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Using the elementary row operation, find the inverse of π΄ for the matrix π΄ is equal to five, three, two, one if possible.
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For a two-by-two matrix, π΄ is equal to π, π, π, π, the inverse of π΄ is given by the formula one over the determinant of π multiplied by π, negative π, negative π, π, where the determinant of π is given by π multiplied by π minus π multiplied by π.
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Notice that this means if the determinant of the matrix π΄ is zero, the inverse cannot exist.
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Thatβs because one over the determinant of π΄ would be one over zero, which we know to be undefined.
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Letβs substitute the values for our matrix π΄ into our formula first for the determinant of π΄.
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π multiplied by π is five multiplied by one, and π multiplied by π is three multiplied by two.
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Evaluating, we get five minus six, which is negative one.
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Since the determinant of π΄ is not zero, the inverse of π΄ does indeed exist.
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Now that we have the determinant, we can substitute everything we know about our matrix into the formula for the inverse of π΄.
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Thatβs one divided by negative one multiplied by one, negative three, negative two, five.
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Remember, we switch the values of the five and the one, and we make three and two negative.
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One divided by negative one is just negative one.
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We can next multiply each of the individual elements in the matrix by negative one.
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And doing so, we can see that the inverse of π΄ is equal to negative one, three, two, negative five.