Cramer's Rule and Matrix Inverse PDF
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This document provides an explanation of Cramer's rule, properties of determinants, and how to construct the inverse matrix based on a given matrix. It also covers eigenvalues, eigenvectors, and eigenspaces in linear algebra. The document appears to be lecture notes or a study document.
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## 3.3 Cramer's Rule and Formula for the Inverse Suppose we want to solve the system Ax = b, where A is an (n x n) matrix, with columns a<sub>1</sub>,... a<sub>n</sub>, A = (a<sub>1</sub>|a<sub>2</sub>|...|a<sub>n</sub>). Make new matrices A<sub>i</sub> by replacing column # i with b: A<sub>1</s...
## 3.3 Cramer's Rule and Formula for the Inverse Suppose we want to solve the system Ax = b, where A is an (n x n) matrix, with columns a<sub>1</sub>,... a<sub>n</sub>, A = (a<sub>1</sub>|a<sub>2</sub>|...|a<sub>n</sub>). Make new matrices A<sub>i</sub> by replacing column # i with b: A<sub>1</sub> = (b|a<sub>2</sub>|...|a<sub>n</sub>), A<sub>2</sub> = (a<sub>1</sub>|b|a<sub>3</sub>|...|a<sub>n</sub>), A<sub>3</sub> = (a<sub>1</sub>|a<sub>2</sub>|...|b). and so on. **Theorem 7 Cramer's Rule** (All these matrices are (n x n)) For any b ∈ R<sup>n</sup> Solution of Ax = b can be Computed by formulas x<sub>1</sub> = det(A<sub>1</sub>)/det(A), x<sub>2</sub> = det(A<sub>2</sub>)/det(A), ... x<sub>n</sub> = det(A<sub>n</sub>)/det(A) I am skipping the proof. **Example** A=( <sup>1</sup>/<sub>3</sub> <sup>2</sup>/<sub>3</sub> ), b = ( <sup>3</sup>/<sub>-5</sub> ) Solve Ax = b. **Cramer's rule yields:** det(A) = -1 - 6 = -7 x<sub>1</sub> = det( <sup>3</sup>/<sub>-5</sub> <sup>2</sup>/<sub>3</sub> )/det (A) = -3 + 10 = 7 = -1 x<sub>2</sub> = det( <sup>1</sup>/<sub>3</sub> <sup>3</sup>/<sub>-5</sub> )/det (A) = -5 - 9 = -14 = 2 So, x = (<sup>-1</sup>/<sub>2</sub>). Check that this works, i.e. Ax = (<sup>3</sup>/<sub>-5</sub>) ## A Couple More Properties of Determinants This is from Section 2 **Theorem 5, Section 2** If A is an (n x n) matrix, then det(A<sup>T</sup>) = det(A). **Theorem 6, Section 2** If A and B are (n x n) matrices, then det(AB) = det(A)det(B) **Theorem 8, Section 3, A formula for the inverse of a matrix** Let A bean (n x n) matrix, and C<sub>ij</sub> be cofactors, i.e. C<sub>ij</sub> = (-1)<sup>i+j</sup> det(A<sub>ij</sub>), where A<sub>ij</sub> is A without column i and row j. Then A<sup>-1</sup> = 1/det(A) [ C<sub>11</sub> C<sub>12</sub> ... C<sub>1n</sub> ]<sup>T </sup> [ C<sub>n1</sub> C<sub>n2</sub> ... C<sub>nn</sub> ]<sup>T</sup> Note the transposition, This matrix is called the adjugate of A. (Not to be confused with the adjoint matrix that is something entirely different.) ## Chapter 5, Section 5.1 Eigenvectors and Eigenvalues **Def** Suppose A is an (n x n) matrix. If X is a non-zero vector such that Ax = λx, where λ is some scalar, then λ is called an eigenvalue of A, and x is an eigenvector of A corresponding to this λ. Note that x = 0 is not an eigenvector. λ is allowed to take any value, including 0. If x is an eigenvector of A, this means that Ax is parallel to x (pointing in the same direction). In other words, A preserves directions of its eigenvectors (but not the length). **Example** Suppose A = ( <sup>5</sup>/<sub>6</sub><sup>2</sup>/<sub>6</sub> ) and x = ( <sup>-5</sup>/<sub>6</sub> ) Compute Ax = ( <sup>5</sup>/<sub>6</sub> <sup>2</sup>/<sub>6</sub> )( <sup>-5</sup>/<sub>6</sub> ) = ( <sup>-25</sup>/<sub>36</sub> <sup>-10</sup>/<sub>36</sub> ) = - <sup>5</sup>/<sub>9</sub> (-5) = - λx. This means λ = -4 is an eigenvalue, and x is the corresponding eigenvector (together, they are an eigen pair). Suppose y = ( <sup>1</sup>/<sub>1</sub> ). Is y an eigenvector? Compute Ay = ( <sup>5</sup>/<sub>6</sub> <sup>2</sup>/<sub>6</sub> )( <sup>1</sup>/<sub>1</sub> ) = ( <sup>5</sup>/<sub>6</sub> <sup>2</sup>/<sub>6</sub> ) = λ( <sup>1</sup>/<sub>1</sub> ) = 4y. So y is an eigenvalue, and y is the eigenvector corresponding to λ = 4. Suppose z = ( <sup>-1</sup>/<sub>1</sub> ). Az = ( <sup>5</sup>/<sub>6</sub> <sup>2</sup>/<sub>6</sub> )( <sup>-1</sup>/<sub>1</sub> ) = ( -<sup>5</sup>/<sub>6</sub> <sup>-2</sup>/<sub>6</sub> ) ≠ λ (<sup>-1</sup>/<sub>1</sub>) Since vectors ( <sup>1</sup>/<sub>1</sub> ) and ( -<sup>5</sup>/<sub>6</sub> <sup>-2</sup>/<sub>6</sub> ) are not parallel, So, z is not an eigenvector. Note that the problem of finding both λ and x, just solving from the equation Ax = λx is more difficult than a linear system ## Suppose we are interested in a particular value of λ. Then, in order to find x, we need to solve the linear system Ax = λx. This can be re-written using identity matrix I: Ax = λx Ax = λIx Ax - λIx = 0 (A - λI)x = 0 This means we are solving a homogeneous system. There are two possibilities: Either there is only a trivial solution (recall x = 0 cannot be an eigenvector) then λ is not an eigenvalue! Or, there is a non-trivial solution. Then, x is an eigenvector. But, when a non-trivial solution exists, it is not unique. There is a whole null space of the matrix A - λI, i.e. there are infinitely many eigenvectors corresponding to eigenvalue λ). These eigenvectors form "an eigenspace" corresponding to the eigenvalue λ. **Example 4** A = ( <sup>4</sup>/<sub>2</sub> <sup>-1</sup>/<sub>-1</sub> <sup>6</sup>/<sub>6</sub> ) Somebody has informed us that the eigenvalue λ = 2. Find the eigenvectors and the eigenspace. The eigenvectors are all the solutions of (A - λI)x = 0 Take λ = 2. Then (A - 2I) = ( <sup>4</sup>/<sub>2</sub> <sup>-1</sup>/<sub>-1</sub> <sup>6</sup>/<sub>6</sub> ) - 2( <sup>1</sup>/<sub>0</sub> <sup>0</sup>/<sub>1</sub> ) = ( <sup>2</sup>/<sub>2</sub> <sup>-1</sup>/<sub>-1</sub> <sup>6</sup>/<sub>6</sub> ) - ( <sup>2</sup>/<sub>0</sub> <sup>0</sup>/<sub>2</sub> ) = ( <sup>2</sup>/<sub>2</sub> <sup>-1</sup>/<sub>-1</sub> <sup>6</sup>/<sub>6</sub>) - ( <sup>2</sup>/<sub>0</sub> <sup>0</sup>/<sub>2</sub> ) = ( <sup>2</sup>/<sub>2</sub> <sup>-1</sup>/<sub>-1</sub> <sup>6</sup>/<sub>6</sub> ) The augmented matrix for (A - 2I)x = 0 is ( <sup>2</sup>/<sub>2</sub> <sup>-1</sup>/<sub>-1</sub> <sup>6</sup>/<sub>6</sub> ) ~ ( <sup>2</sup>/<sub>0</sub> <sup>0</sup>/<sub>0</sub> <sup>6</sup>/<sub>0</sub> ) ~ ( <sup>2</sup>/<sub>0</sub> <sup>0</sup>/<sub>0</sub> <sup>0</sup>/<sub>0</sub> ) This is equivalent to the solution 2x<sub>1</sub> - x<sub>2</sub> + 6x<sub>3</sub> = 0 with X2, X3 being free variables, or x<sub>1</sub> = <sup>1</sup>/<sub>2</sub> x<sub>2</sub> - <sup>3</sup>/<sub>2</sub> x<sub>3</sub>. So, x = ( <sup>1</sup>/<sub>2</sub> x<sub>2</sub> - <sup>3</sup>/<sub>2</sub> x<sub>3</sub> ) = ( <sup>1</sup>/<sub>2</sub> ) x<sub>2</sub> + ( <sup>-3</sup>/<sub>2</sub> ) x<sub>3</sub> So, vectors ( <sup>1</sup>/<sub>2</sub> ) and ( <sup>-3</sup>/<sub>2</sub> ) are eigenvectors corresponding to λ = 2. But most importantly, any linear combination of these two vectors is also an eigenvector corresponding to λ = 2. So, the eigenspace is equal to span{( <sup>1</sup>/<sub>2</sub>), ( <sup>-3</sup>/<sub>2</sub>)}. **Observation** For a given λ, system (A - λI)x = 0 has a non-trivial solution iff matrix A - λI is singular, But a matrix is singular iff its determinant is equal to 0. Therefore, A - λI has eigenvectors iff det(A - λI) = 0. **Theorem 1** If A is a triangular matrix (upper- or lower), the eigenvalues of A are the diagonal entries. **Proof** (For a (3 x 3) matrix) Suppose A = ( <sup>a<sub>11</sub></sup>/<sub>0</sub> <sup>a<sub>12</sub></sup>/<sub>a<sub>22</sub></sub> <sup>a<sub>13</sub></sup>/<sub>a<sub>23</sub></sub>), Then det(A - λI) = det ( <sup>a<sub>11</sub>-λ</sup>/<sub>0</sub> <sup>a<sub>12</sub></sup>/<sub>a<sub>22</sub>-λ</sub> <sup>a<sub>13</sub></sup>/<sub>a<sub>23</sub>-λ</sub> ) = (a<sub>11</sub> - λ)(a<sub>22</sub> - λ)(a<sub>33</sub>- λ) (because det of a triangular matrix is the product of diagonal entries) Solve (a<sub>11</sub> - λ)(a<sub>22</sub> - λ)(a<sub>33</sub>- λ) = 0. There are three possible solutions: λ = a<sub>11</sub>, λ = a<sub>22</sub>, λ = a<sub>33</sub>. Proven.
