1. **Stating the problem:** We have matrices $A$, $B = I_3 + tA$, and $C = I_3 + aA$ where $I_3$ is the $3 \times 3$ identity matrix, and $a,t \in \mathbb{R}$. We want to find $a$ such that $B$ is the inverse of $C$, i.e., $B = C^{-1}$. 2. **Formula and important rules:** If $B = C^{-1}$, then by definition: $$BC = I_3$$ Substitute $B$ and $C$: $$ (I_3 + tA)(I_3 + aA) = I_3 $$ We will expand and simplify this product. 3. **Expanding the product:** $$ (I_3 + tA)(I_3 + aA) = I_3 \cdot I_3 + I_3 \cdot aA + tA \cdot I_3 + tA \cdot aA = I_3 + aA + tA + ta A^2 $$ Combine like terms: $$ I_3 + (a + t)A + ta A^2 = I_3 $$ 4. **Equating to identity:** For the equality to hold: $$ I_3 + (a + t)A + ta A^2 = I_3 \implies (a + t)A + ta A^2 = 0 $$ 5. **Using properties of matrix $A$:** We need to express $A^2$ in terms of $A$ and $I_3$ to solve for $a$. Calculate $A^2$: $$ A = \begin{pmatrix}1 & 3 & 2 \\ 3 & 5 & 6 \\ 2 & 6 & 9\end{pmatrix} $$ Calculate $A^2 = A \times A$: $$ A^2 = \begin{pmatrix} 1\cdot1 + 3\cdot3 + 2\cdot2 & 1\cdot3 + 3\cdot5 + 2\cdot6 & 1\cdot2 + 3\cdot6 + 2\cdot9 \\ 3\cdot1 + 5\cdot3 + 6\cdot2 & 3\cdot3 + 5\cdot5 + 6\cdot6 & 3\cdot2 + 5\cdot6 + 6\cdot9 \\ 2\cdot1 + 6\cdot3 + 9\cdot2 & 2\cdot3 + 6\cdot5 + 9\cdot6 & 2\cdot2 + 6\cdot6 + 9\cdot9 \end{pmatrix} $$ Calculate each element: - First row: $(1 + 9 + 4, 3 + 15 + 12, 2 + 18 + 18) = (14, 30, 38)$ - Second row: $(3 + 15 + 12, 9 + 25 + 36, 6 + 30 + 54) = (30, 70, 90)$ - Third row: $(2 + 18 + 18, 6 + 30 + 54, 4 + 36 + 81) = (38, 90, 121)$ So: $$ A^2 = \begin{pmatrix}14 & 30 & 38 \\ 30 & 70 & 90 \\ 38 & 90 & 121 \end{pmatrix} $$ 6. **Express $A^2$ as a linear combination of $A$ and $I_3$:** Assume: $$ A^2 = \alpha A + \beta I_3 $$ We want to find scalars $\alpha, \beta$ such that this holds. Compare elements: - From $(1,1)$ element: $$ 14 = \alpha \cdot 1 + \beta \cdot 1 = \alpha + \beta $$ - From $(1,2)$ element: $$ 30 = \alpha \cdot 3 + \beta \cdot 0 = 3\alpha $$ - From $(3,3)$ element: $$ 121 = \alpha \cdot 9 + \beta \cdot 1 = 9\alpha + \beta $$ From $(1,2)$: $$ 3\alpha = 30 \implies \alpha = 10 $$ From $(1,1)$: $$ 10 + \beta = 14 \implies \beta = 4 $$ Check $(3,3)$: $$ 9 \times 10 + 4 = 90 + 4 = 94 \neq 121 $$ So $A^2$ is not exactly a linear combination of $A$ and $I_3$. 7. **Try expressing $A^2$ as $pA + qI_3 + rJ$ where $J$ is a matrix of ones:** Since $A$ has a pattern, but this is complicated, let's try a different approach. 8. **Use the fact that $B = C^{-1}$ implies $C B = I_3$:** We have: $$ (I_3 + aA)(I_3 + tA) = I_3 $$ Expanding: $$ I_3 + tA + aA + a t A^2 = I_3 $$ Simplify: $$ (a + t) A + a t A^2 = 0 $$ Rewrite: $$ (a + t) A = - a t A^2 $$ Multiply both sides by $A^{-1}$ if $A$ is invertible: $$ (a + t) I_3 = - a t A $$ But $A$ is not invertible (determinant is zero), so this is invalid. 9. **Check if $A$ is invertible:** Calculate determinant of $A$: $$ \det(A) = 1 \times (5 \times 9 - 6 \times 6) - 3 \times (3 \times 9 - 6 \times 2) + 2 \times (3 \times 6 - 5 \times 2) $$ Calculate each term: - $5 \times 9 - 6 \times 6 = 45 - 36 = 9$ - $3 \times 9 - 6 \times 2 = 27 - 12 = 15$ - $3 \times 6 - 5 \times 2 = 18 - 10 = 8$ So: $$ \det(A) = 1 \times 9 - 3 \times 15 + 2 \times 8 = 9 - 45 + 16 = -20 $$ Since $\det(A) \neq 0$, $A$ is invertible. 10. **Multiply both sides of equation by $A^{-1}$:** $$ (a + t) A + a t A^2 = 0 \implies (a + t) I_3 + a t A = 0 $$ Rearranged: $$ (a + t) I_3 = - a t A $$ Multiply both sides by $A^{-1}$: $$ (a + t) A^{-1} = - a t I_3 $$ 11. **Since $A^{-1}$ is a matrix, the above equality can hold only if $A^{-1}$ is a scalar multiple of $I_3$, which is impossible unless $A$ is scalar multiple of $I_3$. So the only way is that both sides are zero matrices.** 12. **Therefore, the scalar coefficients must satisfy:** $$ (a + t) = 0 \quad \text{and} \quad a t = 0 $$ 13. **Solve the system:** - From $a + t = 0$ we get $a = -t$ - From $a t = 0$ we get either $a=0$ or $t=0$ 14. **Check cases:** - If $a=0$, then from $a = -t$ we get $0 = -t \implies t=0$ - If $t=0$, then $a = -0 = 0$ So the only solution is: $$ a = 0, \quad t = 0 $$ 15. **Conclusion:** The only $a \in \mathbb{R}$ such that $B = I_3 + tA$ is the inverse of $C = I_3 + aA$ is when $a = 0$ and $t = 0$, meaning both $B$ and $C$ are the identity matrix. **Final answer:** $$ \boxed{a = 0} $$