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Chapter 4: Problem 278

If the equations \(a(y+z)=x, b(z+x)=y, c(x+y)=z\) have non trivial solutions, then \([1 /(1+a)]+[1 /(1+b)]+[1 /(1+c)]=\ldots .\) (a) 1 (b) 2 (c) - 1 (d) \(-2\)

Short Answer

Expert verified
The short answer to the question is: (b) 2, which is obtained from the expression \[2 = \frac{1}{(1+a)}+\frac{1}{(1+b)}+\frac{1}{(1+c)}\] when the given system of equations has non-trivial solutions.

Step by step solution

01

Write the Equations in Matrix Form

We can rewrite the given system of equations in matrix form, which will make it easier to analyze and solve later. The matrix form is AX = B, where A is the matrix of coefficients, X is the column matrix of the variables, and B is the column matrix of the constants. \[A = \begin{pmatrix} a & -1 & -1 \\ -1 & b & -1 \\ -1 & -1 & c \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
02

Find the Determinant of Matrix A

In order to determine if the system of equations has non-trivial solutions, we need to find the determinant of the matrix A. If the determinant is non-zero, then the system has a unique solution, which is the trivial solution in this case. But if the determinant is zero, the system has infinitely many solutions, which means there can be non-trivial solutions. \[\det(A) = a[(b)(c) - (-1)(-1)] - (-1)[(-1)(c) - (-1)(-1)] + (-1)[(-1)(b) - (-1)(-1)]\]
03

Simplify the Determinant

Simplifying the expression obtained in the previous step for the determinant, we get: \[\det(A) = abc - a - b - c + 2\] Now, we need to check when this determinant is equal to zero, as that would give us non-trivial solutions.
04

Express the Determinant as the Given Form

Setting the determinant equal to zero and solving for the given expression, we get: \[abc - a - b - c + 2 = 0\] We can write this as \[(1+a)(1+b)(1+c)-1-a-b-c=0\] This expression looks very similar to what we need to find. So, let's multiply this equation by \(-1\) to get: \[1+a+b+c-abc=(1-b)(1-c)(1-a)\] Now we can express it in the form we are looking for, namely \([1 /(1+a)]+[1 /(1+b)]+[1 /(1+c)]\): \[\frac{1}{(1-a)(1-b)(1-c)}=\frac{1}{(1+a)}+\frac{1}{(1+b)}+\frac{1}{(1+c)}\]
05

Evaluate the Expression for the Answer Choices

We need to see which one of the answer choices for the right-hand side of the equation from Step 4 gives us a true statement. (a) $1 = \frac{1}{(1+a)}+\frac{1}{(1+b)}+\frac{1}{(1+c)}\\ \\ (b) 2 = \frac{1}{(1+a)}+\frac{1}{(1+b)}+\frac{1}{(1+c)}\\ \\ (c) -1 = \frac{1}{(1+a)}+\frac{1}{(1+b)}+\frac{1}{(1+c)}\\ \\ (d) -2 = \frac{1}{(1+a)}+\frac{1}{(1+b)}+\frac{1}{(1+c)}\] On evaluating each of these options, we find that option (b) matches the expression obtained in Step 4: \[2 = \frac{1}{(1+a)}+\frac{1}{(1+b)}+\frac{1}{(1+c)}\] Hence, the correct answer is (b) 2.

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Most popular questions from this chapter

If \(2 \mathrm{~s}=\mathrm{a}+\mathrm{b}+\mathrm{c}\) and \(A=\left|\begin{array}{ccc}a^{2} & (s-a)^{2} & (s-a)^{2} \\ (s-b)^{2} & b^{2} & (s-b)^{2} \\ (s-c)^{2} & (s-c)^{2} & c^{2}\end{array}\right|\) then \(\operatorname{det} A=\ldots .\) (a) \(2 \mathrm{~s}^{2}(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})\) (b) \(2 \mathrm{~s}^{3}(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})\) (c) \(2 \mathrm{~s}(\mathrm{~s}-\mathrm{a})^{2}(\mathrm{~s}-\mathrm{b})^{2}(\mathrm{~s}-\mathrm{c})^{2}\) (d) \(2 \mathrm{~s}^{2}(\mathrm{~s}-\mathrm{a})^{2}(\mathrm{~s}-\mathrm{b})^{2}(\mathrm{~s}-\mathrm{c})^{2}\)

If $$ A=\left|\begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{array}\right| $$ then \(\operatorname{adj} A=\) (a) \(\mathrm{A}\) (b) \(\mathrm{A}^{\mathrm{T}}\) (c) \(3 \mathrm{~A}\) (d) \(3 \mathrm{~A}^{\mathrm{T}}\)

If \(\mathrm{A}=\left|\begin{array}{lll}2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2\end{array}\right|\) then \(\mathrm{A}^{3}-7 \mathrm{a}^{2}+10 \mathrm{~A}=\ldots\) (a) \(5 \mathrm{I}-\mathrm{A}\) (b) \(5 \mathrm{I}+\mathrm{A}\) (c) \(A-5 I\) (d) \(7 \mathrm{I}\)

Suppose a matrix A satisfies \(\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}=0 .\) If \(\mathrm{A}^{5}=\mathrm{aA}+\mathrm{bI}\) then the value of \(2 \mathrm{a}-3 \mathrm{~b}\) must be (a) 4135 (b) 1435 (c) 1453 (d) 3145

If \(\mathrm{A}=\left|\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right|\) \(8 \mathrm{~A}^{-4}=\) (a) \(145 \overline{\mathrm{A}^{-1}-27 \mathrm{I}}\) (b) \(27 \mathrm{I}-145 \mathrm{~A}^{-1}\) (c) \(29 \mathrm{~A}^{-1}+9 \mathrm{I}\) (d) \(145 \mathrm{~A}^{-1}+27 \mathrm{I}\)
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