Science:Math Exam Resources/Courses/MATH221/December 2009/Question 02 (b)

MATH221 December 2009

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[hide] Question 02 (b)
Find the determinant of the matrix $B={\begin{bmatrix}1&0&3&0\\2&1&4&-1\\3&2&4&0\\0&3&-1&1\end{bmatrix}}$

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[show] Hint
Science:Math Exam Resources/Courses/MATH221/December 2009/Question 02 (b)/Hint 1

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[show] Solution
Strategy: Use the following determinant properties. The determinant of upper triangular matrix (i.e a matrix in row echelon form in this case) is equal to the product of its diagonal entries. If i $\neq$ j , then subtracting some constant time row i from row j does not change the determinant. Step 1: Perform the following row operations on B: row 2 = row 2 - 2 row 1 row 3 = row 3 - 2 row 1 and we get ${\begin{bmatrix}1&0&3&0\\0&1&-2&-1\\0&2&-5&0\\0&3&-1&1\end{bmatrix}}$ Since the above matrix is still not in upper triangular form, perform the following row operations to it: row 3 = row 3 - 2 row 2 row 4 = row 4 - 3 row 2 Then we get: ${\begin{bmatrix}1&0&3&0\\0&1&-2&0\\0&0&-1&2\\0&0&5&4\end{bmatrix}}$ After performing one more row operation, row 4 = row 4 + 5 row 3, we get the following upper triangular matrix: ${\begin{bmatrix}1&0&3&0\\0&1&-2&0\\0&0&-1&0\\0&0&0&14\end{bmatrix}}$ Step 2 : Let $U'={\begin{bmatrix}1&0&3&0\\0&1&-2&0\\0&0&-1&0\\0&0&0&14\end{bmatrix}}$ Then $det(A)=det(U')=(1)(1)(-1)(14)=-14$ and thus $det(A)=-14$ .