Problem 8
Question
Evaluate each determinant. $$ \left|\begin{array}{rr} {\frac{1}{5}} & {\frac{1}{6}} \\ {-6} & {5} \end{array}\right| $$
Step-by-Step Solution
Verified Answer
The determinant of the given matrix is 2.
1Step 1: Identify the elements of the matrix
The matrix is given as: \( \[ [\frac{1}{5}, \frac{1}{6}], [-6, 5] ] \]. So, a = \(\frac{1}{5}\), b = \(\frac{1}{6}\), c = -6 and d = 5.
2Step 2: Substitute the values into the formula
Substitute these values into the determinant formula: \(det(A) = a*d - b*c = (\frac{1}{5}*5) - (\frac{1}{6}* -6)\).
3Step 3: Simplify the equation
Once we substitute the values into the equation, we can simplify to get the final result. In this case, the equation simplifies to \(1 - (-1) = 2\).
Key Concepts
Determinant of a MatrixMatrix ArithmeticProperties of Determinants
Determinant of a Matrix
The determinant of a matrix is a special number that can be calculated from its elements. In particular, for a 2x2 matrix, the determinant can be found using the simple formula: \( \det(A) = ad - bc \), where \( a, b, c, d \) are the elements of the matrix arranged as \([a, b], [c, d]\). It's a bit like cross-multiplying the elements and then subtracting.For the given matrix \(\left|\begin{array}{cc}\frac{1}{5} & \frac{1}{6} \-6 & 5\end{array}\right|\), you calculate the determinant by multiplying \(\frac{1}{5}\) times 5 and subtract the product of \(\frac{1}{6}\) and -6. Determinants have wide applications in linear algebra, particularly in solving systems of linear equations, finding the inverse of a matrix, and in understanding linear transformations.
Matrix Arithmetic
Matrix arithmetic encompasses operations such as addition, subtraction, and multiplication of matrices. Unlike regular numbers, though, these operations have unique rules when it comes to matrices. For example, matrices must have the same dimensions to be added or subtracted from one another. Moreover, for multiplication to be possible, the number of columns in the first matrix must match the number of rows in the second. The operation conducted in our exercise, however, is not an operation between two matrices but rather a computed value derived from a single matrix—its determinant.\[\text{Remember the key points for matrix operations:}\]
- For addition or subtraction, matrices must be of the same size.
- For multiplication, the inner dimensions must match.
- The product of two matrices is not commutative—\(AB eq BA\) in general.
- The determinant is a scalar value representing a special property of a matrix.
Properties of Determinants
Determinants are not just numbers you calculate; they have significant properties that influence the behavior of matrices. For instance, changing a row or column with another within the matrix affects the determinant sign. Here are some critical properties to keep in mind:
- The determinant of a matrix changes sign if two rows (or columns) are swapped.
- When multiplying a row (or column) of a matrix by a scalar, the determinant is also multiplied by that scalar.
- The determinant of the identity matrix is always 1, regardless of its size.
- If two rows (or columns) are identical or one is a multiple of the other, the determinant is 0 (indicating the matrix is non-invertible).
- The determinant of a product of matrices equals the product of their determinants (when the matrices are square and of the same size).
Other exercises in this chapter
Problem 8
Write the augmented matrix for each system of linear equations. $$ \left\\{\begin{aligned} 4 w+7 x-8 y+z &=3 \\ 5 x+y &=5 \\ w-x-y &=17 \\ 2 w-2 x+11 y &=4 \end
View solution Problem 8
Find values for the variables so that the matrices in each exercise are equal. $$ \left[\begin{array}{rr} {x} & {y+3} \\ {2 z} & {8} \end{array}\right]=\left[\b
View solution Problem 8
Find the products AB and BA to determine.whether \(B\) is the multiplicative inverse of \(A\). $$ A=\left[\begin{array}{rrr} {-2} & {1} & {-1} \\ {-5} & {2} & {
View solution Problem 8
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} x+y-10 z &=-4 \\ x &-7
View solution