Problem 46
Question
If \(x \neq 0, y \neq 0, z \neq 0\) and \(\left|\begin{array}{ccc}1+x & 1 & 1 \\\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right|=0\), then \(x^{-1}+y^{-1}+z^{-1}\) is equal to (A) \(-1\) (B) \(-2\) (C) \(-3\) (D) None of these
Step-by-Step Solution
Verified Answer
None of the given options match; the value is \(-\frac{11}{6}\).
1Step 1: Write the Determinant
Identify the components of the given determinant: \( \begin{vmatrix} 1+x & 1 & 1 \ 1+y & 1+2y & 1 \ 1+z & 1+z & 1+3z \end{vmatrix} = 0 \).
2Step 2: Simplify Using Row Operations
Apply row operations to simplify the determinant. Subtract the first row from the second and third rows to make calculations simpler: \[ R_2 \rightarrow R_2 - R_1 \] and \[ R_3 \rightarrow R_3 - R_1 \]. This results in:\[\begin{vmatrix}1+x & 1 & 1 \y-x & y & 0 \z-x & z & 2z \end{vmatrix}= 0.\]
3Step 3: Expand the Determinant
Compute the determinant by expanding along the first row:\[(1+x)(y \cdot 2z - 0) - (1)(y(z-x)) + (1)(y(z-x)) = 0.\]Simplify to obtain \((1+x)yz = 0\).
4Step 4: Analyze the Simplified Expression
Since none of \(x, y,\) or \(z\) are zero, \((1+x)yz = 0\) implies \(1+x = 0\). This leads to \(x = -1\).
5Step 5: Solve for Individual Values
As we know \(x = -1\), substitute back into the matrix and recompute or logically deduce: for the equation to hold true, similar deductions are applied to solve for \(y = -2\) and \(z = -3\).
6Step 6: Calculate Required Expression
Compute \(x^{-1} + y^{-1} + z^{-1}\):\(\frac{-1}{1} + \frac{-1}{2} + \frac{-1}{3} = -1 - \frac{1}{2} - \frac{1}{3}.\)Continue calculation:\(= -\frac{6}{6} - \frac{3}{6} - \frac{2}{6} = -\frac{11}{6}.\)
7Step 7: Validate Options and Conclude
Compare this result \(-\frac{11}{6}\) with the given options. None of the options match the calculated value.
Key Concepts
Matrix AlgebraRow OperationsInverse Operations
Matrix Algebra
Matrix algebra is a branch of mathematics involving matrices, which are arrays of numbers handled using algebraic operations. Matrices are crucial in solving systems of equations and in transforming geometric data. In this context, a matrix determinant is a special numerical value that is calculated from its elements.
To understand determinants, visualize them as providing a "volume" or "scaling factor" information about a transformation represented by the matrix.
Calculating determinants through matrix algebra helps determine properties like the invertibility of a matrix. If the determinant is zero, as in the given exercise, the matrix is singular, which means it does not have an inverse, indicating dependencies among its rows or columns.
To understand determinants, visualize them as providing a "volume" or "scaling factor" information about a transformation represented by the matrix.
Calculating determinants through matrix algebra helps determine properties like the invertibility of a matrix. If the determinant is zero, as in the given exercise, the matrix is singular, which means it does not have an inverse, indicating dependencies among its rows or columns.
Row Operations
In matrix algebra, row operations are a vital tool for simplifying complex problems, especially when calculating determinants. Row operations allow us to manipulate the rows of a matrix to achieve a simpler equivalent form, making calculations easier.
There are primarily three types of row operations:
There are primarily three types of row operations:
- Swapping two rows
- Multiplying a row by a non-zero scalar
- Adding or subtracting a scalar multiple of one row from another row
Inverse Operations
Inverse operations are key to solving matrix equations and finding solutions to systems of linear equations. For a matrix to have an inverse, its determinant must be non-zero.
In the given problem, the determinant is zero, meaning the matrix does not have an inverse. This indicates that there is a linear dependency among the matrix rows, which prevents the inverse from existing. Solving the problem involves simplifying the determinant's expression and using the resulting equations to solve for the variables.
Recognizing when a zero determinant renders a matrix non-invertible is important to understand system constraints, leading to adjustments in solving the problem by focusing on logical deductions from simplified expressions.
In the given problem, the determinant is zero, meaning the matrix does not have an inverse. This indicates that there is a linear dependency among the matrix rows, which prevents the inverse from existing. Solving the problem involves simplifying the determinant's expression and using the resulting equations to solve for the variables.
Recognizing when a zero determinant renders a matrix non-invertible is important to understand system constraints, leading to adjustments in solving the problem by focusing on logical deductions from simplified expressions.
Other exercises in this chapter
Problem 43
If \(f_{j}=\sum_{i=0}^{2} a_{i j} x^{i}, j=1,2,3\) and if \(f_{j}^{\prime} f_{j}^{\prime \prime}\) denote \(\frac{d f_{j}}{d x}, \frac{d^{2} f_{j}}{d x^{2}}\) r
View solution Problem 45
\((b+c)(y+z)-a x=b-c\), \((c+a)(z+x)-b y=c-a\), \((a+b)(x+y)-c z=a-b\), where \(a+b+c \neq 0\), then \(x=\) (A) \(\frac{c-b}{a+b+c}\) (B) \(\frac{a-c}{a+b+c}\)
View solution Problem 47
If \(2 s=a+b+c\) and \(\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
View solution Problem 48
Let \(\alpha, \beta\) be the roots of the equation \(a x^{2}+b x+c=0\). Let \(s_{n}=\alpha^{n}+\beta^{n}\) for \(n \geq 1\). Then, the value of the determinant
View solution