Question Video: Finding the Point of Intersection of Three Planes Using a Matrix Inverse Mathematics

Three planes are defined by the following equations: โˆ’3๐‘ฅ โˆ’ ๐‘ฆ โˆ’ 2๐‘ง โˆ’ 2 = 0, 4๐‘ฅ โˆ’ ๐‘ฆ โˆ’ 3๐‘ง + 1 = 0, and โˆ’4๐‘ฅ + 3๐‘ฆ + ๐‘ง โˆ’ 4 = 0. Find their point of intersection.

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Video Transcript

Three planes are defined by the following equations: negative three ๐‘ฅ minus ๐‘ฆ minus two ๐‘ง minus two is equal to zero, four ๐‘ฅ minus ๐‘ฆ minus three ๐‘ง plus one is equal to zero, and negative four ๐‘ฅ plus three ๐‘ฆ plus ๐‘ง minus four is equal to zero. Find their point of intersection.

Weโ€™re given the equations for three planes and asked for their point of intersection. This is the point where all three planes have the same ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-values. What we have, in fact, is a system of three linear equations which we want to solve for ๐‘ฅ, ๐‘ฆ, and ๐‘ง. In order to solve this system, weโ€™d like to write this is as a matrix equation so that we have a three-by-three matrix multiplying a three-by-one matrix is equal to a three-by-one matrix. And in order to get this in the correct form, we move our constants to the right-hand side.

Adding two to both sides of our first equation, we have negative three ๐‘ฅ minus ๐‘ฆ minus two is ๐‘ง is equal to two. Similarly, for our second equation, we subtract one from both sides. And for our third equation, we add four. We can then create the three-by-three matrix ๐ด from the coefficients of ๐‘ฅ, ๐‘ฆ, and ๐‘ง on the left-hand side. Our matrix ๐ฎ is then the column matrix with elements ๐‘ฅ, ๐‘ฆ, and ๐‘ง. And our matrix ๐ฏ on the right-hand side is the column matrix with elements two, negative one, and four. And this then gives us our matrix equation ๐ด multiplied by ๐ฎ is equal to ๐ฏ. Assuming that ๐ด is nonsingular, that is, it has an inverse, weโ€™re going to use the inverse matrix method to solve this system of equations.

And to begin with, we use the fact that for a nonsingular ๐‘›-by-๐‘› matrix ๐ด, ๐ด inverse multiplied by ๐ด is equal to ๐ด multiplied by ๐ด inverse is equal to the identity matrix. Thatโ€™s the ๐‘›-by-๐‘› matrix with all elements equal to zero except for the leading diagonal, where the elements are all one. Multiplying our equation through on the left by ๐ด inverse, we have ๐ด inverse ๐ด๐ฎ is equal to ๐ด inverse ๐ฏ. We know that ๐ด inverse ๐ด is the identity matrix so that on our left we have ๐ผ multiplied by ๐ฎ. And in fact, ๐ผ multiplied by ๐ฎ is equal to ๐ฎ.

So now weโ€™ve isolated the column matrix ๐ฎ on the left-hand side which contains our three unknowns ๐‘ฅ, ๐‘ฆ, and ๐‘ง. On our right-hand side then, we have ๐ด inverse multiplied by ๐ฏ. This means to solve our equation, we need to find ๐ด inverse. To do this, weโ€™re going to use the adjoint method. That is, for a nonsingular ๐‘›-by-๐‘› matrix ๐ด, ๐ด inverse is equal to one over the determinant of ๐ด multiplied by the adjoint matrix of ๐ด. Weโ€™re therefore going to need to find both the determinant of ๐ด and the adjoint matrix of ๐ด.

Now making some room, letโ€™s begin by calculating the determinant of the matrix ๐ด. To do this, letโ€™s expand along the first row of ๐ด. This gives us the determinant of ๐ด is equal to the element ๐‘Ž one one multiplied by the determinant of the two-by-two matrix minor ๐ด one one minus the element ๐‘Ž one two multiplied by the determinant of its matrix minor plus the element ๐‘Ž one three multiplied by the determinant of its matrix minor.

Remember that the matrix minor ๐ด ๐‘–๐‘— is the two-by-two matrix obtained by removing row ๐‘– and column ๐‘— from matrix ๐ด. For our first term then, we have the element ๐‘Ž one one, which is negative three, multiplied by the determinant of the two-by-two matrix obtained by removing row one and column one from our matrix ๐ด. That is the determinant of the two-by-two matrix with elements negative one, negative three, three, and one.

And similarly, for our second and third terms where we have negative the element ๐‘Ž one two, so thatโ€™s negative negative one, plus the element ๐‘Ž one three, which is negative two, multiplying the determinant of each of their respective matrix minors. So weโ€™re going to need to calculate these two-by-two determinants. And we recall that for a two-by-two matrix ๐‘€ with elements ๐‘Ž, ๐‘, ๐‘, and ๐‘‘, its determinant is ๐‘Ž๐‘‘ minus ๐‘๐‘.

For our first term then, this determinant is negative one multiplied by one minus negative three multiplied by three and similarly for our second and third terms. In our first term then inside our parentheses, we have negative one plus nine, which is eight. Our second term evaluates to negative eight, and our third term, negative two times eight so that the determinant of our matrix ๐ด is negative 48.

Now making some room and making a note of our determinant, letโ€™s find the adjoint of matrix ๐ด remembering that this is the transpose of the matrix of cofactors. This, of course, means that we must work out the cofactors, where the cofactor ๐ถ ๐‘–๐‘— is equal to negative one raised to the power ๐‘– plus ๐‘— multiplied by the determinant of the matrix minor ๐ด ๐‘–๐‘—. Note that the factor negative one raised to the power ๐‘– plus ๐‘— gives us the sign or parity of the cofactor. For a three-by-three matrix, this is positive, negative, positive, negative, positive, negative, and positive, negative, positive.

In fact, weโ€™ve already seen three of our cofactors when calculating the determinant of the matrix ๐ด. ๐ถ one one is the positive determinant of the matrix minor ๐ด one one. That is minus one, minus negative nine, which is equal to eight. Similarly, ๐ถ one two is the negative determinant of the matrix minor ๐ด one two, and ๐ถ one three is the positive determinant of the matrix minor ๐ด one three. And all three of our first row of cofactors evaluate to eight.

So letโ€™s write out now the remaining cofactors, being careful to use the correct signs. Our second row of cofactors evaluate to negative five, negative 11, and 13. And our third row of cofactors evaluate to one, negative 17, and seven. Our adjoint matrix is then the transpose of the matrix with elements eight, eight, eight, negative five, negative 11, 13, one, negative 17, and seven.

Making some room, we recall that the transpose of the matrix is the matrix obtained by swapping the rows and columns. Hence, in our case, our first row with elements eight, eight, eight becomes our first column. Similarly, our second row becomes our second column, and our third row becomes our third column. Now remember, itโ€™s actually the inverse of the matrix ๐ด that weโ€™re looking for, and the inverse of ๐ด is one over the determinant of ๐ด multiplied by its adjoint matrix. In our case, since our determinant is negative 48, we have one over negative 48 multiplied by the adjoint matrix is the inverse of our matrix ๐ด.

So now letโ€™s put this into our equation. We have ๐ด inverse ๐ฏ is equal to ๐ฎ, which is the same as ๐ฎ is equal to ๐ด inverse ๐ฏ. And swapping our sides for consistency, we multiply out the right-hand side to solve our equation. Using matrix multiplication on the right-hand side then for our first row, we have eight multiplied by two plus negative five multiplied by negative one plus one multiplied by four and similarly for our second and third rows. And evaluating each of our rows, we have negative one over 48 multiplied by the column matrix with elements 25, negative 41, and 31.

And now equating this with our column matrix on the left-hand side with elements ๐‘ฅ, ๐‘ฆ, and ๐‘ง, we have ๐‘ฅ is equal to negative 25 over 48, we have ๐‘ฆ is equal to negative 41 multiplied by negative one over 48, which is 41 over 48, and ๐‘ง is negative 31 over 48. The point of intersection then of the three planes defined by the given equations has the coordinates ๐‘ฅ is equal to negative 25 over 48, ๐‘ฆ is 41 over 48, and ๐‘ง is negative 31 over 48.

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