Delirious Determinant

Consider all 3x3 determinants whose each element can be $1$ or $0$ .Find the number of such determinants whose value is $0$ when evaluated.

#Algebra #Determinant #CosinesGroup #Goldbach'sConjurersGroup #TorqueGroup

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let $R_1, R_2, R_3$ represent the rows of the matrix. If the determinant is zero, then the rows must be linearly dependent. Therefore there exist integers $a,b,c$ such that $a R_1 + b R_2 + c R_3 = 0$ . So I guess we can look at several mutually exclusive cases:

$(1)$ All of the rows are zero. There is just 1 matrix like this.

$(2)$ Exactly 2 of the rows are zero. We choose which two ( $\binom{3}{2}$ ways), and there are 7 possibilities for the remaining nonzero row. (the remaining row can't be zero because we have already counted the case where all 3 rows are zero). Therefore there are 21 ways in this case.

$(3)$ Exactly one of the rows is zero. We choose which one ( $\binom{3}{1}$ ways), and there are $7*7$ possibilities for the remaining two nonzero rows. Therefore there are 147 ways in this case.

$(4)$ None of the rows are zero, but the rows are all equal to each other. Well, there are 7 possibilities for what the row looks like, so there are 7 ways.

$(5)$ Exactly 2 of the rows are equal to each other. There are $\binom{3}{2}$ ways to choose which two, 7 ways to decide what the two equal rows look like, and 6 ways remaining to decide how the last one looks (because we want it to be different). Therefore, there are 126 ways in this case.

$(6)$ The rows are all different, but two of them add to make the third row.

First we choose which two rows add to make the third one ( $\binom{3}{2}$ ways). Suppose that $R_a + R_b = R_c$ where $a,b,c$ are the numbers $1,2,3$ in some order.

Now we'll need to split this case into a few subcases:

$(6a)$ $R_a$ has exactly one '1'. There are 3 possibilities: $(1,0,0),(0,1,0),(0,0,1)$ . Note that if $R_{a,k} = 1$ , then we must have $R_{b,k} = 0$ , because the entries can only be 0 and 1 and we assumed that $R_a + R_b = R_c$ . For the other two digits in $R_b$ , there are 3 possibilities (we can't have $R_b = (0,0,0)$ since that possibility was covered in the previous cases). Since $R_a + R_b = R_c$ , there is only 1 choice for what $R_c$ is. This gives a total of 9 for this subcase.

$(6b)$ $R_a$ has exactly two '1's. There are three possibilities: $(1,1,0),(0,1,1),(1,0,1)$ . Now $R_b$ can only be $(0,0,1),(1,0,0),(0,1,0)$ respectively (i,e, only one choice for $R_b$ ). Again since $R_a + R_b = R_c$ , there is only 1 choice for what $R_c$ is. This gives a total of 3 for this subcase.

Now $R_a$ can't be all zeros since this case was covered earlier. It also can't be all ones, because that would force $R_b$ to be all zeros.

Therefore, the total of the subcases is 12, and multiplying that by $\binom{3}{2}$ gives 36 for case $(6)$ .

Hence, the total of all the cases is $1 + 21 + 147 + 7 + 126 + 36 = 328$ .

Ariel Gershon - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$