One of the possible values of x which will satisfy equation...

Key Concepts

Factorization and Root-Finding

Once a polynomial is derived from a determinant condition, techniques such as factorization and substitution are used to solve for the variable. This process involves breaking down the polynomial into simpler factors and verifying which candidate solutions satisfy the original equation.

Polynomial Equations from Determinants

When the elements of a matrix include a variable, setting the determinant equal to zero leads to a polynomial equation. This polynomial encapsulates the conditions under which the matrix becomes singular, and solving it yields the critical values of the variable.

Singular Matrices

A matrix is called singular if its determinant equals zero, meaning it does not have an inverse. Identifying when a matrix becomes singular is crucial because it often indicates the presence of dependent equations in a system or special values of parameters that lead to non-unique solutions.

Matrix Determinants

A determinant is a scalar value computed from a square matrix that reflects important properties such as invertibility and volume scaling in linear transformations. For a 3×3 matrix, the determinant is calculated using systematic methods (like cofactor expansion), which break the calculation down into simpler, smaller determinants.

Transcript

00:01 So in the given question we have a determinant that is given as x a b a x b a x b a d x and the value of this determinant is given as 0 and what we are told to do is we are told to find one of the possible values of x of x which will satisfy this given equation, right? we have to find one value of x which would satisfy this equation.

00:43 So what we can do over here is we can take this determinant and simplify it first, right? so what we are going to do is we have to perform a row operation that is a column operation which is c1 changes to c1 plus c2 plus c3 and when we do this what we would have is we would get the determinant x plus a plus b a b x plus a plus b x plus a plus b x b x plus a plus b b x plus a plus b b x so now from this determination we can take the common factor x plus a plus a plus b outside so x plus a plus b times the determinant 1 1 1 a b x b and b x right so in the next step what we are going to do is we can evaluate this determinant by teaching x plus a plus b times we have 1 times x squared minus b square minus we have a x minus a x minus a x minus b square then we have plus plus we have a b minus b x so this is what we have as the determinant and now what we can do is x plus a plus b times x squared minus b squared minus a x plus b squared plus ab minus b x plus b squared as equal to 0 right so we have x plus a plus a plus a plus b plus b squared times x square minus a plus b x plus a b so this is what we have as the determinant of the given a determinant in the given question right so now we need to find a possible value of x which would satisfy this equation that is the determinant of this as equal to zero right so so this determinant would be equal to 0 when x is equal to minus a minus b, right? and this we can write it as minus a plus b, right? so when x is equal to minus of a plus b, x plus a plus b would be equal to 0, which would make this equation, which would make this expression evaluate to 0, right? so the required answer is x, one possible solution of x that would satisfy this equation is x equal to minus of a plus b and in the question the option that gives us this as the answer is option c...

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