Question

Find the determinant of matrix with row one 9,1,3 row two 4,6,7 row three 2,-5,10

Answer 1

733

Answer 2

The given matrix is $A = \begin{pmatrix} 9 & 1 & 3 \\ 4 & 6 & 7 \\ 2 & -5 & 10 \end{pmatrix}.$

The determinant of a 3x3 matrix $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ is given by the formula: $det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$

Using this formula for the given matrix: $a=9, b=1, c=3$ $d=4, e=6, f=7$ $g=2, h=-5, i=10$

$det(A) = 9 \times ((6)(10) - (7)(-5)) - 1 \times ((4)(10) - (7)(2)) + 3 \times ((4)(-5) - (6)(2))$ $det(A) = 9 \times (60 - (-35)) - 1 \times (40 - 14) + 3 \times (-20 - 12)$ $det(A) = 9 \times (60 + 35) - 1 \times (26) + 3 \times (-32)$ $det(A) = 9 \times (95) - 26 - 96$ $det(A) = 855 - 26 - 96$ $det(A) = 855 - (26 + 96)$ $det(A) = 855 - 122$ $det(A) = 733$

The determinant of the matrix is 733.

Alternative method:

Calculate the determinant of the 3x3 matrix $\begin{pmatrix} 9 & 1 & 3 \\ 4 & 6 & 7 \\ 2 & -5 & 10 \end{pmatrix}$ using the cofactor expansion method along the first row. $det(A) = 9 \times \det \begin{pmatrix} 6 & 7 \\ -5 & 10 \end{pmatrix} - 1 \times \det \begin{pmatrix} 4 & 7 \\ 2 & 10 \end{pmatrix} + 3 \times \det \begin{pmatrix} 4 & 6 \\ 2 & -5 \end{pmatrix}$ $det(A) = 9 \times ((6)(10) - (7)(-5)) - 1 \times ((4)(10) - (7)(2)) + 3 \times ((4)(-5) - (6)(2))$ $det(A) = 9 \times (60 + 35) - 1 \times (40 - 14) + 3 \times (-20 - 12)$ $det(A) = 9 \times 95 - 1 \times 26 + 3 \times (-32)$ $det(A) = 855 - 26 - 96$ $det(A) = 855 - 122$ $det(A) = 733$