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In this video, weβll learn how to check whether a two-by-two matrix has an inverse and then find its inverse if possible.
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For matrices, thereβs no such thing as division.
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We can add, subtract, and multiply them, but we canβt divide matrices.
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There is, however, a related concept which is called inversion.
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And itβs extremely useful for helping us to find the inverse of a matrix and to solve matrix equations.
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We say that an π-by-π matrix, thatβs a square matrix, is invertible if there exists a second π-by-π square matrix such that the product of the matrix and its inverse is πΌ, the identity matrix.
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In the two-by-two case, thatβs the matrix one, zero, zero, one.
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Now, remember, multiplying matrices is not commutative.
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It canβt be done in any order.
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However, when multiplying a matrix by its inverse in any order, weβll always get the identity matrix.
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Note, though, that not all matrices do have an inverse.
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Weβre going to look at the process for finding the inverse of a two-by-two matrix and see what criteria our matrices need to meet to ensure that they do have an inverse.
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We say that for a two-by-two matrix π, π, π, π, its inverse is found by multiplying one over the determinant of π΄ by the two-by-two matrix π, negative π, negative π, π.
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Where the determinant of π΄ is the product of the top-left and bottom-right elements minus the product of the top-right and bottom-left elements.
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Notice also that to get from the two-by-two matrix π΄ to the two-by-two matrix the inverse of π΄, we swap these elements and we change the sign of these.
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So, what does this latter part of the formula mean for finding the inverse of a matrix π΄?
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Well, we know that one divided by zero is undefined.
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We say then that if the determinant of the matrix is zero, it is not invertible; it doesnβt have an inverse.
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Itβs also worth noting that only square matrices are invertible.
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So, letβs have a look at some examples.
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Is the matrix three, one, negative three, negative one invertible?
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Remember, for a two-by-two matrix π΄ given by π, π, π, π, the inverse of π΄ is one over the determinant of π΄ times π, negative π, negative π, π, where the determinant of π΄ is equal to ππ minus ππ.
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We say that the determinant of the matrix doesnβt exist.
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In other words, π΄ is not invertible if its determinant is equal to zero.
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So, all we need to do to establish whether a matrix is invertible is evaluate its determinant and see if thatβs equal to zero.
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So, letβs find the determinant of the matrix three, one, negative three, negative one.
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And we use these bars either side of the matrix to represent its determinant.
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The determinant is found by subtracting the product of the top-right and bottom-left elements from the product of the top left and bottom right.
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So here, π times π is three multiplied by negative one.
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We then subtract π multiplied by π.
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Thatβs one multiplied by negative three.
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And so, our determinant is three multiplied by negative one minus one multiplied by negative three.
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Thatβs negative three minus negative three, which is negative three plus three, which is equal to zero.
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So, the determinant of our matrix three, one, negative three, negative one is zero.
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We can therefore say that the matrix doesnβt have an inverse.
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And therefore, the answer to this question is no, itβs not invertible.
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In our next example, weβll extend this idea about the criteria for matrix invertibility.
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Given that the matrix seven, one, negative seven, π is invertible, what must be true of π?
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Remember, for a square matrix π΄ equals π, π, π, π to be invertible, its determinant must not be equal to zero.
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Now, for this matrix, its determinant is ππ minus ππ.
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So, letβs find an expression for the determinant of our matrix.
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Itβs the product of the top-left and bottom-right elements, thatβs seven times π or seven π, minus the product of the top-right and bottom-left elements.
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Thatβs one multiplied by negative seven.
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So, the determinant of our matrix is seven π minus negative seven, which we can write as seven π plus seven.
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Weβre told that our matrix is invertible, so its determinant cannot be equal to zero.
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In other words, seven π plus seven cannot be equal to zero.
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We need to find the values of π such that this expression is not equal to zero.
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So, weβre going to solve the inequation.
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Weβll solve it like solving a normal equation.
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But instead of our answer being π is equal to some constant, we know π will not be equal to the result.
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Letβs begin by subtracting seven from both sides.
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When we do, we find that seven π will not be equal to negative seven.
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Next, we divide through by seven, and we find π cannot be equal to negative one.
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So, if π is equal to negative one, the matrix does not have an inverse.
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So, for it to be invertible, π cannot be equal to negative one.
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In our next example, weβll look at how to establish whether two matrices are multiplicative inverses of one another.
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Are the matrices one, two, three, four and one, one-half, one-third, and one-quarter multiplicative inverses of each other?
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Letβs remind ourselves what we mean by the phrase multiplicative inverses.
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We say that an π-by-π matrix, thatβs a square matrix, is invertible if there exists a second π-by-π matrix such that the product of the matrix and its inverse in any order is πΌ, where πΌ is the identity matrix.
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In the case of two-by-two identity matrices, thatβs one, zero, zero, one.
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So, we could answer this question in one of two ways.
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We could find the inverse of each matrix and check whether it matches the original of the other matrix.
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Alternatively, we could find the product of the matrices and see if we get the identity matrix.
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Letβs look at the latter method.
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Weβre going to multiply one, two, three, four by one, one-half, a third, a quarter.
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And we remind ourselves that to do so, we begin by finding the dot product of the elements in the first row of the first matrix by the elements in the first column of the second.
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So, thatβs the dot product of one, two and one, one-third.
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Thatβs one multiplied by one plus two multiplied by one-third.
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Thatβs five-thirds.
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Next, we find the dot product of the elements in the first row of our first matrix and of the second column in our second.
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So, thatβs one multiplied by a half plus two multiplied by one-quarter, which is one.
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Now, we find the dot product of the elements in the second row of our first matrix and the first column of our second.
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Thatβs three times one plus four times one-third, which is 13 over three.
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Finally, the dot product of three, four and a half, one-quarter.
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Thatβs three times a half plus four times a quarter, which is five over two.
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And so, we see that when we multiply the matrix one, two, three, four by the matrix one, one-half, a third, a quarter, we get a two-by-two matrix of five-thirds, one, thirteen-thirds, and five over two.
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This is quite clearly not equal to the identity matrix one, zero, zero, one.
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And so, we can say no, the two matrices are not multiplicative inverses of each other.
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Weβll now calculate the multiplicative inverse of a two-by-two matrix.
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Find the multiplicative inverse of the matrix π΄ equals negative four, negative 10, three, five, if possible.
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Remember, to find the multiplicative inverse or the inverse of a two-by-two matrix π΄ equals π, π, π, π, we multiply one over the determinant of π΄ by the matrix π, negative π, negative π, π.
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Remember that this means that if the determinant of π΄ is equal to zero, weβre performing the calculation one divided by zero, which is undefined.
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And this means the inverse does not exist.
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So, letβs begin by calculating the determinant of our matrix π΄ equals negative four, negative 10, three, five.
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To find its determinant, we multiply the element in the top left by the element in the bottom right.
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And then we subtract the product of the elements in the top right by the bottom left.
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So, the determinant of our matrix is negative four times five minus negative 10 times three.
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Thatβs negative 20 minus negative 30, which is negative 20 plus 30.
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And so we see that the determinant of our matrix π΄ is 10.
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Thatβs quite clearly not equal to zero.
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So, the inverse of our matrix π΄ does exist.
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The formula tells us to multiply one over the determinant of π΄, so thatβs one over 10, by a two-by-two matrix which is found by switching the elements in the top left and bottom right and then changing the sign of the other two elements.
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So, the inverse of π΄ is one-tenth times five, 10, negative three, negative four.
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And we know that we can multiply a matrix by a scalar by simply multiplying each of its individual elements.
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One-tenth times five is five-tenths.
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One-tenth times 10 is ten-tenths.
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One-tenth times negative three is negative three-tenths.
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And one-tenth times negative four is negative four-tenths.
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All thatβs left to evaluate the multiplicative inverse of our matrix π΄ is to simplify each of these fractions.
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And so, we see that the inverse of our matrix π΄ is one-half, one, negative three over 10, negative two-fifths.
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In our next example, weβll consider how we combine this process with another operation on matrices.
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Consider the matrices π΄ and π΅.
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Determine the inverse of π΄ plus π΅.
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Here, π΄ is given as the matrix negative three, negative two, negative five, negative seven.
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And π΅ is negative one, two, eight, nine.
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This expression here means to find the inverse of the sum of these two matrices.
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So, letβs simply begin by calculating the sum of π΄ and π΅.
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We do this by adding the individual elements on each row and each column.
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So, the element in the first row and first column is negative three plus negative one, which is negative four.
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The second element in the first row is negative two plus two, which is zero.
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We then add negative five and eight to get three.
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And finally, we add negative seven and nine to get two.
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So, the sum of our two matrices is the two-by-two matrix negative four, zero, three, two.
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We now need to find the inverse of this matrix.
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And so, we recall that for a two-by-two matrix π, π, π, π, its inverse is one over the determinant of π΄ times π, negative π, negative π, π.
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Now, the determinant of π΄ is actually ππ minus ππ.
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Itβs the product of the elements in the top left and bottom right of our matrix minus the product of the elements in the top right and bottom left.
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Note, of course, this means that if the determinant is equal to zero, weβre performing the calculation one divided by zero, which is undefined.
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And so, our matrix does not have an inverse.
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So, letβs begin by calculating the determinant of our matrix π΄ plus π΅.
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Itβs negative four times two minus zero times three, which is simply negative eight.
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So, we know the inverse of the sum of our two matrices does exist, and weβre now ready to work it out.
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Itβs one over the determinant of our matrix, so one over negative eight.
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And then we swap the element on the top left with the element on the bottom right and change the sign of the other two elements.
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So, the inverse of π΄ plus π΅ is one over negative eight times two, zero, negative, three, negative four.
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Now, of course, one divided by negative eight is the same as negative one-eighth.
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And we know that we can multiply a matrix by a scalar by multiplying each of its individual elements.
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Negative one-eighth multiplied by two is negative one-quarter or negative 0.25.
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Negative one-eighth multiplied by zero is, of course, zero.
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Then, we multiply negative one-eighth by negative three.
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Well, a negative multiplied by a negative is a positive.
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So, we get three-eighths which is 0.375.
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Finally, we multiply negative four by negative one-eighth, and we get one-half or 0.5.
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And so, the inverse of our matrix π΄ plus π΅ is negative 0.25, zero, 0.375, and 0.5.
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Note, of course, that we could check our solution by ensuring that when we find the product of the matrix π΄ plus π΅ and its inverse, we get the identity matrix.
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Thatβs the matrix one, zero, zero, one.
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In our final example, weβll look at what we mean when we say a matrix is singular and how this definition can help us to solve problems.
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Find the set of real values of π₯ that make the two-by-two matrix π₯ minus three, eight, two, π₯ plus three singular.
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Letβs begin by defining the word singular when it comes to matrices.
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We say that a matrix is singular if itβs not invertible; it doesnβt have an inverse.
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We know that a matrix is invertible if its determinant is not equal to zero, and the converse is also true.
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So, in other words, a matrix is singular if its determinant is equal to zero.
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In this case then, we need to find the set of real values of π₯ such that the determinant of our matrix is equal to zero.
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The determinant of a two-by-two matrix π, π, π, π is ππ minus ππ.
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We subtract the product of the elements in the top right and bottom left from the product of those in the top left and bottom right.
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So, in this case, thatβs π₯ minus three times π₯ plus three minus eight times two.
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If we distribute these parentheses, we get π₯ times π₯, which is π₯ squared, plus three π₯ minus three π₯ minus three times three, which is nine.
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That simplifies to π₯ squared minus nine.
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And eight multiplied by two is 16.
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So the determinant of our matrix is π₯ squared minus nine minus 16, which is π₯ squared minus 25.
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Weβre trying to find the set of values of π₯ that make our matrix singular.
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In other words, which values of π₯ make the determinant zero?
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So, letβs set our expression for the determinant equal to zero and solve for π₯.
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That is, π₯ squared minus 25 equals zero.
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Adding 25 to both sides of this equation gives us π₯ squared equals 25.
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And then weβll take the square root of both sides of our equations, remembering to take the positive and negative square root of 25.
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That gives us π₯ is equal to positive or negative five.
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We can use these squiggly brackets to help us represent the set of values that make our matrix singular.
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They are negative five and five.
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Note that at this stage, we could check our solutions by substituting each value of π₯ into our original matrix and then checking that the determinant is indeed equal to zero.
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Weβll now summarize the key points from this video.
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We saw that an π-by-π matrix, a square matrix, is invertible if there exists a second matrix such that the product of that matrix and its inverse is πΌ, the identity matrix.
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We saw that if a two-by-two matrix is defined as π, π, π, π, then its inverse is one over the determinant of π times the matrix π, negative π, negative π, π.
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And we, of course, calculate the determinant of π΄ by finding the product of ππ and subtracting the product of ππ.
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Finally, we saw that a matrix is said to be singular if it doesnβt have an inverse.
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In other words, if the determinant of that matrix is equal to zero.