Question Video: Using Three-by-Three Determinants to Find an Unknown Value | Nagwa Question Video: Using Three-by-Three Determinants to Find an Unknown Value | Nagwa

Question Video: Using Three-by-Three Determinants to Find an Unknown Value Mathematics • First Year of Secondary School

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Determine the value of π‘˜ that makes π‘₯ = 4 a root of the equation |9, π‘₯ βˆ’ 3, βˆ’4 and π‘₯ + 8, βˆ’2, π‘˜ and 9π‘₯ βˆ’ 3, π‘₯ βˆ’ 7, βˆ’5| = 0.

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Video Transcript

Determine the value of π‘˜ that makes π‘₯ equal four a root of the equation the determinant of the matrix with elements nine, π‘₯ minus three, negative four, π‘₯ plus eight, negative two, π‘˜, nine π‘₯ minus three, π‘₯ minus seven, negative five is equal to zero.

If π‘₯ equals four is a root of the equation, then our determinant must equal zero for π‘₯ is equal to four. So let’s begin by substituting π‘₯ equal four into our equation. In our first row, this then gives us elements nine, four minus three, and negative four, where four minus three is equal to one. In our second row, this gives us elements four plus eight, negative two, and π‘˜, where four plus eight is equal to 12. In our third row, we have nine times four minus three, four minus seven, and negative five, that is, 33, negative three, and negative five.

So now to find the value of π‘˜, we need to find the determinant of our new matrix. To do this, we recall that, for an 𝑛-by-𝑛 matrix 𝐴 with elements π‘Ž 𝑖𝑗, then the matrix minor uppercase 𝐴 subscript 𝑖𝑗 is equal to the matrix 𝐴 minus row 𝑖 and column 𝑗. So, for example, if we have a three-by-three matrix, the matrix minor 𝐴 one two is the matrix 𝐴 minus row one and column two, that is, the two-by-two matrix with elements π‘Ž two one, π‘Ž two three, π‘Ž three one, and π‘Ž three three.

Now, recall also that to find the determinant of the 𝑛-by-𝑛 matrix 𝐴 by expanding along the 𝑖th row, we use the formula the determinant of 𝐴 is equal to the sum from 𝑗 is one to 𝑛 of negative one raised to the power 𝑖 plus 𝑗 multiplied by the element π‘Ž subscript 𝑖𝑗 multiplied by the determinant of the matrix minor 𝐴 𝑖𝑗. In our case, our matrix is a three-by-three matrix. Therefore, 𝑛 is equal to three.

If we expand along the first row, we have 𝑖 equal to one so that our determinant is negative one raised to the power one plus one, which is two, multiplied by the element π‘Ž one one multiplied by the determinant of the matrix minor 𝐴 one one plus negative one raised to the power one plus two multiplied by the element π‘Ž one two multiplied by the determinant of the matrix minor 𝐴 one two plus negative one raised to the power one plus three multiplied by the element π‘Ž one three multiplied by the determinant of the matrix minor 𝐴 one three.

Negative one raised to the power two is positive one. The element π‘Ž one one is equal to nine. The matrix minor 𝐴 one one is the result of deleting row one and column one, that is, the matrix with elements negative two, π‘˜, negative three, and negative five, so that our first term is one multiplied by nine multiplied by the determinant of the two-by-two matrix with elements negative two, π‘˜, negative three, and negative five. Negative one raised to the power three is negative one. The element π‘Ž one two is equal to one. And the matrix minor 𝐴 one two is the two-by-two matrix resulting in deleting row one and column two. That’s the two-by-two matrix with elements 12, π‘˜, 33, and negative five. And finally, negative one raised to the power four is equal to one. Element π‘Ž one three is negative four. And the matrix minor 𝐴 one three has elements 12, negative two, 33, and negative three.

The determinant of our matrix is therefore nine multiplied by the determinant of the two-by-two matrix with elements negative two, π‘˜, negative three, and negative five minus the determinant of the two-by-two matrix with elements 12, π‘˜, 33, negative five minus four times the determinant of the two-by-two matrix with elements 12, negative two, 33, and negative three.

Now making some room and rewriting our determinant, our next step is to calculate the three two-by-two determinants. And recall that a two-by-two matrix 𝑀 with elements π‘Ž, 𝑏, 𝑐, 𝑑 has determinant π‘Žπ‘‘ minus 𝑏𝑐. For our first term, this then gives us nine multiplied by negative two times negative five minus negative three π‘˜.

Our second term is negative 12 times negative five minus 33π‘˜. And our third term is negative four times 12 times negative three minus negative two times 33. Evaluating inside our parentheses and then multiplying them out, we have the determinant of our matrix is 90 plus 27π‘˜ plus 60 plus 33π‘˜ minus 120. And collecting like terms, this is 60π‘˜ plus 30.

But remember, we’re trying to determine the value of π‘˜ that makes this equal to zero. So we need to solve this equation for π‘˜. Now making some room, if we subtract 30 from both sides, we have 60π‘˜ is equal to negative 30. And now dividing both sides by 60, we have π‘˜ is negative 30 over 60. Now dividing both numerator and denominator by 30, we have π‘˜ is equal to negative one over two. Therefore, the value of π‘˜ that makes π‘₯ equal four a root of the given equation is π‘˜ is negative one over two.

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