Problem 230
Question
For what value of \(k\) the matrix \(A=\left[\begin{array}{cc}2 & k \\ 3 & 5\end{array}\right]\) has no inverse.
Step-by-Step Solution
Verified Answer
The value of 'k' for which the matrix does not have an inverse is \(k = \frac{2*5}{3} = \frac{10}{3}\).
1Step 1: Write out the matrix and its determinant
The matrix \[A=\begin{pmatrix}2 & k \ 3 & 5\end{pmatrix}\] has the determinant \(2*5 - 3*k\).
2Step 2: Set the determinant equal to zero
A matrix is non-invertible if its determinant equals zero. So, we set up the following equation \(2*5 - 3*k = 0\)
3Step 3: Solve the equation for k
To find the value of 'k' that makes the equation true, solve the equation \(2*5 - 3*k = 0\). We subtract \(2*5\) from both sides resulting in \(-3*k = -(2*5)\). Now dividing by \(-3\) from both sides, we get \(k = \frac{2*5}{3}\)
Key Concepts
DeterminantsNon-Invertible MatricesAlgebraic Equations
Determinants
In linear algebra, the determinant is a scalar value that can be computed from the elements of a square matrix. It provides important information about the matrix, including whether it is invertible or not. The determinant of a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) can be calculated using the formula \( ad - bc \).
For our specific exercise, we are given the matrix \( A = \begin{pmatrix} 2 & k \ 3 & 5 \end{pmatrix} \) and the determinant is determined by the formula \( 2 \cdot 5 - 3 \cdot k \). This value is crucial because if the determinant of the matrix is zero, the matrix does not have an inverse and is thus termed 'non-invertible' or 'singular'. This property is not just a mathematical curiosity; it has practical implications in various fields such as physics, engineering, and computer science, where matrices are used to solve systems of linear equations and to represent linear transformations.
For our specific exercise, we are given the matrix \( A = \begin{pmatrix} 2 & k \ 3 & 5 \end{pmatrix} \) and the determinant is determined by the formula \( 2 \cdot 5 - 3 \cdot k \). This value is crucial because if the determinant of the matrix is zero, the matrix does not have an inverse and is thus termed 'non-invertible' or 'singular'. This property is not just a mathematical curiosity; it has practical implications in various fields such as physics, engineering, and computer science, where matrices are used to solve systems of linear equations and to represent linear transformations.
Non-Invertible Matrices
Non-invertible matrices, also known as singular matrices, are square matrices that do not have an inverse. A square matrix is invertible if there exists another matrix that when multiplied by the original matrix, results in the identity matrix. However, not all matrices meet this criterion. For a matrix to be invertible, its determinant must be non-zero.
The concept from the exercise illustrates this well. The determinant \( 2 \cdot 5 - 3 \cdot k \) set to zero indicates that the matrix \( A \) will not have an inverse when \( k = \frac{10}{3} \). Therefore, the value of \( k \) is crucial to determining the invertibility of the matrix. In real-world applications, non-invertible matrices indicate situations where systems of equations have either no solution or an infinite number of solutions, which is why it's essential to identify them.
The concept from the exercise illustrates this well. The determinant \( 2 \cdot 5 - 3 \cdot k \) set to zero indicates that the matrix \( A \) will not have an inverse when \( k = \frac{10}{3} \). Therefore, the value of \( k \) is crucial to determining the invertibility of the matrix. In real-world applications, non-invertible matrices indicate situations where systems of equations have either no solution or an infinite number of solutions, which is why it's essential to identify them.
Algebraic Equations
Algebraic equations are mathematical statements that demonstrate the equality between two expressions. They consist of variables and constants and can range from simple linear equations to more complex polynomial equations. Solving algebraic equations involves finding the value or values of the variables that make the equation true.
In our matrix inversion problem, the algebraic equation \( 2 \cdot 5 - 3 \cdot k = 0 \) was used to find the value of \( k \) that would result in a zero determinant, and therefore a non-invertible matrix. This equation is a linear one, with \( k \) being the variable. Algebraic equations are the basis for solving a wide array of problems in mathematics, and understanding how to manipulate and solve these equations is fundamental for students.
In our matrix inversion problem, the algebraic equation \( 2 \cdot 5 - 3 \cdot k = 0 \) was used to find the value of \( k \) that would result in a zero determinant, and therefore a non-invertible matrix. This equation is a linear one, with \( k \) being the variable. Algebraic equations are the basis for solving a wide array of problems in mathematics, and understanding how to manipulate and solve these equations is fundamental for students.
Other exercises in this chapter
Problem 228
If \(A=\left[\begin{array}{ll}5 & 2 \\ 3 & 1\end{array}\right]\), then find \(A^{-1}\).
View solution Problem 229
Find \(A^{-1}\), if the matrix \(A\) is given by \(A=\left[\begin{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]\)
View solution Problem 231
If \(A=\left[\begin{array}{cc}2 x & 0 \\ x & x\end{array}\right]\) and \(A^{-1}=\left[\begin{array}{cc}1 & 0 \\ -1 & 2\end{array}\right]\), then find the value
View solution Problem 232
If \(A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]\) be such that \(A^{-1}=k A\), then find the value of \(k\).
View solution