You are watching: The determinant of a is the product of the diagonal entries in a

What I have so far:

We will prove this by induction because that an $n$ $\times$ $n$ matrix. Because that the case of a $2 \times 2$ matrix, allow A=$ \left( \beginarrayccca_11 & a_12 \\0 & a_22 \endarray \right)$. For this reason det($A$)=$a_11a_22$ and also the explain is true because that the situation of a $2 \times 2$ matrix. Currently suppose that this declare is true because that an $n$ $\times$ $n$ matrix. We will display that it also is true for an $(n + 1)$ $\times$ $(n + 1)$ matrix. Allow A =$ \left( \beginarrayccca_11 & a_12 & \cdots & a_1(n+1)\\0 & a_22 & \cdots & a_2(n+1) \\\vdots & \cdots & & \vdots\\0 & 0 & \cdots & a_(n+1)(n+1)\endarray \right)$.

I don"t understand what to execute after this.

Using the cofactor growth along the an initial column we acquire $$|A|=(-1)^1+1a_11 \left |\beginmatrix a_22&a_23 & \cdots & a_2(n+1) \\ 0 &a_33 & \cdots & a_3(n+1) \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & a_n+1n+1 \endmatrix\right |$$ (Note that remaining terms in the growth are zero.)

Using the induction hypothesis (you know det the the $n\times n$ matrix), we gain

$|A|$=$a_11a_22a_33\cdots a_n+1n+1$

Use the Laplace development to complete the induction. If you take it the young $A_12$ because that example, you"ll get the determinant of an top triangle matrix through top-left entry $0$, i beg your pardon by induction is $0$. The only minor that the peak row i m sorry is no $0$ is $A_11$.

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