Linear Algebra Examples
[4233][4233]
Step 1
Step 1.1
Find the eigenvalues.
Step 1.1.1
Set up the formula to find the characteristic equation p(λ)p(λ).
p(λ)=determinant(A-λI2)p(λ)=determinant(A−λI2)
Step 1.1.2
The identity matrix or unit matrix of size 22 is the 2×22×2 square matrix with ones on the main diagonal and zeros elsewhere.
[1001][1001]
Step 1.1.3
Substitute the known values into p(λ)=determinant(A-λI2)p(λ)=determinant(A−λI2).
Step 1.1.3.1
Substitute [4233][4233] for AA.
p(λ)=determinant([4233]-λI2)p(λ)=determinant([4233]−λI2)
Step 1.1.3.2
Substitute [1001][1001] for I2I2.
p(λ)=determinant([4233]-λ[1001])p(λ)=determinant([4233]−λ[1001])
p(λ)=determinant([4233]-λ[1001])p(λ)=determinant([4233]−λ[1001])
Step 1.1.4
Simplify.
Step 1.1.4.1
Simplify each term.
Step 1.1.4.1.1
Multiply -λ−λ by each element of the matrix.
p(λ)=determinant([4233]+[-λ⋅1-λ⋅0-λ⋅0-λ⋅1])p(λ)=determinant([4233]+[−λ⋅1−λ⋅0−λ⋅0−λ⋅1])
Step 1.1.4.1.2
Simplify each element in the matrix.
Step 1.1.4.1.2.1
Multiply -1−1 by 11.
p(λ)=determinant([4233]+[-λ-λ⋅0-λ⋅0-λ⋅1])p(λ)=determinant([4233]+[−λ−λ⋅0−λ⋅0−λ⋅1])
Step 1.1.4.1.2.2
Multiply -λ⋅0−λ⋅0.
Step 1.1.4.1.2.2.1
Multiply 00 by -1−1.
p(λ)=determinant([4233]+[-λ0λ-λ⋅0-λ⋅1])p(λ)=determinant([4233]+[−λ0λ−λ⋅0−λ⋅1])
Step 1.1.4.1.2.2.2
Multiply 00 by λλ.
p(λ)=determinant([4233]+[-λ0-λ⋅0-λ⋅1])p(λ)=determinant([4233]+[−λ0−λ⋅0−λ⋅1])
p(λ)=determinant([4233]+[-λ0-λ⋅0-λ⋅1])p(λ)=determinant([4233]+[−λ0−λ⋅0−λ⋅1])
Step 1.1.4.1.2.3
Multiply -λ⋅0−λ⋅0.
Step 1.1.4.1.2.3.1
Multiply 00 by -1−1.
p(λ)=determinant([4233]+[-λ00λ-λ⋅1])p(λ)=determinant([4233]+[−λ00λ−λ⋅1])
Step 1.1.4.1.2.3.2
Multiply 00 by λλ.
p(λ)=determinant([4233]+[-λ00-λ⋅1])p(λ)=determinant([4233]+[−λ00−λ⋅1])
p(λ)=determinant([4233]+[-λ00-λ⋅1])p(λ)=determinant([4233]+[−λ00−λ⋅1])
Step 1.1.4.1.2.4
Multiply -1−1 by 11.
p(λ)=determinant([4233]+[-λ00-λ])p(λ)=determinant([4233]+[−λ00−λ])
p(λ)=determinant([4233]+[-λ00-λ])p(λ)=determinant([4233]+[−λ00−λ])
p(λ)=determinant([4233]+[-λ00-λ])p(λ)=determinant([4233]+[−λ00−λ])
Step 1.1.4.2
Add the corresponding elements.
p(λ)=determinant[4-λ2+03+03-λ]p(λ)=determinant[4−λ2+03+03−λ]
Step 1.1.4.3
Simplify each element.
Step 1.1.4.3.1
Add 22 and 00.
p(λ)=determinant[4-λ23+03-λ]p(λ)=determinant[4−λ23+03−λ]
Step 1.1.4.3.2
Add 33 and 00.
p(λ)=determinant[4-λ233-λ]p(λ)=determinant[4−λ233−λ]
p(λ)=determinant[4-λ233-λ]p(λ)=determinant[4−λ233−λ]
p(λ)=determinant[4-λ233-λ]p(λ)=determinant[4−λ233−λ]
Step 1.1.5
Find the determinant.
Step 1.1.5.1
The determinant of a 2×22×2 matrix can be found using the formula |abcd|=ad-cb∣∣∣abcd∣∣∣=ad−cb.
p(λ)=(4-λ)(3-λ)-3⋅2p(λ)=(4−λ)(3−λ)−3⋅2
Step 1.1.5.2
Simplify the determinant.
Step 1.1.5.2.1
Simplify each term.
Step 1.1.5.2.1.1
Expand (4-λ)(3-λ)(4−λ)(3−λ) using the FOIL Method.
Step 1.1.5.2.1.1.1
Apply the distributive property.
p(λ)=4(3-λ)-λ(3-λ)-3⋅2p(λ)=4(3−λ)−λ(3−λ)−3⋅2
Step 1.1.5.2.1.1.2
Apply the distributive property.
p(λ)=4⋅3+4(-λ)-λ(3-λ)-3⋅2p(λ)=4⋅3+4(−λ)−λ(3−λ)−3⋅2
Step 1.1.5.2.1.1.3
Apply the distributive property.
p(λ)=4⋅3+4(-λ)-λ⋅3-λ(-λ)-3⋅2p(λ)=4⋅3+4(−λ)−λ⋅3−λ(−λ)−3⋅2
p(λ)=4⋅3+4(-λ)-λ⋅3-λ(-λ)-3⋅2p(λ)=4⋅3+4(−λ)−λ⋅3−λ(−λ)−3⋅2
Step 1.1.5.2.1.2
Simplify and combine like terms.
Step 1.1.5.2.1.2.1
Simplify each term.
Step 1.1.5.2.1.2.1.1
Multiply 44 by 33.
p(λ)=12+4(-λ)-λ⋅3-λ(-λ)-3⋅2p(λ)=12+4(−λ)−λ⋅3−λ(−λ)−3⋅2
Step 1.1.5.2.1.2.1.2
Multiply -1−1 by 44.
p(λ)=12-4λ-λ⋅3-λ(-λ)-3⋅2p(λ)=12−4λ−λ⋅3−λ(−λ)−3⋅2
Step 1.1.5.2.1.2.1.3
Multiply 33 by -1−1.
p(λ)=12-4λ-3λ-λ(-λ)-3⋅2p(λ)=12−4λ−3λ−λ(−λ)−3⋅2
Step 1.1.5.2.1.2.1.4
Rewrite using the commutative property of multiplication.
p(λ)=12-4λ-3λ-1⋅-1λ⋅λ-3⋅2p(λ)=12−4λ−3λ−1⋅−1λ⋅λ−3⋅2
Step 1.1.5.2.1.2.1.5
Multiply λλ by λλ by adding the exponents.
Step 1.1.5.2.1.2.1.5.1
Move λλ.
p(λ)=12-4λ-3λ-1⋅-1(λ⋅λ)-3⋅2p(λ)=12−4λ−3λ−1⋅−1(λ⋅λ)−3⋅2
Step 1.1.5.2.1.2.1.5.2
Multiply λλ by λλ.
p(λ)=12-4λ-3λ-1⋅-1λ2-3⋅2p(λ)=12−4λ−3λ−1⋅−1λ2−3⋅2
p(λ)=12-4λ-3λ-1⋅-1λ2-3⋅2p(λ)=12−4λ−3λ−1⋅−1λ2−3⋅2
Step 1.1.5.2.1.2.1.6
Multiply -1−1 by -1−1.
p(λ)=12-4λ-3λ+1λ2-3⋅2p(λ)=12−4λ−3λ+1λ2−3⋅2
Step 1.1.5.2.1.2.1.7
Multiply λ2λ2 by 11.
p(λ)=12-4λ-3λ+λ2-3⋅2p(λ)=12−4λ−3λ+λ2−3⋅2
p(λ)=12-4λ-3λ+λ2-3⋅2p(λ)=12−4λ−3λ+λ2−3⋅2
Step 1.1.5.2.1.2.2
Subtract 3λ3λ from -4λ−4λ.
p(λ)=12-7λ+λ2-3⋅2p(λ)=12−7λ+λ2−3⋅2
p(λ)=12-7λ+λ2-3⋅2p(λ)=12−7λ+λ2−3⋅2
Step 1.1.5.2.1.3
Multiply -3−3 by 22.
p(λ)=12-7λ+λ2-6p(λ)=12−7λ+λ2−6
p(λ)=12-7λ+λ2-6p(λ)=12−7λ+λ2−6
Step 1.1.5.2.2
Subtract 66 from 1212.
p(λ)=-7λ+λ2+6p(λ)=−7λ+λ2+6
Step 1.1.5.2.3
Reorder -7λ−7λ and λ2λ2.
p(λ)=λ2-7λ+6p(λ)=λ2−7λ+6
p(λ)=λ2-7λ+6p(λ)=λ2−7λ+6
p(λ)=λ2-7λ+6p(λ)=λ2−7λ+6
Step 1.1.6
Set the characteristic polynomial equal to 00 to find the eigenvalues λλ.
λ2-7λ+6=0λ2−7λ+6=0
Step 1.1.7
Solve for λλ.
Step 1.1.7.1
Factor λ2-7λ+6λ2−7λ+6 using the AC method.
Step 1.1.7.1.1
Consider the form x2+bx+cx2+bx+c. Find a pair of integers whose product is cc and whose sum is bb. In this case, whose product is 66 and whose sum is -7−7.
-6,-1−6,−1
Step 1.1.7.1.2
Write the factored form using these integers.
(λ-6)(λ-1)=0(λ−6)(λ−1)=0
(λ-6)(λ-1)=0(λ−6)(λ−1)=0
Step 1.1.7.2
If any individual factor on the left side of the equation is equal to 00, the entire expression will be equal to 00.
λ-6=0λ−6=0
λ-1=0λ−1=0
Step 1.1.7.3
Set λ-6λ−6 equal to 00 and solve for λλ.
Step 1.1.7.3.1
Set λ-6λ−6 equal to 00.
λ-6=0λ−6=0
Step 1.1.7.3.2
Add 66 to both sides of the equation.
λ=6λ=6
λ=6λ=6
Step 1.1.7.4
Set λ-1λ−1 equal to 00 and solve for λλ.
Step 1.1.7.4.1
Set λ-1λ−1 equal to 00.
λ-1=0λ−1=0
Step 1.1.7.4.2
Add 11 to both sides of the equation.
λ=1λ=1
λ=1λ=1
Step 1.1.7.5
The final solution is all the values that make (λ-6)(λ-1)=0(λ−6)(λ−1)=0 true.
λ=6,1λ=6,1
λ=6,1λ=6,1
λ=6,1λ=6,1
Step 1.2
The eigenvector is equal to the null space of the matrix minus the eigenvalue times the identity matrix where NN is the null space and II is the identity matrix.
εA=N(A-λI2)εA=N(A−λI2)
Step 1.3
Find the eigenvector using the eigenvalue λ=6λ=6.
Step 1.3.1
Substitute the known values into the formula.
N([4233]-6[1001])N([4233]−6[1001])
Step 1.3.2
Simplify.
Step 1.3.2.1
Simplify each term.
Step 1.3.2.1.1
Multiply -6−6 by each element of the matrix.
[4233]+[-6⋅1-6⋅0-6⋅0-6⋅1][4233]+[−6⋅1−6⋅0−6⋅0−6⋅1]
Step 1.3.2.1.2
Simplify each element in the matrix.
Step 1.3.2.1.2.1
Multiply -6−6 by 11.
[4233]+[-6-6⋅0-6⋅0-6⋅1][4233]+[−6−6⋅0−6⋅0−6⋅1]
Step 1.3.2.1.2.2
Multiply -6−6 by 00.
[4233]+[-60-6⋅0-6⋅1][4233]+[−60−6⋅0−6⋅1]
Step 1.3.2.1.2.3
Multiply -6−6 by 00.
[4233]+[-600-6⋅1][4233]+[−600−6⋅1]
Step 1.3.2.1.2.4
Multiply -6−6 by 11.
[4233]+[-600-6][4233]+[−600−6]
[4233]+[-600-6][4233]+[−600−6]
[4233]+[-600-6][4233]+[−600−6]
Step 1.3.2.2
Add the corresponding elements.
[4-62+03+03-6][4−62+03+03−6]
Step 1.3.2.3
Simplify each element.
Step 1.3.2.3.1
Subtract 66 from 44.
[-22+03+03-6][−22+03+03−6]
Step 1.3.2.3.2
Add 22 and 00.
[-223+03-6][−223+03−6]
Step 1.3.2.3.3
Add 33 and 00.
[-2233-6][−2233−6]
Step 1.3.2.3.4
Subtract 66 from 33.
[-223-3][−223−3]
[-223-3][−223−3]
[-223-3][−223−3]
Step 1.3.3
Find the null space when λ=6λ=6.
Step 1.3.3.1
Write as an augmented matrix for Ax=0Ax=0.
[-2203-30][−2203−30]
Step 1.3.3.2
Find the reduced row echelon form.
Step 1.3.3.2.1
Multiply each element of R1R1 by -12−12 to make the entry at 1,11,1 a 11.
Step 1.3.3.2.1.1
Multiply each element of R1R1 by -12−12 to make the entry at 1,11,1 a 11.
[-12⋅-2-12⋅2-12⋅03-30][−12⋅−2−12⋅2−12⋅03−30]
Step 1.3.3.2.1.2
Simplify R1R1.
[1-103-30][1−103−30]
[1-103-30][1−103−30]
Step 1.3.3.2.2
Perform the row operation R2=R2-3R1R2=R2−3R1 to make the entry at 2,12,1 a 00.
Step 1.3.3.2.2.1
Perform the row operation R2=R2-3R1R2=R2−3R1 to make the entry at 2,12,1 a 00.
[1-103-3⋅1-3-3⋅-10-3⋅0][1−103−3⋅1−3−3⋅−10−3⋅0]
Step 1.3.3.2.2.2
Simplify R2R2.
[1-10000][1−10000]
[1-10000][1−10000]
[1-10000][1−10000]
Step 1.3.3.3
Use the result matrix to declare the final solution to the system of equations.
x-y=0x−y=0
0=00=0
Step 1.3.3.4
Write a solution vector by solving in terms of the free variables in each row.
[xy]=[yy][xy]=[yy]
Step 1.3.3.5
Write the solution as a linear combination of vectors.
[xy]=y[11][xy]=y[11]
Step 1.3.3.6
Write as a solution set.
{y[11]|y∈R}{y[11]∣∣∣y∈R}
Step 1.3.3.7
The solution is the set of vectors created from the free variables of the system.
{[11]}{[11]}
{[11]}{[11]}
{[11]}{[11]}
Step 1.4
Find the eigenvector using the eigenvalue λ=1λ=1.
Step 1.4.1
Substitute the known values into the formula.
N([4233]-[1001])N([4233]−[1001])
Step 1.4.2
Simplify.
Step 1.4.2.1
Subtract the corresponding elements.
[4-12-03-03-1][4−12−03−03−1]
Step 1.4.2.2
Simplify each element.
Step 1.4.2.2.1
Subtract 11 from 44.
[32-03-03-1][32−03−03−1]
Step 1.4.2.2.2
Subtract 00 from 22.
[323-03-1][323−03−1]
Step 1.4.2.2.3
Subtract 00 from 33.
[3233-1][3233−1]
Step 1.4.2.2.4
Subtract 11 from 33.
[3232][3232]
[3232][3232]
[3232][3232]
Step 1.4.3
Find the null space when λ=1λ=1.
Step 1.4.3.1
Write as an augmented matrix for Ax=0Ax=0.
[320320][320320]
Step 1.4.3.2
Find the reduced row echelon form.
Step 1.4.3.2.1
Multiply each element of R1R1 by 1313 to make the entry at 1,11,1 a 11.
Step 1.4.3.2.1.1
Multiply each element of R1R1 by 1313 to make the entry at 1,11,1 a 11.
[332303320][332303320]
Step 1.4.3.2.1.2
Simplify R1R1.
[1230320][1230320]
[1230320][1230320]
Step 1.4.3.2.2
Perform the row operation R2=R2-3R1R2=R2−3R1 to make the entry at 2,12,1 a 00.
Step 1.4.3.2.2.1
Perform the row operation R2=R2-3R1R2=R2−3R1 to make the entry at 2,12,1 a 00.
[12303-3⋅12-3(23)0-3⋅0]⎡⎢⎣12303−3⋅12−3(23)0−3⋅0⎤⎥⎦
Step 1.4.3.2.2.2
Simplify R2R2.
[1230000][1230000]
[1230000][1230000]
[1230000][1230000]
Step 1.4.3.3
Use the result matrix to declare the final solution to the system of equations.
x+23y=0x+23y=0
0=00=0
Step 1.4.3.4
Write a solution vector by solving in terms of the free variables in each row.
[xy]=[-2y3y][xy]=[−2y3y]
Step 1.4.3.5
Write the solution as a linear combination of vectors.
[xy]=y[-231][xy]=y[−231]
Step 1.4.3.6
Write as a solution set.
{y[-231]|y∈R}{y[−231]∣∣
∣∣y∈R}
Step 1.4.3.7
The solution is the set of vectors created from the free variables of the system.
{[-231]}{[−231]}
{[-231]}{[−231]}
{[-231]}{[−231]}
Step 1.5
The eigenspace of AA is the list of the vector space for each eigenvalue.
{[11],[-231]}{[11],[−231]}
{[11],[-231]}{[11],[−231]}
Step 2
Define PP as a matrix of the eigenvectors.
P=[1-2311]P=[1−2311]
Step 3
Step 3.1
The inverse of a 2×22×2 matrix can be found using the formula 1ad-bc[d-b-ca]1ad−bc[d−b−ca] where ad-bcad−bc is the determinant.
Step 3.2
Find the determinant.
Step 3.2.1
The determinant of a 2×22×2 matrix can be found using the formula |abcd|=ad-cb∣∣∣abcd∣∣∣=ad−cb.
1⋅1--231⋅1−−23
Step 3.2.2
Simplify the determinant.
Step 3.2.2.1
Simplify each term.
Step 3.2.2.1.1
Multiply 11 by 11.
1--231−−23
Step 3.2.2.1.2
Multiply --23−−23.
Step 3.2.2.1.2.1
Multiply -1−1 by -1−1.
1+1(23)1+1(23)
Step 3.2.2.1.2.2
Multiply 2323 by 11.
1+231+23
1+231+23
1+231+23
Step 3.2.2.2
Write 11 as a fraction with a common denominator.
33+2333+23
Step 3.2.2.3
Combine the numerators over the common denominator.
3+233+23
Step 3.2.2.4
Add 33 and 22.
5353
5353
5353
Step 3.3
Since the determinant is non-zero, the inverse exists.
Step 3.4
Substitute the known values into the formula for the inverse.
P-1=153[123-11]P−1=153[123−11]
Step 3.5
Multiply the numerator by the reciprocal of the denominator.
P-1=1(35)[123-11]P−1=1(35)[123−11]
Step 3.6
Multiply 35 by 1.
P-1=35[123-11]
Step 3.7
Multiply 35 by each element of the matrix.
P-1=[35⋅135⋅2335⋅-135⋅1]
Step 3.8
Simplify each element in the matrix.
Step 3.8.1
Multiply 35 by 1.
P-1=[3535⋅2335⋅-135⋅1]
Step 3.8.2
Cancel the common factor of 3.
Step 3.8.2.1
Cancel the common factor.
P-1=[3535⋅2335⋅-135⋅1]
Step 3.8.2.2
Rewrite the expression.
P-1=[3515⋅235⋅-135⋅1]
P-1=[3515⋅235⋅-135⋅1]
Step 3.8.3
Combine 15 and 2.
P-1=[352535⋅-135⋅1]
Step 3.8.4
Multiply 35⋅-1.
Step 3.8.4.1
Combine 35 and -1.
P-1=[35253⋅-1535⋅1]
Step 3.8.4.2
Multiply 3 by -1.
P-1=[3525-3535⋅1]
P-1=[3525-3535⋅1]
Step 3.8.5
Move the negative in front of the fraction.
P-1=[3525-3535⋅1]
Step 3.8.6
Multiply 35 by 1.
P-1=[3525-3535]
P-1=[3525-3535]
P-1=[3525-3535]
Step 4
Use the similarity transformation to find the diagonal matrix D.
D=P-1AP
Step 5
Substitute the matrices.
[3525-3535][4233][1-2311]
Step 6
Step 6.1
Multiply [3525-3535][4233].
Step 6.1.1
Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is 2×2 and the second matrix is 2×2.
Step 6.1.2
Multiply each row in the first matrix by each column in the second matrix.
[35⋅4+25⋅335⋅2+25⋅3-35⋅4+35⋅3-35⋅2+35⋅3][1-2311]
Step 6.1.3
Simplify each element of the matrix by multiplying out all the expressions.
[185125-3535][1-2311]
[185125-3535][1-2311]
Step 6.2
Multiply [185125-3535][1-2311].
Step 6.2.1
Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is 2×2 and the second matrix is 2×2.
Step 6.2.2
Multiply each row in the first matrix by each column in the second matrix.
[185⋅1+125⋅1185(-23)+125⋅1-35⋅1+35⋅1-35(-23)+35⋅1]
Step 6.2.3
Simplify each element of the matrix by multiplying out all the expressions.
[6001]
[6001]
[6001]