Problem 47
Question
CHALLENGE For which values of \(a, b, c,\) and \(d\) will \(A=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]=A^{-1} ?\)
Step-by-Step Solution
Verified Answer
Not possible with given condition.
1Step 1: Understand the Problem
We need to determine the values of \(a\), \(b\), \(c\), and \(d\) such that matrix \(A\) is equal to its inverse. This occurs when \(A \cdot A^{-1} = I\), where \(I\) is the identity matrix.
2Step 2: Determine the Condition for the Inverse
A 2x2 matrix \(A\) has an inverse if its determinant is not zero. The determinant of \(A\) is \(ad - bc\). Thus, we need \(ad - bc eq 0\).
Key Concepts
2x2 MatrixDeterminantIdentity MatrixInverse Matrix
2x2 Matrix
A 2x2 matrix is simply a square matrix with two rows and two columns. This means it is composed of four elements arranged in a grid. For example:
- First row: elements \(a\) and \(b\)
- Second row: elements \(c\) and \(d\)
Determinant
The determinant is a special number that can be calculated from a square matrix. For a 2x2 matrix like:\[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]the determinant is calculated using the formula \(ad - bc\).
- It is crucial because a matrix only has an inverse if its determinant is non-zero.
- A zero determinant indicates that the matrix cannot be inverted, meaning the matrix is singular.
Identity Matrix
The identity matrix is a special kind of matrix in which all the elements in the leading diagonal are ones, and all other elements are zeros. For a 2x2 identity matrix:\[\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]
- The identity matrix acts like "1" in matrix multiplication. It doesn't change the other matrix it is multiplied with.
- Multiplying any matrix by the identity matrix results in the same matrix.
Inverse Matrix
An inverse matrix is a matrix that, when multiplied by the original matrix, results in the identity matrix. For a 2x2 matrix \(A\) with elements \(a, b, c,\) and \(d\), it has an inverse if and only if the determinant \(ad - bc\) is not zero.
- The formula to find the inverse is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]
- It is vital to remember that not all matrices have inverses. Only those with a non-zero determinant are invertible.
- The inverse matrix is analogous to the reciprocal of numbers in arithmetic.
Other exercises in this chapter
Problem 46
Find the value of each expression. $$ \frac{1}{2}(34) $$
View solution Problem 47
Find each product, if possible. \(\left[\begin{array}{rr}{5} & {-4} \\ {8} & {3}\end{array}\right] \cdot\left[\begin{array}{l}{5} \\ {1}\end{array}\right]\)
View solution Problem 47
ACT/SAT Triangle \(A B C\) has vertices with coordinates \(A(-4,2), B(-4,-3)\) and \(C(3,-2) .\) After a dilation, triangle \(A^{\prime} B^{\prime} C^{\prime}\)
View solution Problem 47
Perform the indicated matrix operations. If the matrix does not exist, write impossible. 3\(\left[\begin{array}{rr}{4} & {-2} \\ {-1} & {7}\end{array}\right]\)
View solution