row.

A matrix is in echelon form if

(i) first non-zero entry (leading entry) of each row is 1. (ii) The leading 1 of each row is to the right of the leading 1 of the previous row. (iii) Zero rows are all in the bottom.

Scalar Multiplication Operation Output

(rA)ᵢⱼ = raᵢⱼ (m×n) matrix

All equal Elementary Matrix of A (EA)...

1. A Rᵢ<-->Rⱼ ->B 2. A cRᵢ -> B 3. A Rᵢ+cRⱼ -> B

Solving a system of linear equations

1. Find augmented matrix [A/b] 2. Find the reduced echelon form [c/d] of [A/b] 3. From the system associated [c/d] , get solutions (if any) a) A row of [c/d] is [00...0|1] - inconsistent system - no solution b) No row of [c/d] is [00...0|1] - consistent system = r is equal to # of leading ones is less than the number of unknowns = n --> INFINITELY MANY SOLUTIONS or # of leading ones = # of unknowns which gives a unique solution To solve a system of linear equations, you first create an augmented matrix and then find its reduced echelon form. From there, you can determine if the system has no solution, infinitely many solutions, or a unique solution. It's like organizing a jigsaw puzzle to find the right fit for each piece.

Difference between echelon and reduced row echelon...

A matrix in echelon form can contain any values as its non 0 entries, while the reduced echelon form can only contain ones.

Find the coefficient matrix of: x₁-x₂+4x₃=4 -2x₁+x₂-x₃=-2

A= [1. -1. 4] [-2. 1. -1]

For the following, compute (i) det (A), (ii) adj A and (iii) A^-1 (b) A=[1 2] [ 2 −4]

1. First we will calculate the determinant. det(A)=(1)×(−4)−2×2=−8 2. For the second part we will calculate the adjoint of the matrix. *Calculate the minor for a. The minor is the determinant with row 1 and column 1 deleted - 4 *Calculate the minor for a. The minor is the determinant with row1 and column 2 deleted = 2 *Calculate the minor for a. The minor is the determinant with row2 and column 1 deleted = 2 *Calculate the minor for a. The minor is the determinant with row2 and column 2 deleted = 1 Transpose the matrix -4 -2 -2 1 3. To find the inverse use the formula: 1/(ad-bc) [d -b] [-c a] A^−1=1/det(A)*adj(A) =−1/8 [−4 2] [2 1] = [-1/8*-4 -1/8*-2] = [1/2 1/4] [-1/8*-2 -1/8*1] [1/4 -1/8]

For the following, compute (i) det (A), (ii) adj A and (iii) A^-1(a) A= [−1 4] [ 2−3]

1. First we will calculate the determinant.det(A)=(−1)×(−3)−4×2=−5 2. For the second part we will calculate the adjoint of the matrix.adj(A)=[(−1)²(−3) (−1)³(2)]^T [(−1)³(4) (−1)⁴(−1)]=[−3−4] [−2−1] 3. Finally we will calculate the inverse.A^−1=1/det(A)*adj(A)=−1/5 [−3−4] [−2−1]= [3/5 4/5] [2/5 1/5]; [+ -], [- +]

For the following, compute (i) det (A), (ii) adj A and (iii) A^-1(b) Find the determinant A= [1 2] [ 2 −4]

1. First, we will calculate the determinant.det(A)=(1)×(-4)+ 2(-2)=-8

Reduced Row Echelon Form (RREF)

1. In REF 2. All leading entries are 1 3. In a column with a leading 1, all other elements are 0. not unique

x₂ + 2x₃ + x₄ + 3x₅ = 1 x₁+ x₂ -x₅ = 1 x₁ -x₃ + x₄ + x₅ = 0 What are the correct row operations for this augmented matrix?

1. R₁ <-> R₂ 2. R₃ = R₃ - R₁ 3. R₃ = R₃ + R₂ 4. R₂ = R₂ - 2R₃ 5. R₁= R₁-R₂

1.1 - 3d - x₁+x₂=1 x₁-x₂=1 -x₁+3x₂=3, write an augmented matrix using Guassian elimination, verify the correct steps for row operations...?

1. R₂=R₂-R₁, to make the entry at 2,1 a 0. 2. R₃=R₃+R₁, to make the entry at 3,1 a 0. 3. Multiply each element of R₂*-½ to make an entry at 2,2 a 1. 4. R₃=R₃-4R₂ to make the entry at 3,2 a 0. 5. R₃*¼ 6. R₁=R₁- R₃ to make the entry at 1,3 a 0. 7. R₁=R₁-R₂ to make the entry at 1,2 a 0.

Steps for Guass- Jordan Elimination...

1. Swap rows so that all rows with zero entries are on the bottom of the matrix. 2. Swap rows so that the row with the largest left-most digit is on the top of the matrix. 3. Multiply the top row by a scalar that converts the top row's leading entry into 1 (If the leading entry of the top row is a, then multiply it by 1/a to get 1). 4. Add multiples of the top row to the rows below it to create zeros below the leading entry. 5. Repeat steps 2-4 for the remaining rows until the matrix is in reduced row echelon form.

[2. 3. -6] [1. 2. -1] [-2 -3. -6], determine the null space... ​

1. Write as an augmented matrix. [2. 3. -6| 0] [1. 2. -1| 0] [-2 -3. -6|0] 2. Find the reduced row echelon form (attached) [1 0 15| 0] [0 1 -8| 0] [0 0 0| 0] 3. Use the result matrix for system equations x+15z=0 y-8z=0 0=0 4. Write a solution vector [x] [-15z] [y] = [8z] [z] [z] 5. Write a linear combination of vectors. [x] [-15] [y] = z [8] [z] [1] 6. Write a solution set {z[-15]| } { [8] | z∈R} { [1] | } Span (-15, 8, 1)^T

Inverse of a matrix formula

1/(ad-bc) [d -b] [-c a]

Matrix Addition

2. Output (A+B)ᵢⱼ = aᵢⱼ+bᵢⱼ (m×n) matrix 1. Input A=(aᵢⱼ), B=(bᵢⱼ) both are (m×n) matrices

Is 2x₁-3x₂+½x₃=0 a linear equation...?

2x₁-3x₂+½x₃=0 is linear equation.

Pₙ

= { p: p= aₒ+a₁x+.. +ax^n-1 is a polynomial with degree lesser than n} Zero for Addition: 0= 0+0x+0x²+...+0x^n-1

R^n

= { vector x: vector x = [x₁] and x₁, x₂, ..., xₙ [x₂] ..... [xₙ] are all real numbers} Zero for Addition: 0 = [0] [0] [0]

A^-1 is called the inverse of

A where the number n is 1/n as the matrix A is to A^-1

For the following, compute (ii) adj A (d) A= [2 0 0] [ 0 3 0] [ 0 0 1]

For the second part we will calculate the adjoint of the matrix. adj(A)= *Calculate the minor for element a - delete row 1 and column 1 - a = 3 * Calculate the minor for element a₁₂ - delete row 1 and column 2 = 0 * Calculate the minor for element a₁₃ - delete row one and column 3 = 0 * Calculate the minor for element a₂₁ - delete row two and column 1 = 0 * Calculate the minor for element a₂₂ - delete row two and column 2 = 2 * Calculate the minor for element a₂₃ - delete row two and column 3 = 0 * Calculate the minor for element a₃₁ - delete row three and column 1 = 0 * Calculate the minor for element a₃₂ - delete row three and column 2 = 0 * Calculate the minor for element a₃₃ - delete row three and column 3 = 6

Gaussian elimination

Row-reduce until you get an x = number and substitute until you get an actual number. - an algorithm used for row operations that can operate similar to elementary linear operations.

Reduced Echelon Form

Same as echelon form, except all leading entries are 1; each leading 1 is the only non-zero entry in its row; there is only one unique reduced echelon form for every matrix, unique

Determine the (2,2) entry of A^-1 by computing a quotient of the two determinants 14 1 51 3 1 2 1

Step 1 - Find the determinant of Matrix A, denoted a det (A).det(A) = 2(3*2-4*1) - 1(0*2 - 4*1)+2(0*1-3*1)=2*(6-4)-1(0-4)+2(0-3)=4+4-6=2 Step 2 - Find the mayrix A₂ٖٖٖ,₂ by removing the 2nd row and 2nd column from A.[2. 2][1. 2] = 2(2)-1(2)=4-2=2 Step 3 - Find the (2,2) entry of A^-1:(A^-1)₂,₂ = det(A₂,₂/det(A) =2/2 =1

[1. 4. 3.] × [3. 2] [0. 1. 4] × [1. 1] [0. 0. 2] × [4. 5], correct steps for multiplying both matrices

Step 1 - Multiply each row in the 1st matrix by each column in the 2nd matrix [1•3+4•1+3•4. 1•2+4•1+3•5] [0•3+1•1+4•4. 0•2+1•1+4•5] [0•3+0•1+2•1. 0•2+0•1+2•5] Step 2- Simplify by multiplying out all expressions [19. 21] [17. 21] [8. 10]

Use the Cramer's Rule to solve each of the following systems: a) 2x₁ + 3x₂ = 2 8x₁ - 6x₂ = 2

Step 1 - Represent the system of equations in a matrix format.[2 3] [ x] = [2][8 -6] [y] = [2] Step 2 - Determine the determinant2* -6 - 8*3 = -36 Step 3 - Since the determinant is not 0, the system can be solved using Cramer's Rule, x = Dx/DReplace column 1 of the coefficient matrix that corresponds to the x-coefficient of the system with [2] [2]Substitute -36 for D and -18 for Dx in the formula, x= -18/-36, x = 1/2 Step 4 - Next, find the value of y. Replace column 1 of the coefficient matrix that corresponds to the x-coefficient of the system with [2]--> [2 2] = 2*2 - 8*12 = -36 [2] [8 2]Substitute -36 for D and -12 for Dx in the formula, x= -12/-36, x = 1/3

Use the Cramer's Rule to solve each of the following systems: c)6x₁ + 3x₂ = 5 5x₁ +7x₂ = 12

Step 1 - Represent the system of equations in a matrix format.[6 3] [ x] = [5][5 7] [y] = [12] Step 2 - Determine the determinant2* -7 - 5*3 = 27 Step 3 - Since the determinant is not 0, the system can be solved using Cramer's Rule, x = Dx/DReplace column 1 of the coefficient matrix that corresponds to the x-coefficient of the system with [5 3] [12 7]Substitute 27 for D and 57 for Dx in the formula, x= -1/27 Step 4 - Next, find the value of y. Replace column 1 of the coefficient matrix that corresponds to the x-coefficient of the system with [2]--> [6 5] = 6*12 - 5*5= 47 [2] [5 12]Substitute 47 for D and 27 for Dx in the formula, y= 47/27, x = -1/27

For Matrix: [1,2,4] R1 [-2,-4,4] R2, What are the correct row operations?

Step 1 - perform the row operation R(2) = R(2)+2R(1), to make the entry at 2,1 a 0. Step 2 - Multiply each element of R2 by 1/12 to make the entry at 2,3 a 1. Step 3 - Perform the row operation R1 = R1-4R2 to make the entry at 1,2 a 0.

For Matrix: [1,1,4] R1 [1,-1,2] R2, What is are the row operations?

Step 1 - perform the row operation R(2) = R(2)- R(1), to make the entry at 2,1 a 0. Step 2 - Divide Row 2 by -2; Row 2 = Row 2 by -½ to make the entry at 2,2 a 1. Step 3 - Perform the row operation: Subtract Row 1 multiplied by 1 from Row 2.

Determine whether the following vectors are linearly independent in R³? [1] [0] [2] [0] [0] [0] [1] [1] [-1]

Step 1- Determine if Ax=0 has any nontrivial solutions Step 2- Write an augmented matrix where Ax=0 [1] [0] [2]|[0] [0] [0] [0]|[0] [1] [1] [-1]|[0] Step 3 - Find the reduced row echelon (attached) [1] [0] [2]|[0] [0] [0] [0]|[0] [1] [1] [-1]|[0] Step 4 - Remove rows that are all zeros. [1. 0. 2| 0] [0. 1.-3| 0] Step 5 - Write the matrix as a system of linear equations. x+2z = 0 y-3z=0 Step 6 - linearly dependent

Which of the sets are spanning sets for R³ for {(1,0,1)^T, (0,0,1)^T, (2,0,-1)^T}

Step 1: Consider β= {(1,0,1)^T, (0,0,1)^T, (2,0,-1)}, Then, |β|=3 = dim(R³) A= [1 0. 1] [1. 0. 1] [0. 1. 0] => detA. [0. 1. 0] = 1 ≠0 [0. 1. 0] [0. 1. 1] Then A is invertible and beta is linearly independent. The lineraly dependent set of vectors is equal to the dimension of the space is a basis. Therefore this is a spending set for R³.

Which of the sets that follow are spanning sets for R³ {(1,2,4)^T, (2,1,4)^T, (4,4,0)^T}

Step 1: Consider β= {(1,2,4)^T, (2,1,4)^T, (0,1,0)}, Then, |β|=3 = dim(R³) A= [1 2. 0] [2. 1. 4] => detA. = 4 ≠0 [4. 4. 0] Then A is invertible and beta is linearly independent. The linearly dependent set of vectors is equal to the dimension of the space is a basis. Therefore this is a spending set for R³.

There are three types of elementary row operations of Gaussian Elimination:

Swapping two rows, Multiplying a row by a nonzero number, Adding a multiple of one row to another row.

Transpose of a Matrix

Switching the rows to columns - imagine it kinda swinging up/down a 90 degree angle

Find the augmented matrix of: x₁-x₂+4x₃=4 -2x₁+x₂-x₃=-2

[A|b]= [1. -1. 4| 4] [-2. 1. -1| -2]

A= [2. 4. -2] [-1. 4. 7] [1. -3. 4], find the elementary matrix...?

Use row operation of: R₂<->B₃ = [2. 4. -2] [1. -3 4] [-1. 4 7] det(B) = - det(A)

lead variables

Variables corresponding to the first nonzero elements in each row of the reduced matrix. are associated with leading ones

is [0,1] R1 [0,0] R2 in echelon form?

Yes, [0,1] R1 [0,0] R2 is in row echelon form. There is a leading one and zeros are on the bottom.

is [1,-1,2] R1 [0,1,3] R2 in echelon form?

Yes, [1,-1,2] R1 [0,1,3] R2 is in row echelon form. The leading entry is a one, to the right of the leading entry is a one, and zeros are on the bottom row

2. 1. 2. 2 -3 3 1 3 2 1. -1. -4 1. -3. 2. 0 Determine the minors (a₄₁&a₄₂&a₄₃&a₄₄):

The minors are: a₄₁: row 1 and column 1 are deleted; a₄₂: row 4 and column 2 deleted; a₄₃= with row 4 and column 4 deleted; and a₄₄ row 4 and column 4 deleted

If A is a triangular matrix, then det A is

The product of the entries on the main diagonal

Linear Independence

Two or more vectors are said to be linearly dependent if none of them can be written as a linear combination of the others.

1. 0. 4 0. 2. 0 0. 0. -1, is triangular...?

Yes, this matrix is triangular. There are zeros below the diagonal. Det(A) is the product of diagonal entries. (1)-2-(-1) = -2

2x₁-x₂=3 3x₁+4x₂=-3, what is the matrix equation for the system...?

[2. -1] [x₁] = [3] [3. 4] [x₂] =[-3]

The (i , j)-cofactor of a matrix A is obtained

by deleting from A its ith row and jth column, Aᵢⱼ= (-1)^irjdet(Mᵢⱼ)

The (i,j) minor of a matrix A is obtained by

deleting row I and column j from A,

determinant of A is denoted as:

det(A) or |A|

Let A,B be an (n×n) matrix, det(AB) =

det(A)*det(B)

Matrix A is non-invertible iff

det(A)= 0, nonsingular matrix, inverse matrix does not exist

A matrix is invertible if

det(A)≠0, if there exists another Matrix B of the same dimension such that AB=BA=I where I is the Identity Matrix

Relate determinant A and determinant B...

det(B) = 1/10 × ⅙ ×½ det(A) B is triangular, so det(B) = [1. 2. -1] [0. 1. 1] == (1)(1)(1) = 1 [0. 0. 1] So det(A) = 1×2×6×10 = 120

det(A^T)=

detA

Free variables are

has infinitely many solutions

Scalar Multiplication Operation Inputs

i) A=(aᵢⱼ) (m×n) matrix ii) r scalar

A vector space V is a set endowed with:

i) addition operation ii) scalar Multiplication