Q5Numerical1 Mark13 Apr 2025Let A=[a1b1c1a2b2c2a3b3c3]A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}A=a1a2a3b1b2b3c1c2c3 and det(A)=5\det(A) = 5det(A)=5. Let B=[3a1−6b13c1a2−2b2c2a3−2b3c3]B = \begin{bmatrix} 3a_1 & -6b_1 & 3c_1 \\ a_2 & -2b_2 & c_2 \\ a_3 & -2b_3 & c_3 \end{bmatrix}B=3a1a2a3−6b1−2b2−2b33c1c2c3. Find det(B)\det(B)det(B).Your answerPress Enter to check.
