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I am working on some practice problems and I'm unsure where to begin this problem. It starts off by giving $\det(X)= 1$ for the following matrix $X$:$$ \begin{matrix} a & 1 & d \\ b & 1 & e \\ c & 1 & f \end{matrix} $$ and the $\det(Y)= 4$ for the following matrix $Y$:

$$ \begin{matrix} a & 1 & d \\ b & 2 & e \\ c & 3 & f \end{matrix} $$

Knowing this, the text asked me to solve some other matrices, which I did with relative ease. For some reason, this last one I cannot solve: Find the determinant of: $$ \begin{matrix} a & 5 & d \\ b & 7 & e \\ c & 9 & f \end{matrix} $$

I have tried various rules and identities, but I can't seem to get to this form using scalar multiples or anything like that.

Thanks so much for any help/suggestions/solutions.

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The determinant is multilinear i.e. linear for every column. Notice that the second column of the desired determinant is the sum of $3$ times the second column of the first determinant with twice the second column of the second determinant: hence the desired determinant is $$3\times 1+2\times 4=11$$

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While the answers provided are correct, it's (very) important to know that determinants of some matrices are not always the determinants of linear combinations of other matrices; that is, it's not true for ANY matrices $A, B, C$ that $$\det(A) = m \det(B) + n \det(C),$$ where $m,n$ are scalars. There is a condition for that equality to be true, and thankfully the matrices in your problem meet this condition.

I would imagine that most linear algebra books have a section regarding determinant properties, and one you should probably commit to memory goes something like this:

Let $A, B, C$ be $n\times n$ matrices that differ only in a single row or column; call this row or column the $i^\text{th}$ row or column. If this $i^\text{th}$ row or column in $C$ can be obtained by adding scalar multiples of $A$ and $B$, then it is true that $\det(A) = m \det(B) + n \det(C),$ where $m,n$ are scalars.

In your case, note that the matrices differ by one column only, so the equality holds.

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  • $\begingroup$ Note the use of \times in $n\times n$ and \det in $a\det A$. The latter not only prevents italicization but also results in proper spacing between $a$ and $\det$ and between $\det$ and $A$. I edited accordingly. $\endgroup$ – Michael Hardy Oct 18 '14 at 22:34
  • $\begingroup$ Thanks @MichaelHardy. I was hoping someone would edit to improve it; I wasn't sure what the syntax was. $\endgroup$ – daOnlyBG Oct 18 '14 at 22:35
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How do you write $(5,7,9)$ as a linear combination of $(1,1,1)$ and $(1,2,3)$?

It's not hard to see that it's $$ (5,7,9)=3(1,1,1) + 2(1,2,3). $$ So $$ \det \begin{bmatrix} a & 5 & d \\ b & 7 & e \\ c & 9 & f \end{bmatrix} = 3\det \begin{bmatrix} a & 1 & d \\ b & 1 & e \\ c & 1 & f \end{bmatrix} + 2\det \begin{bmatrix} a & 1 & d \\ b & 2 & e \\ c & 3 & f \end{bmatrix}. $$

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