2.2 Matrices
- Syllabus
- 9231–2028–2029
- Topic
- 2.2
- Level
- A2
Thesystema_{11}x+a_{12}y+a_{13}z=b_1,\quad a_{21}x+a_{22}y+a_{23}z=b_2,\quad a_{31}x+a_{32}y+a_{33}z=b_3isexactlyA\mathbf x=\mathbf b,\qquad A=(a_{ij}),\quad\mathbf x=\begin{pmatrix}x\y\z\end{pmatrix},\quad\mathbf b=\begin{pmatrix}b_1\b_2\b_3\end{pmatrix}.
\begin{aligned}2x-y+z&=4\x+3y-2z&=5\3x+y+z&=6\end{aligned}\quad\Longleftrightarrow\quad \begin{pmatrix}2&-1&1\1&3&-2\3&1&1\end{pmatrix}\begin{pmatrix}x\y\z\end{pmatrix}=\begin{pmatrix}4\5\6\end{pmatrix}.
To reverse the process, multiply each row of A by the unknown column and equate it to the corresponding entry of b. If the problem supplies quantities rather than x,y,z, define the three unknowns first and keep that column order unchanged throughout.
The constants form b, not an extra coefficient column inside A. A row represents one equation; a column represents one unknown across all equations.
For Ax=b with three unknowns:
| Algebraic evidence | Solutions | Plane geometry |
|---|---|---|
| detA=0 | One, x=A−1b | Three planes meet at one point |
| detA=0 and rankA=rank[A∣b]<3 | Infinitely many | Common line when rank 2; a common plane/more freedom when rank is lower |
| rankA<rank[A∣b] | None (inconsistent row such as 0=1) | The three planes have no common point |
First use detA to test uniqueness. If it is zero, row-reduce the augmented matrix [A∣b]. A zero row on both sides removes a constraint and leaves free parameter(s); a zero coefficient row with a non-zero right side is a contradiction. Solve every consistent case and state the geometric interpretation.
Theequationsx+y+z=1,\quad2x+2y+2z=2,\quad x-y+z=0have one dependent row and two independent constraints, so they are consistent with one free parameter: three planes share a line. Replacing the second right side by 3 creates $0=1$ after elimination, so there is no common point.
Singular means ‘not uniquely solvable’, not automatically ‘no solution’. Determinant alone cannot distinguish a common line from inconsistency; the augmented system must be examined.
A non-zero vector e is an eigenvector of A with eigenvalue λ when Ae=λe. Rearranging gives (A−λI)e=0. A non-zero solution exists exactly when A−λI is singular, so det(A−λI)=0 is the characteristic equation.
All non-zero scalar multiples of e represent the same eigendirection. A positive eigenvalue preserves direction, a negative one reverses it, and eigenvalue zero collapses that direction; the zero vector itself is excluded because it would satisfy every lambda.
If $A\mathbf e=\lambda\mathbf e$, thenA^2\mathbf e=A(\lambda\mathbf e)=\lambda A\mathbf e=\lambda^2\mathbf e.Repeating(orinducting)givesA^n\mathbf e=\lambda^n\mathbf e,so e is an eigenvector of $A^n$ with eigenvalue $\lambda^n$.
The characteristic equation finds eigenvalues, not eigenvectors. Each root lambda must still be substituted into (A−λI)e=0 to obtain a non-zero eigendirection.
For a 2x2 or 3x3 matrix A: form det(A−λI)=0 without row-operating A first; solve for the real distinct eigenvalues; for each lambda solve (A−λI)e=0; choose any convenient non-zero scalar representative; finally verify Ae=λe.
ForA=\begin{pmatrix}1&1&0\0&2&1\0&0&3\end{pmatrix},\det(A-\lambda I)=(1-\lambda)(2-\lambda)(3-\lambda),sotherealdistincteigenvaluesare1,2and3.
| lambda | Solve (A−λI)e=0 | One eigenvector |
|---|---|---|
| 1 | y=z=0 | (1,0,0)T |
| 2 | y=x,z=0 | (1,1,0)T |
| 3 | y=2x,z=2x | (1,2,2)T |
This syllabus restricts computation here to real distinct eigenvalues. Diagonal entries reveal eigenvalues immediately only for triangular matrices; the characteristic determinant remains the general method.
If $A\mathbf q_i=\lambda_i\mathbf q_i$ and the eigenvectors form a basis, setQ=(\mathbf q_1\ \mathbf q_2\ \cdots),\qquad D=\operatorname{diag}(\lambda_1,\lambda_2,\ldots).Then $AQ=QD$, soA=QDQ^{-1}.
Adjacentfactorscancel:(QDQ^{-1})^n=QD^nQ^{-1},and $D^n$ is obtained by raising each diagonal eigenvalue to n. The order of columns in Q must match the order of entries in D.
Forthepreviousmatrix,Q=\begin{pmatrix}1&1&1\0&1&2\0&0&2\end{pmatrix},\quad D=\operatorname{diag}(1,2,3),\quad Q^{-1}=\begin{pmatrix}1&-1&1/2\0&1&-1\0&0&1/2\end{pmatrix}. HenceA^n=\begin{pmatrix}1&2^n-1&(3^n-2\cdot2^n+1)/2\0&2^n&3^n-2^n\0&0&3^n\end{pmatrix}.
Check n=0 gives I and n=1 gives A. Do not power Q and Q−1 separately; their cancellation is what makes only D receive the exponent.
Cayley−Hamiltonstates:ifp(\lambda)=\lambda^m+c_{m-1}\lambda^{m-1}+\cdots+c_1\lambda+c_0isthecharacteristicpolynomialofA,thenp(A)=A^m+c_{m-1}A^{m-1}+\cdots+c_1A+c_0I=0.
Rearrange the identity to replace the highest power, then repeat to reduce successive powers to degree below m. If A is non-singular, multiply by a suitable negative power or isolate the constant-I term to express A−1 as a polynomial in A.
ForA=\begin{pmatrix}2&1\1&1\end{pmatrix},p(\lambda)=\lambda^2-3\lambda+1,soA^2-3A+I=0. ThereforeA^3=3A^2-A=8A-3I,and multiplying the identity by $A^{-1}$ givesA^{-1}=3I-A=\begin{pmatrix}1&-1\-1&2\end{pmatrix}.
Replace scalar 1 by I and scalar powers by matrix powers; order is harmless only because every term is a polynomial in the same A. This objective is Cayley-Hamilton—not merely the trace/product eigenvalue check.