# Matrix \( A \) \[ A = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 4 & 2 \\ 3 & 5 & 6 \endpmatrix} ] --- ## 1. Determinant of \( A \) The determinant of a \( 3 \times 3 \) matrix \[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \] is \[ \det(A) = aei + bfg + cdh - ceg - bdi - afh \] For \( A \): \[ \begin{align*} \det(A) &= 2(4 \cdot 6 - 2 \cdot 5) - 1(1 \cdot 6 - 2 \cdot 3) + 3(1 \cdot 5 - 4 \cdot 3) \\ &= 2(24 - 10) - 1(6 - 6) + 3(5 - 12) \\ &= 2(14) - 1() + 3(-7) \\ &= 28 - - 21 \\ &= 7 \end{align*} \] **Determinant:** \[ \boxed{7} \] --- ## 2. Eigenvalues and Eigenvectors ### Characteristic Polynomial Solve \( \det(A - \lambda I) = \): \[ A - \lambda I = \begin{pmatrix} 2-\lambda & 1 & 3 \\ 1 & 4-\lambda & 2 \\ 3 & 5 & 6-\lambda \end{pmatrix} \] Calculate determinant: \[ \begin{align*} &\left| \begin{matrix} 2-\lambda & 1 & 3 \\ 1 & 4-\lambda & 2 \\ 3 & 5 & 6-\lambda \end{matrix} \right| \\ = \ & (2-\lambda)[(4-\lambda)(6-\lambda) - (2)(5)] \\ &- 1[1(6-\lambda) - 2 \cdot 3] \\ &+ 3[1 \cdot 5 - (4-\lambda) \cdot 3] \\ \end{align*} \] Expand: \[ (4-\lambda)(6-\lambda) = 24 - 4\lambda - 6\lambda + \lambda^2 = 24 - 10\lambda + \lambda^2 \] So, \[ (2-\lambda)[(24 - 10\lambda + \lambda^2) - 10] = (2-\lambda)(14 - 10\lambda + \lambda^2) \] Other terms: \[ -1[(6-\lambda) - 6] = -1[ - \lambda] = \lambda \] \[ +3[5 - 3(4-\lambda)] = 3[5 - 12 + 3\lambda] = 3[-7 + 3\lambda] = -21 + 9\lambda \] Sum all: \[ (2-\lambda)(14 - 10\lambda + \lambda^2) + \lambda - 21 + 9\lambda = \] Expand \( (2-\lambda)(14 - 10\lambda + \lambda^2) \): \[ = 2(14 - 10\lambda + \lambda^2) - \lambda(14 - 10\lambda + \lambda^2) \] \[ = 28 - 20\lambda + 2\lambda^2 - 14\lambda + 10\lambda^2 - \lambda^3 \] \[ = 28 - 34\lambda + 12\lambda^2 - \lambda^3 \] So, the characteristic polynomial: \[ -\lambda^3 + 12\lambda^2 - 34\lambda + 28 + \lambda - 21 + 9\lambda = \] \[ -\lambda^3 + 12\lambda^2 - 34\lambda + 28 + 10\lambda - 21 = \] \[ -\lambda^3 + 12\lambda^2 - 24\lambda + 7 = \] Or \[ \boxed{\lambda^3 - 12\lambda^2 + 24\lambda - 7 = } \] ### Eigenvalues This cubic does not factor nicely, but let's check for rational roots using the Rational Root Theorem. Possible rational roots are \( \pm1, \pm7 \). Test \( \lambda = 1 \): \[ 1 - 12 + 24 - 7 = 6 \neq \] Test \( \lambda = 7 \): \[ 343 - 12(49) + 24(7) - 7 = 343 - 588 + 168 - 7 = (343 + 168) - (588 + 7) = 511 - 595 = -84 \neq \] Test \( \lambda = 7 \) (wait, that's plugging \( 7^3 \), but the polynomial is \( \lambda^3 - 12\lambda^2 + 24\lambda - 7 \), so with \( \lambda = 7 \): \[ 343 - 12(49) + 24(7) - 7 = 343 - 588 + 168 - 7 = (343 + 168) - (588 + 7) = 511 - 595 = -84 \neq \] Try \( \lambda = 1 \): \[ 1^3 - 12(1)^2 + 24(1) - 7 = 1 - 12 + 24 - 7 = 6 \] Try \( \lambda = 7 \): \[ 343 - 588 + 168 - 7 = -84 \] Try \( \lambda = \): \[ - + - 7 = -7 \] Try \( \lambda = -1 \): \[ -1 - 12 + (-24) - 7 = -1 - 12 - 24 - 7 = -44 \] Try \( \lambda = .3 \): \[ (.3)^3 - 12(.3)^2 + 24(.3) - 7 \approx .027 - 1.08 + 7.2 - 7 \approx .027 - 1.08 + .2 = -.853 \] No obvious rational roots, so the eigenvalues are **real but irrational**. They can be found numerically. #### **Numerical Solution** Let’s use the cubic formula or estimate roots. Let’s graphically estimate between 1 and 7: - At \( \lambda = 1 \), value is 6 - At \( \lambda = 2 \): \[ 8 - 48 + 48 - 7 = 8 - 48 + 48 - 7 = 8 - 7 = 1 \] So at 1 → 6, at 2 → 1, so root in (1,2). At \( \lambda = 3 \): \[ 27 - 108 + 72 - 7 = 27 - 108 + 72 - 7 = (27 + 72) - (108 + 7) = 99 - 115 = -16 \] So at 2 → 1, at 3 → -16, so root between 2 and 3. Try \( \lambda = 2.1 \): \[ (2.1)^3 - 12(2.1^2) + 24(2.1) - 7 \approx 9.261 - 52.92 + 50.4 - 7 \approx 9.261 + 50.4 - 52.92 - 7 \approx 59.661 - 59.92 = -.259 \] At 2.1 → -.26, so root between 2 and 2.1. Try \( \lambda = 2.05 \): \[ (2.05)^3 - 12(2.05^2) + 24(2.05) - 7 \approx 8.617 - 50.43 + 49.2 - 7 \approx 8.617 + 49.2 - 50.43 - 7 \approx 57.817 - 57.43 = .387 \] At 2.05 → .387 So at 2 → 1, at 2.05 → .39, at 2.1 → -.26 So between 2.05 and 2.1 Let’s interpolate: \[ \lambda_1 \approx 2.075 \] **Similarly for other roots:** At \( \lambda = 10 \): \[ 100 - 120 + 240 - 7 = 100 - 120 + 240 - 7 = 104 - 1207 = -167 \] At \( \lambda = 6 \): \[ 216 - 432 + 144 - 7 = 216 + 144 - 432 - 7 = 360 - 439 = -79 \] At \( \lambda = 8 \): \[ 512 - 768 + 192 - 7 = 512 + 192 - 768 - 7 = 704 - 775 = -71 \] So the third root is much higher. Alternatively, we can use a cubic root calculator or the cubic formula. For now, the eigenvalues are the roots of \[ \boxed{\lambda^3 - 12\lambda^2 + 24\lambda - 7 = } \] --- ### Eigenvectors For each eigenvalue \(\lambda\), solve: \[ (A - \lambda I)\mathbf{x} = \] Plug in each \(\lambda\) found above (numerically) to \(A - \lambda I\) and solve for the nullspace. This is standard but tedious by hand since \(\lambda\) is irrational. --- ## 3. Is \( A \) invertible? A matrix is invertible if and only if its determinant is nonzero. \[ \det(A) = 7 \neq \] **Conclusion:** \[ \boxed{A \text{ is invertible}} \] --- ## 4. Inverse of \( A \) The inverse of a \( 3 \times 3 \) matrix \( A \) is: \[ A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) \] ### Compute Cofactors Let’s denote the minors and cofactors (for entry \( a_{ij} \)): - \( C_{11} = \begin{vmatrix} 4 & 2 \\ 5 & 6 \end{vmatrix} = 4 \cdot 6 - 2 \cdot 5 = 24 - 10 = 14 \) - \( C_{12} = -\begin{vmatrix} 1 & 2 \\ 3 & 6 \end{vmatrix} = -[1 \cdot 6 - 2 \cdot 3] = -[6 - 6] = \) - \( C_{13} = \begin{vmatrix} 1 & 4 \\ 3 & 5 \end{vmatrix} = 1 \cdot 5 - 4 \cdot 3 = 5 - 12 = -7 \) Second row: - \( C_{21} = -\begin{vmatrix} 1 & 3 \\ 5 & 6 \end{vmatrix} = -[1 \cdot 6 - 3 \cdot 5] = -[6 - 15] = 9 \) - \( C_{22} = \begin{vmatrix} 2 & 3 \\ 3 & 6 \end{vmatrix} = 2 \cdot 6 - 3 \cdot 3 = 12 - 9 = 3 \) - \( C_{23} = -\begin{vmatrix} 2 & 1 \\ 3 & 5 \end{vmatrix} = -[2 \cdot 5 - 1 \cdot 3] = -[10 - 3] = -7 \) Third row: - \( C_{31} = \begin{vmatrix} 1 & 3 \\ 4 & 2 \end{vmatrix} = 1 \cdot 2 - 3 \cdot 4 = 2 - 12 = -10 \) - \( C_{32} = -\begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = -[2 \cdot 2 - 3 \cdot 1] = -[4 - 3] = -1 \) - \( C_{33} = \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} = 2 \cdot 4 - 1 \cdot 1 = 8 - 1 = 7 \) So the cofactor matrix is: \[ C = \begin{pmatrix} 14 & & -7 \\ 9 & 3 & -7