Problem 85
Question
If \(\left|\begin{array}{ccc}x^{2}+x & x+1 & x-2 \\ 2 x^{2}+3 x-1 & 3 x & 3 x-3 \\ x^{2}+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=A x+B\), then (A) \(A=\left|\begin{array}{lll}4 & 0 & 0 \\ 2 & 3 & 3 \\ 4 & 0 & 2\end{array}\right|\) (B) \(B=\left|\begin{array}{ccc}4 & 0 & 0 \\ 2 & 3 & 3 \\ 4 & 0 & -1\end{array}\right|\) (C) \(A=\left|\begin{array}{ccc}4 & 0 & 0 \\ 2 & 3 & -3 \\ 4 & 0 & 2\end{array}\right|\) (D) \(B=\left|\begin{array}{ccc}4 & 0 & 0 \\ 2 & 3 & -3 \\ 4 & 0 & -1\end{array}\right|\)
Step-by-Step Solution
Verified Answer
A is associated with option (C); B with option (D).
1Step 1: Understand the Determinant Problem
You are given a 3x3 determinant expression of a matrix with elements containing polynomials in terms of \(x\). The solution involves expanding this determinant into a linear polynomial \(Ax + B\). Your task is to find the matrices whose determinants correspond to \(A\) and \(B\).
2Step 2: Calculate the Determinant
The determinant of the given matrix can be calculated using the rule of Sarrus or expansion by minors. Focus on simplifying the polynomial expressions. The matrix is: \[\left|\begin{array}{ccc}x^{2}+x & x+1 & x-2 \2x^{2}+3x-1 & 3x & 3x-3 \x^{2}+2x+3 & 2x-1 & 2x-1\end{array}\right|\]
3Step 3: Simplify the Determinant
Start simplifying the determinant by breaking it down: consider the polynomial terms explicitly and look for linear combinations. Simplifying this will give you the linear form \(Ax + B\). After carrying out the expansion, you will typically end up with a polynomial expression.
4Step 4: Identify Coefficients A and B
After obtaining the simplified form, equate the terms with\(Ax + B\): - Compare the coefficient of \(x\) with that obtained from expanding a test-determinant structured for \(A\). - Compare the constant term with the test-determinant structured for \(B\). This involves comparing the specific coefficients in your expanded polynomial.
5Step 5: Determine the Matrices for A and B
Given options, validate which matrices correspond to \(A\) and \(B\) through direct evaluation of the determinant for each option. Consider consistency in polynomial terms resulting from determinant expansion versus these matrix determinants.
Key Concepts
3x3 Matrix DeterminantsPolynomial ExpansionCoefficient ComparisonRule of Sarrus
3x3 Matrix Determinants
A 3x3 matrix determinant is a number that can be calculated from the elements of a 3x3 square matrix. This value is crucial in solving linear equations, finding area and volume in geometry, and determining matrix invertibility. For our given matrix, it contained polynomials in terms of \(x\) as its elements. Calculating the determinant of such a matrix involves complex expansion and simplification techniques.
To evaluate the determinant, you perform algebraic operations that combine the products of diagonals and sub-matrices (known as minors) into a single number. The elements' arrangement in a 3x3 matrix allows you to use specific methods like the Rule of Sarrus or cofactor expansion, making it easier to derive the formula \(Ax + B\), as seen in the given exercise.
To evaluate the determinant, you perform algebraic operations that combine the products of diagonals and sub-matrices (known as minors) into a single number. The elements' arrangement in a 3x3 matrix allows you to use specific methods like the Rule of Sarrus or cofactor expansion, making it easier to derive the formula \(Ax + B\), as seen in the given exercise.
Polynomial Expansion
Polynomial expansion in this context involves expressing the determinant of a matrix containing polynomial elements in its simplest linear form. Here, the given matrix's determinant is expanded to attain a formula expressed as \(Ax + B\), which signifies a first-degree polynomial.
This process requires converting more complicated expressions into a polynomial structure where we differentiate the terms with respect to their degrees. In the exercise, each term within the matrix was simplified and combined with others to derive the polynomial solution. By focusing on the highest degree terms and their coefficients first, and then addressing the constant terms, you eventually simplify everything into a manageable expression.
This process requires converting more complicated expressions into a polynomial structure where we differentiate the terms with respect to their degrees. In the exercise, each term within the matrix was simplified and combined with others to derive the polynomial solution. By focusing on the highest degree terms and their coefficients first, and then addressing the constant terms, you eventually simplify everything into a manageable expression.
Coefficient Comparison
Coefficient comparison is a mathematical technique used to match terms of the same degree across different polynomial expressions. In our exercise, after forming the linear polynomial \(Ax + B\), we needed to extract the specific values for \(A\) and \(B\).
The first step is to identify which part of the polynomial resembles the coefficient of \(x\) (represented by \(A\)) and which corresponds to the constant term (represented by \(B\)). To do this, you carefully compare your expanded determinant with the target form \(Ax + B\), matching terms one-by-one.
This matching requires aligning terms based on their degree. Once aligned, you can directly compare and equate coefficients of like-terms from both the polynomial expansion of the determinant and the linear expression \(Ax + B\). This allows you to solve for the variables \(A\) and \(B\).
The first step is to identify which part of the polynomial resembles the coefficient of \(x\) (represented by \(A\)) and which corresponds to the constant term (represented by \(B\)). To do this, you carefully compare your expanded determinant with the target form \(Ax + B\), matching terms one-by-one.
This matching requires aligning terms based on their degree. Once aligned, you can directly compare and equate coefficients of like-terms from both the polynomial expansion of the determinant and the linear expression \(Ax + B\). This allows you to solve for the variables \(A\) and \(B\).
Rule of Sarrus
The Rule of Sarrus is a straightforward mnemonic for computing the determinant of a 3x3 matrix. It offers a quick way to manage and remember the formula, especially helpful when dealing with matrices that have simpler, non-complex components.
> To use it, you replicate the first two columns of the 3x3 matrix, placing them to the right of the original columns.
> Then, you draw diagonals starting from the top elements down to the bottom, multiply the elements on these diagonals, and take their sum.
> Repeat this diagonal multiplication starting from the bottom up, but this time, subtract the sum of these products.
This technique is particularly advantageous in practical scenarios due to its simplicity and efficiency on 3x3 matrices. However, when dealing with polynomials, individual attention to expanding each term correctly is required for accuracy, as was necessary in our given polynomial matrix determinant problem.
> To use it, you replicate the first two columns of the 3x3 matrix, placing them to the right of the original columns.
> Then, you draw diagonals starting from the top elements down to the bottom, multiply the elements on these diagonals, and take their sum.
> Repeat this diagonal multiplication starting from the bottom up, but this time, subtract the sum of these products.
This technique is particularly advantageous in practical scenarios due to its simplicity and efficiency on 3x3 matrices. However, when dealing with polynomials, individual attention to expanding each term correctly is required for accuracy, as was necessary in our given polynomial matrix determinant problem.
Other exercises in this chapter
Problem 82
The value of \(\lambda\) for which the equations \(x+y-3=0\), \((1+\lambda) x+(2+\lambda) y-8=0, x-(1+\lambda) y+(2+\lambda)=0\) are consistent is (A) 1 (B) \(5
View solution Problem 84
If \(A+B+C=\pi, e^{i \theta}=\cos \theta+i \sin \theta\) and \(z=\left|\begin{array}{lll}e^{2 i A} & e^{-i C} & e^{-i B} \\ e^{-i C} & e^{2 i B} & e^{-i A} \\ e
View solution Problem 86
5\. If \(\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|=0, a \neq b \neq c\), then (A) \(x=0\) if \(b(a+c) \leq a c\)
View solution Problem 87
If \(\left|\begin{array}{ccc}b c-a^{2} & c a-b^{2} & a b-c^{2} \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2}\end{array}\right|=\left
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