(B) We know that $\det(AB) = \det(A) \cdot \det(B)$.Given $\det(AB) = 0$, we have $\det(A) \cdot \det(B) = 0$.Calculating $\det(A) = (x+2)^2 - 9x = x^

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(B) We know that $\det(AB) = \det(A) \cdot \det(B)$.Given $\det(AB) = 0$, we have $\det(A) \cdot \det(B) = 0$.Calculating $\det(A) = (x+2)^2 - 9x = x^2 + 4x + 4 - 9x = x^2 - 5x + 4 = (x-1)(x-4)$.Calculating $\det(B) = x(x+2) - 0 = x(x+2)$.Thus, the equation becomes $(x-1)(x-4) \cdot x(x+2) = 0$.Setting each factor to zero, we get $x-1=0, x-4=0, x=0, x+2=0$.Therefore, the solutions are $x = 1, 4, 0, -2$.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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Let $A = \begin{bmatrix} x+2 & 3x \\ 3 & x+2 \end{bmatrix}$ and $B = \begin{bmatrix} x & 0 \\ 5 & x+2 \end{bmatrix}$. Then all solutions of the equation $\det(AB) = 0$ are:

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A
$1, -1, 0, 2$
B
$1, 4, 0, -2$
C
$1, -1, 4, 3$
D
$-1, 4, 0, 3$

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3 and 4 .Determinants and Matrices

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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Let $A = \begin{bmatrix} x+2 & 3x \\ 3 & x+2 \end{bmatrix}$ and $B = \begin{bmatrix} x & 0 \\ 5 & x+2 \end{bmatrix}$. Then all solutions of the equation $\det(AB) = 0$ are:

Similar Questions

If $P, Q$ and $R$ are $3 \times 3$ matrices such that $\begin{bmatrix} 3x^2+x+3 & 2x^2-x+4 & 7x^2+8x+5 \\ 5x^2+3x+2 & 4x^2-2x-1 & 7x^2+5x+8 \\ 3x^2+2x+5 & 4x^2-x-2 & 3x^2+8x+7 \end{bmatrix} = Px^2+Qx+R$,then $\det R = $

The value of $x$ obtained from the equation $\left| \begin{array}{ccc} x + \alpha & \beta & \gamma \\ \gamma & x + \beta & \alpha \\ \alpha & \beta & x + \gamma \end{array} \right| = 0$ is:

If $a, b, c$ are sides of $\triangle ABC$ and $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$,then $\sin^2 A + \sin^2 B + \sin^2 C = $ . . . . . . .

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