Linear Algebra Examples

$[\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}]$

Step 1

Find the eigenvectors.

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Step 1.1

Find the eigenvalues.

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Step 1.1.1

Set up the formula to find the characteristic equation $p (λ)$ .

$p (λ) = determinant (A - λ I_{2})$

Step 1.1.2

The identity matrix or unit matrix of size $2$ is the $2 \times 2$ square matrix with ones on the main diagonal and zeros elsewhere.

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Step 1.1.3

Substitute the known values into $p (λ) = determinant (A - λ I_{2})$ .

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Step 1.1.3.1

Substitute $[\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] - λ I_{2})$

Step 1.1.3.2

Substitute $[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$ for $I_{2}$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 1.1.4

Simplify.

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Step 1.1.4.1

Simplify each term.

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Step 1.1.4.1.1

Multiply $- λ$ by each element of the matrix.

$p (λ) = determinant ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.1.4.1.2

Simplify each element in the matrix.

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Step 1.1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 1.1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 1.1.4.1.2.4

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 1.1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 4 - λ & 2 + 0 \\ 3 + 0 & 3 - λ \end{array}]$

Step 1.1.4.3

Simplify each element.

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Step 1.1.4.3.1

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 2 \\ 3 + 0 & 3 - λ \end{array}]$

Step 1.1.4.3.2

Add $3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 2 \\ 3 & 3 - λ \end{array}]$

Step 1.1.5

Find the determinant.

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Step 1.1.5.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (4 - λ) (3 - λ) - 3 \cdot 2$

Step 1.1.5.2

Simplify the determinant.

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Step 1.1.5.2.1

Simplify each term.

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Step 1.1.5.2.1.1

Expand $(4 - λ) (3 - λ)$ using the FOIL Method.

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Step 1.1.5.2.1.1.1

Apply the distributive property.

$p (λ) = 4 (3 - λ) - λ (3 - λ) - 3 \cdot 2$

Step 1.1.5.2.1.1.2

Apply the distributive property.

$p (λ) = 4 \cdot 3 + 4 (- λ) - λ (3 - λ) - 3 \cdot 2$

Step 1.1.5.2.1.1.3

Apply the distributive property.

$p (λ) = 4 \cdot 3 + 4 (- λ) - λ \cdot 3 - λ (- λ) - 3 \cdot 2$

Step 1.1.5.2.1.2

Simplify and combine like terms.

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Step 1.1.5.2.1.2.1

Simplify each term.

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Step 1.1.5.2.1.2.1.1

Multiply $4$ by $3$ .

$p (λ) = 12 + 4 (- λ) - λ \cdot 3 - λ (- λ) - 3 \cdot 2$

Step 1.1.5.2.1.2.1.2

Multiply $- 1$ by $4$ .

$p (λ) = 12 - 4 λ - λ \cdot 3 - λ (- λ) - 3 \cdot 2$

Step 1.1.5.2.1.2.1.3

Multiply $3$ by $- 1$ .

$p (λ) = 12 - 4 λ - 3 λ - λ (- λ) - 3 \cdot 2$

Step 1.1.5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 12 - 4 λ - 3 λ - 1 \cdot - 1 λ \cdot λ - 3 \cdot 2$

Step 1.1.5.2.1.2.1.5

Multiply $λ$ by $λ$ by adding the exponents.

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Step 1.1.5.2.1.2.1.5.1

Move $λ$ .

$p (λ) = 12 - 4 λ - 3 λ - 1 \cdot - 1 (λ \cdot λ) - 3 \cdot 2$

Step 1.1.5.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 12 - 4 λ - 3 λ - 1 \cdot - 1 λ^{2} - 3 \cdot 2$

Step 1.1.5.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 12 - 4 λ - 3 λ + 1 λ^{2} - 3 \cdot 2$

Step 1.1.5.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 12 - 4 λ - 3 λ + λ^{2} - 3 \cdot 2$

Step 1.1.5.2.1.2.2

Subtract $3 λ$ from $- 4 λ$ .

$p (λ) = 12 - 7 λ + λ^{2} - 3 \cdot 2$

Step 1.1.5.2.1.3

Multiply $- 3$ by $2$ .

$p (λ) = 12 - 7 λ + λ^{2} - 6$

Step 1.1.5.2.2

Subtract $6$ from $12$ .

$p (λ) = - 7 λ + λ^{2} + 6$

Step 1.1.5.2.3

Reorder $- 7 λ$ and $λ^{2}$ .

$p (λ) = λ^{2} - 7 λ + 6$

Step 1.1.6

Set the characteristic polynomial equal to $0$ to find the eigenvalues $λ$ .

$λ^{2} - 7 λ + 6 = 0$

Step 1.1.7

Solve for $λ$ .

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Step 1.1.7.1

Factor $λ^{2} - 7 λ + 6$ using the AC method.

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Step 1.1.7.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 7$ .

$- 6, - 1$

Step 1.1.7.1.2

Write the factored form using these integers.

$(λ - 6) (λ - 1) = 0$

Step 1.1.7.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$λ - 6 = 0$

$λ - 1 = 0$

Step 1.1.7.3

Set $λ - 6$ equal to $0$ and solve for $λ$ .

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Step 1.1.7.3.1

Set $λ - 6$ equal to $0$ .

$λ - 6 = 0$

Step 1.1.7.3.2

Add $6$ to both sides of the equation.

$λ = 6$

Step 1.1.7.4

Set $λ - 1$ equal to $0$ and solve for $λ$ .

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Step 1.1.7.4.1

Set $λ - 1$ equal to $0$ .

$λ - 1 = 0$

Step 1.1.7.4.2

Add $1$ to both sides of the equation.

$λ = 1$

Step 1.1.7.5

The final solution is all the values that make $(λ - 6) (λ - 1) = 0$ true.

$λ = 6, 1$

Step 1.2

The eigenvector is equal to the null space of the matrix minus the eigenvalue times the identity matrix where $N$ is the null space and $I$ is the identity matrix.

$ε_{A} = N (A - λ I_{2})$

Step 1.3

Find the eigenvector using the eigenvalue $λ = 6$ .

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Step 1.3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] - 6 [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 1.3.2

Simplify.

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Step 1.3.2.1

Simplify each term.

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Step 1.3.2.1.1

Multiply $- 6$ by each element of the matrix.

$[\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - 6 \cdot 1 & - 6 \cdot 0 \\ - 6 \cdot 0 & - 6 \cdot 1 \end{array}]$

Step 1.3.2.1.2

Simplify each element in the matrix.

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Step 1.3.2.1.2.1

Multiply $- 6$ by $1$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - 6 & - 6 \cdot 0 \\ - 6 \cdot 0 & - 6 \cdot 1 \end{array}]$

Step 1.3.2.1.2.2

Multiply $- 6$ by $0$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - 6 & 0 \\ - 6 \cdot 0 & - 6 \cdot 1 \end{array}]$

Step 1.3.2.1.2.3

Multiply $- 6$ by $0$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - 6 & 0 \\ 0 & - 6 \cdot 1 \end{array}]$

Step 1.3.2.1.2.4

Multiply $- 6$ by $1$ .

$[\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] + [\begin{array}{c} - 6 & 0 \\ 0 & - 6 \end{array}]$

Step 1.3.2.2

Add the corresponding elements.

$[\begin{array}{c} 4 - 6 & 2 + 0 \\ 3 + 0 & 3 - 6 \end{array}]$

Step 1.3.2.3

Simplify each element.

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Step 1.3.2.3.1

Subtract $6$ from $4$ .

$[\begin{array}{c} - 2 & 2 + 0 \\ 3 + 0 & 3 - 6 \end{array}]$

Step 1.3.2.3.2

Add $2$ and $0$ .

$[\begin{array}{c} - 2 & 2 \\ 3 + 0 & 3 - 6 \end{array}]$

Step 1.3.2.3.3

Add $3$ and $0$ .

$[\begin{array}{c} - 2 & 2 \\ 3 & 3 - 6 \end{array}]$

Step 1.3.2.3.4

Subtract $6$ from $3$ .

$[\begin{array}{c} - 2 & 2 \\ 3 & - 3 \end{array}]$

Step 1.3.3

Find the null space when $λ = 6$ .

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Step 1.3.3.1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} - 2 & 2 & 0 \\ 3 & - 3 & 0 \end{array}]$

Step 1.3.3.2

Find the reduced row echelon form.

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Step 1.3.3.2.1

Multiply each element of $R_{1}$ by $- \frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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Step 1.3.3.2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 2 & - \frac{1}{2} \cdot 0 \\ 3 & - 3 & 0 \end{array}]$

Step 1.3.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 \\ 3 & - 3 & 0 \end{array}]$

Step 1.3.3.2.2

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 1.3.3.2.2.1

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 0 \\ 3 - 3 \cdot 1 & - 3 - 3 \cdot - 1 & 0 - 3 \cdot 0 \end{array}]$

Step 1.3.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 0 \\ 0 & 0 & 0 \end{array}]$

Step 1.3.3.3

Use the result matrix to declare the final solution to the system of equations.

$x - y = 0$

$0 = 0$

Step 1.3.3.4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} y \\ y \end{array}]$

Step 1.3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} 1 \\ 1 \end{array}]$

Step 1.3.3.6

Write as a solution set.

${y [\begin{array}{c} 1 \\ 1 \end{array}] | y \in R}$

Step 1.3.3.7

The solution is the set of vectors created from the free variables of the system.

${[\begin{array}{c} 1 \\ 1 \end{array}]}$

Step 1.4

Find the eigenvector using the eigenvalue $λ = 1$ .

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Step 1.4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] - [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 1.4.2

Simplify.

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Step 1.4.2.1

Subtract the corresponding elements.

$[\begin{array}{c} 4 - 1 & 2 - 0 \\ 3 - 0 & 3 - 1 \end{array}]$

Step 1.4.2.2

Simplify each element.

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Step 1.4.2.2.1

Subtract $1$ from $4$ .

$[\begin{array}{c} 3 & 2 - 0 \\ 3 - 0 & 3 - 1 \end{array}]$

Step 1.4.2.2.2

Subtract $0$ from $2$ .

$[\begin{array}{c} 3 & 2 \\ 3 - 0 & 3 - 1 \end{array}]$

Step 1.4.2.2.3

Subtract $0$ from $3$ .

$[\begin{array}{c} 3 & 2 \\ 3 & 3 - 1 \end{array}]$

Step 1.4.2.2.4

Subtract $1$ from $3$ .

$[\begin{array}{c} 3 & 2 \\ 3 & 2 \end{array}]$

Step 1.4.3

Find the null space when $λ = 1$ .

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Step 1.4.3.1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} 3 & 2 & 0 \\ 3 & 2 & 0 \end{array}]$

Step 1.4.3.2

Find the reduced row echelon form.

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Step 1.4.3.2.1

Multiply each element of $R_{1}$ by $\frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

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Step 1.4.3.2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{3}{3} & \frac{2}{3} & \frac{0}{3} \\ 3 & 2 & 0 \end{array}]$

Step 1.4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & 0 \\ 3 & 2 & 0 \end{array}]$

Step 1.4.3.2.2

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 1.4.3.2.2.1

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & 0 \\ 3 - 3 \cdot 1 & 2 - 3 (\frac{2}{3}) & 0 - 3 \cdot 0 \end{array}]$

Step 1.4.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & 0 \\ 0 & 0 & 0 \end{array}]$

Step 1.4.3.3

Use the result matrix to declare the final solution to the system of equations.

$x + \frac{2}{3} y = 0$

$0 = 0$

Step 1.4.3.4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{2 y}{3} \\ y \end{array}]$

Step 1.4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} - \frac{2}{3} \\ 1 \end{array}]$

Step 1.4.3.6

Write as a solution set.

${y [\begin{array}{c} - \frac{2}{3} \\ 1 \end{array}] | y \in R}$

Step 1.4.3.7

The solution is the set of vectors created from the free variables of the system.

${[\begin{array}{c} - \frac{2}{3} \\ 1 \end{array}]}$

Step 1.5

The eigenspace of $A$ is the list of the vector space for each eigenvalue.

${[\begin{array}{c} 1 \\ 1 \end{array}], [\begin{array}{c} - \frac{2}{3} \\ 1 \end{array}]}$

Step 2

Define $P$ as a matrix of the eigenvectors.

$P = [\begin{array}{c} 1 & - \frac{2}{3} \\ 1 & 1 \end{array}]$

Step 3

Find the inverse of $P$ .

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Step 3.1

The inverse of a $2 \times 2$ matrix can be found using the formula $\frac{1}{a d - b c} [\begin{array}{c} d & - b \\ - c & a \end{array}]$ where $a d - b c$ is the determinant.

Step 3.2

Find the determinant.

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Step 3.2.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$1 \cdot 1 - - \frac{2}{3}$

Step 3.2.2

Simplify the determinant.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Multiply $1$ by $1$ .

$1 - - \frac{2}{3}$

Step 3.2.2.1.2

Multiply $- - \frac{2}{3}$ .

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Step 3.2.2.1.2.1

Multiply $- 1$ by $- 1$ .

$1 + 1 (\frac{2}{3})$

Step 3.2.2.1.2.2

Multiply $\frac{2}{3}$ by $1$ .

$1 + \frac{2}{3}$

Step 3.2.2.2

Write $1$ as a fraction with a common denominator.

$\frac{3}{3} + \frac{2}{3}$

Step 3.2.2.3

Combine the numerators over the common denominator.

$\frac{3 + 2}{3}$

Step 3.2.2.4

Add $3$ and $2$ .

$\frac{5}{3}$

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.

$P^{- 1} = \frac{1}{\frac{5}{3}} [\begin{array}{c} 1 & \frac{2}{3} \\ - 1 & 1 \end{array}]$

Step 3.5

Multiply the numerator by the reciprocal of the denominator.

$P^{- 1} = 1 (\frac{3}{5}) [\begin{array}{c} 1 & \frac{2}{3} \\ - 1 & 1 \end{array}]$

Step 3.6

Multiply $\frac{3}{5}$ by $1$ .

$P^{- 1} = \frac{3}{5} [\begin{array}{c} 1 & \frac{2}{3} \\ - 1 & 1 \end{array}]$

Step 3.7

Multiply $\frac{3}{5}$ by each element of the matrix.

$P^{- 1} = [\begin{array}{c} \frac{3}{5} \cdot 1 & \frac{3}{5} \cdot \frac{2}{3} \\ \frac{3}{5} \cdot - 1 & \frac{3}{5} \cdot 1 \end{array}]$

Step 3.8

Simplify each element in the matrix.

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Step 3.8.1

Multiply $\frac{3}{5}$ by $1$ .

$P^{- 1} = [\begin{array}{c} \frac{3}{5} & \frac{3}{5} \cdot \frac{2}{3} \\ \frac{3}{5} \cdot - 1 & \frac{3}{5} \cdot 1 \end{array}]$

Step 3.8.2

Cancel the common factor of $3$ .

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Step 3.8.2.1

Cancel the common factor.

$P^{- 1} = [\begin{array}{c} \frac{3}{5} & \frac{3}{5} \cdot \frac{2}{3} \\ \frac{3}{5} \cdot - 1 & \frac{3}{5} \cdot 1 \end{array}]$

Step 3.8.2.2

Rewrite the expression.

$P^{- 1} = [\begin{array}{c} \frac{3}{5} & \frac{1}{5} \cdot 2 \\ \frac{3}{5} \cdot - 1 & \frac{3}{5} \cdot 1 \end{array}]$

Step 3.8.3

Combine $\frac{1}{5}$ and $2$ .

$P^{- 1} = [\begin{array}{c} \frac{3}{5} & \frac{2}{5} \\ \frac{3}{5} \cdot - 1 & \frac{3}{5} \cdot 1 \end{array}]$

Step 3.8.4

Multiply $\frac{3}{5} \cdot - 1$ .

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Step 3.8.4.1

Combine $\frac{3}{5}$ and $- 1$ .

$P^{- 1} = [\begin{array}{c} \frac{3}{5} & \frac{2}{5} \\ \frac{3 \cdot - 1}{5} & \frac{3}{5} \cdot 1 \end{array}]$

Step 3.8.4.2

Multiply $3$ by $- 1$ .

$P^{- 1} = [\begin{array}{c} \frac{3}{5} & \frac{2}{5} \\ \frac{- 3}{5} & \frac{3}{5} \cdot 1 \end{array}]$

Step 3.8.5

Move the negative in front of the fraction.

$P^{- 1} = [\begin{array}{c} \frac{3}{5} & \frac{2}{5} \\ - \frac{3}{5} & \frac{3}{5} \cdot 1 \end{array}]$

Step 3.8.6

Multiply $\frac{3}{5}$ by $1$ .

$P^{- 1} = [\begin{array}{c} \frac{3}{5} & \frac{2}{5} \\ - \frac{3}{5} & \frac{3}{5} \end{array}]$

Step 4

Use the similarity transformation to find the diagonal matrix $D$ .

$D = P^{- 1} A P$

Step 5

Substitute the matrices.

$[\begin{array}{c} \frac{3}{5} & \frac{2}{5} \\ - \frac{3}{5} & \frac{3}{5} \end{array}] [\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}] [\begin{array}{c} 1 & - \frac{2}{3} \\ 1 & 1 \end{array}]$

Step 6

Simplify.

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Step 6.1

Multiply $[\begin{array}{c} \frac{3}{5} & \frac{2}{5} \\ - \frac{3}{5} & \frac{3}{5} \end{array}] [\begin{array}{c} 4 & 2 \\ 3 & 3 \end{array}]$ .

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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 $2 \times 2$ and the second matrix is $2 \times 2$ .

Step 6.1.2

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} \frac{3}{5} \cdot 4 + \frac{2}{5} \cdot 3 & \frac{3}{5} \cdot 2 + \frac{2}{5} \cdot 3 \\ - \frac{3}{5} \cdot 4 + \frac{3}{5} \cdot 3 & - \frac{3}{5} \cdot 2 + \frac{3}{5} \cdot 3 \end{array}] [\begin{array}{c} 1 & - \frac{2}{3} \\ 1 & 1 \end{array}]$

Step 6.1.3

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} \frac{18}{5} & \frac{12}{5} \\ - \frac{3}{5} & \frac{3}{5} \end{array}] [\begin{array}{c} 1 & - \frac{2}{3} \\ 1 & 1 \end{array}]$

Step 6.2

Multiply $[\begin{array}{c} \frac{18}{5} & \frac{12}{5} \\ - \frac{3}{5} & \frac{3}{5} \end{array}] [\begin{array}{c} 1 & - \frac{2}{3} \\ 1 & 1 \end{array}]$ .

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Step 6.2.1

Step 6.2.2

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} \frac{18}{5} \cdot 1 + \frac{12}{5} \cdot 1 & \frac{18}{5} (- \frac{2}{3}) + \frac{12}{5} \cdot 1 \\ - \frac{3}{5} \cdot 1 + \frac{3}{5} \cdot 1 & - \frac{3}{5} (- \frac{2}{3}) + \frac{3}{5} \cdot 1 \end{array}]$

Step 6.2.3

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} 6 & 0 \\ 0 & 1 \end{array}]$

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