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Given that the matrix negative one, one, negative one, negative one times π₯, π¦ is equal to negative seven, five, find π₯, π¦.
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Now, in fact, what weβve been given is a matrix form of a system of linear equations.
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And so imagine weβre given a matrix equation π΄π equals π΅, where π΄ is the two-by-two matrix negative one, one, negative one, negative one.
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Capital π is the column matrix π₯, π¦.
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And π΅ is the column matrix negative seven, five.
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We can multiply both sides by the inverse of π΄ providing it exists.
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This gives us the inverse of π΄ times π΄π is equal to the inverse of π΄ times π΅.
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And of course, we remember that matrix multiplication is not commutative.
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We have to do it in this order.
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But of course, the inverse of a matrix times itself is equal to πΌ, the identity matrix.
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So this equation becomes πΌπ equals the inverse of π΄ times π΅.
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But in fact, πΌπ is simply equal to π since multiplying any matrix by an identity matrix of the appropriate size just leaves the original matrix.
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So we can say that π is equal to the inverse of π΄ times π΅.
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And this gives us a way to solve our matrix equation.
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We begin by finding the inverse of the two-by-two matrix negative one, one, negative one, negative one and then weβll multiply that by the column matrix negative seven, five.
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So, given the π΄ weβve defined, how do we calculate the inverse?
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Well, given a general two-by-two matrix π, π, π, π, its inverse is one over the determinant of the original matrix times the matrix with elements π, negative π, negative π, π.
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Essentially, we switch the elements in the top left and bottom right.
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And then we change the sign of the remaining two elements.
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And of course, the determinant of a two-by-two matrix is ππ minus ππ.
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Essentially, we subtract the product of the top-right and bottom-left elements from the product of the top left and bottom right.
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And this inverse will not exist if the determinant is equal to zero.
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In this case, we say that the matrix is not invertible.
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So it makes sense to begin by calculating the determinant of our matrix π΄.
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Itβs the product of the top left and bottom right, so negative one times negative one minus the product of the top right and bottom left.
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So we subtract one times negative one.
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Negative one times negative one is one.
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So this actually becomes one plus one, which is equal to two.
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So since the determinant is not equal to zero, the inverse of π΄ exists, and we can carry on with the next step.
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We calculate one over the determinant.
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So thatβs one over two.
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Then we switch the elements in the top left and bottom right.
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And we change the sign of the remaining two.
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So we get that the inverse of π΄ is one-half times negative one, negative one, one, negative one.
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And so, given that weβve defined π to be the matrix π₯, π¦ and π΅ to be the matrix negative seven, five, the solution to our equation is one-half times the matrix negative one, negative one, one, negative one times the column matrix negative seven, five.
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Now, of course, we could distribute that one-half across the two-by-two matrix or we can complete that step at the end.
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And in fact, since distributing a half across the two-by-two matrix will make it very fraction heavy, we will in fact do this step at the end.
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Letβs clear some space and multiply the pair of matrices.
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Weβre multiplying a two-by-one matrix by a two-by-two matrix.
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And this will, in fact, result in a two-by-one matrix.
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To find the first element in this result, we find the dot product of the elements in the first row with the elements in our column matrix, thatβs negative one times negative seven plus negative one times five, which is two.
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Letβs repeat this process to find the next element.
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This time, itβs the dot product of the elements in the second row with the elements in this column matrix.
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So this time, itβs one times negative seven plus negative one times five, which is negative 12.
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Substituting this into our earlier expression for π₯, π¦ and we see π₯, π¦ is equal to one-half times the matrix two, negative 12.
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All thatβs left is to distribute one-half across this matrix.
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A half of two is one and a half of negative 12 is negative six.
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And so, the matrix π₯, π¦ is the matrix one, negative six.