Determinants of $3\times 3$ Matrix

Question

Suppose that $a,b,c,d,e,f$ are numbers such that $$\det\left(\begin{matrix} a&1&d\\b&1&e\\ c&1&f \end{matrix}\right)=7$$ and $$\det\left(\begin{matrix} a&1&d\\b&2&e\\ c&3&f \end{matrix}\right)=11.$$

How do you find the determinant of the Matrix $$\begin{pmatrix} a&3&d\\b&5&e\\ c&7&f \end{pmatrix}?$$

Any suggestions on how to approach the question would be greatly appreciated.

Answer 1

Using multilinearity of the determinant (by columns), we get:

$$\begin{vmatrix} a&3&d\\b&5&e\\ c&7&f \end{vmatrix}=\begin{vmatrix} a&1+2&d\\b&1+4&e\\ c&1+6&f \end{vmatrix}=\begin{vmatrix} a&1&d\\b&1&e\\ c&1&f \end{vmatrix}+2\begin{vmatrix} a&1&d\\b&2&e\\ c&3&f \end{vmatrix}=7+2\cdot 11=29$$

Answer 2

HINT Use that the determinant is multilinear:

• Multiplying a column in the matrix by a scalar multiplies the determinant by the same amount. In symbols: $\det [a_1, \cdots, ca_r, \cdots, a_n] = c \det A$.
• It is linear in the columns. Explicitly (using Wikipedia's notation), we have $\det\begin{bmatrix} a_1, & \ldots, & b a_j + c v, & \ldots, a_n \end{bmatrix} = b \det(A) + c \det\begin{bmatrix} a_1, & \ldots, & v, & \ldots, a_n \end{bmatrix}$.