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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.

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Did you describe the matrices by lines or columns? – Git Gud Jan 26 '13 at 18:30
Matrices equal to numbers? I thought this was only true for $1\times 1$ matrices... – Asaf Karagila Jan 26 '13 at 18:30
Judging by the title, he probably means the determinant of the matrix :) – N. S. Jan 26 '13 at 18:31
@Anna-Banana Write the formulas for the three determinants and then the problem is easy... – N. S. Jan 26 '13 at 18:32
@GitGud does it really matter? – Tim Seguine Jan 26 '13 at 18:33
up vote 1 down vote accepted

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

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

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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 $\det\begin{bmatrix} a_1, & \ldots, & b a_j + c v, & \ldots, a_n \end{bmatrix} = b \det(A) + c \det\begin{bmatrix} a_1, & \ldots, & v, & \ldots, a_n \end{bmatrix}$.
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