1. **State the problem:** We are given a determinant of a 3x3 matrix and asked to show that it equals zero using properties of determinants. The matrix is: $$\begin{vmatrix} b+c+b & a+b \\ 9 & 1 \\ b & c \\ 1 & 1 \end{vmatrix} = 0$$ However, the matrix appears to have 4 rows and 2 columns, which is not a square matrix and thus its determinant is not defined. Assuming the problem intends a 3x3 matrix, let's clarify the matrix structure. 2. **Interpreting the matrix:** From the description, it seems the matrix is: $$\begin{bmatrix} b+c+b & a+b & ? \\ 9 & 1 & ? \\ b & c & ? \\ 1 & 1 & ? \end{bmatrix}$$ But this is 4 rows and 3 columns, which is invalid for determinant. 3. **Assuming the intended matrix is 3x3:** Let's consider the matrix: $$\begin{bmatrix} b+c+b & a+b & 0 \\ 9 & 1 & 0 \\ b & c & 1 \end{bmatrix}$$ 4. **Simplify the first row:** $$b+c+b = 2b + c$$ So the matrix is: $$\begin{bmatrix} 2b + c & a + b & 0 \\ 9 & 1 & 0 \\ b & c & 1 \end{bmatrix}$$ 5. **Calculate the determinant:** Using the formula for 3x3 determinant: $$\det = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$$ Substitute values: $$= (2b + c)(1 \times 1 - 0 \times c) - (a + b)(9 \times 1 - 0 \times b) + 0(9 \times c - 1 \times b)$$ Simplify: $$= (2b + c)(1) - (a + b)(9) + 0$$ $$= 2b + c - 9a - 9b$$ $$= -9a - 7b + c$$ 6. **Set determinant equal to zero:** $$-9a - 7b + c = 0$$ 7. **Conclusion:** The determinant equals zero if and only if: $$c = 9a + 7b$$ This shows the relationship between $a$, $b$, and $c$ for the determinant to be zero. **Summary:** We used the determinant formula and properties to simplify and find the condition for the determinant to be zero.