If \( A \operatorname{adj} A = \begin{bmatrix} -5 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{bmatrix} \), then the value of \( |A| \) is:
-5
Understanding the Matrix Problem: Finding the Determinant \(|A|\)
The problem asks us to find the value of the determinant of a matrix \( A \), denoted by \( |A| \), given the equation involving the matrix \( A \) and its adjoint \( \operatorname{adj} A \).
The given equation is:
\( A \operatorname{adj} A = \begin{bmatrix} -5 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{bmatrix} \)
Key Property of Adjoint Matrix and Determinant
A fundamental property of square matrices relates the matrix, its adjoint, and its determinant. For any square matrix \( A \) of order \( n \), the following relationship holds:
\( A (\operatorname{adj} A) = (\operatorname{adj} A) A = |A| I_n \)
where \( |A| \) is the determinant of \( A \) and \( I_n \) is the identity matrix of order \( n \).
Analyzing the Given Matrix Equation
First, let's look at the matrix on the right-hand side of the given equation:
\( \begin{bmatrix} -5 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{bmatrix} \)
This is a diagonal matrix. We can factor out the scalar \( -5 \) from this matrix:
\( \begin{bmatrix} -5 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{bmatrix} = -5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
The matrix \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) is the identity matrix of order 3, denoted by \( I_3 \). This tells us that the original matrix \( A \) must be a 3x3 matrix, as the identity matrix \( I \) in the property \( A (\operatorname{adj} A) = |A| I \) must be of the same order as \( A \).
So, the given equation becomes:
\( A \operatorname{adj} A = -5 I_3 \)
Solving for the Determinant \(|A|\)
Now we can compare this equation with the fundamental property:
\( A (\operatorname{adj} A) = |A| I_n \)
Substituting \( n=3 \), we have:
\( A (\operatorname{adj} A) = |A| I_3 \)
By comparing \( A \operatorname{adj} A = -5 I_3 \) and \( A \operatorname{adj} A = |A| I_3 \), we can directly equate the scalar coefficients multiplying the identity matrix \( I_3 \).
Therefore, we get:
\( |A| I_3 = -5 I_3 \)
Since \( I_3 \) is a non-zero matrix, we can conclude that the scalar coefficients must be equal.
\( |A| = -5 \)
Conclusion on the Determinant Value
Based on the property \( A \operatorname{adj} A = |A| I \) and the given equation, the determinant of matrix \( A \) is found to be \( -5 \).
Revision Table: Key Matrix Concepts
| Concept | Description | Property/Formula |
|---|---|---|
| Determinant of a Matrix \(|A|\) | A scalar value that can be computed for a square matrix. It provides information about the matrix, such as whether it is invertible. | For 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), \( |A| = ad - bc \). For higher orders, involves cofactor expansion. |
| Adjoint of a Matrix \( \operatorname{adj} A \) | The transpose of the cofactor matrix of \( A \). | Defined for square matrices. Used in finding the inverse of a matrix. |
| Identity Matrix \( I \) | A square matrix with ones on the main diagonal and zeros elsewhere. It behaves like the number 1 in matrix multiplication. | \( A I = I A = A \) for any matrix \( A \) compatible for multiplication. |
| Relation between \( A \), \( \operatorname{adj} A \), and \( |A| \) | The product of a matrix and its adjoint (in either order) is equal to the determinant of the matrix multiplied by the identity matrix. | \( A (\operatorname{adj} A) = (\operatorname{adj} A) A = |A| I \) |
Additional Information: Adjoint Matrix Properties and Inverse
The adjoint matrix is closely related to the inverse of a matrix. If a square matrix \( A \) is invertible (i.e., \( |A| \neq 0 \)), its inverse \( A^{-1} \) can be calculated using the adjoint matrix and the determinant:
\( A^{-1} = \frac{1}{|A|} \operatorname{adj} A \)
This formula comes directly from the property \( A (\operatorname{adj} A) = |A| I \). If \( |A| \neq 0 \), we can multiply both sides by \( \frac{1}{|A|} \):
\( \frac{1}{|A|} A (\operatorname{adj} A) = \frac{1}{|A|} (|A| I) \)
\( A \left( \frac{1}{|A|} \operatorname{adj} A \right) = I \)
By definition of matrix inverse, \( A A^{-1} = I \). Comparing this with the equation above, we see that \( A^{-1} = \frac{1}{|A|} \operatorname{adj} A \).
In our specific problem, since \( |A| = -5 \) (which is not zero), the matrix \( A \) is invertible. If \( |A| \) were 0, the matrix \( A \) would be singular and would not have an inverse.
Similar Questions
If \( B \) is a non-singular \( 4 \times 4 \) matrix and \( A \) is its adjoint such that \( |A| = 125 \), then \( |B| \) is:
Value of determinant
\[ \begin{vmatrix} a - b & b - c & c - a \\ b - c & c - a & a - b \\ c - a & a - b & b - c \end{vmatrix} \] is:
If \( A \) is a square matrix of order 3 such that \( |2 \operatorname{adj} A| = 288 \), then the value of \( |A| \) is:
If \( A \) is a square matrix of order 3 and \( |A| = -3 \), then the value of \( |2AA^T| \) is:
Important Questions from Determinants
An even number is the determinant of which of the following matrices?
(A) \(\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}\)
(B) \(\begin{bmatrix} 13 & -1 \\ -1 & 15 \end{bmatrix}\)
(C) \(\begin{bmatrix} 16 & -1 \\ -11 & 15 \end{bmatrix}\)
(D) \(\begin{bmatrix} 6 & -12 \\ 11 & 15 \end{bmatrix}\)
Choose the correct answer from the options given below:
Two pipes A and B together can fill a tank in 40 minutes. Pipe A is twice as fast as pipe B. Pipe A alone can fill the tank in :
There are 6 cards numbered 1 to 6, one number on one card. Two cards are drawn at random without replacement.
Let X denote the sum of the numbers on the two cards drawn.
Then P(X > 3) is:
A random variable X has the following probability distribution:
X | -2 | -1 | 0 | 1 | 2 -------------------------------------------- P(X) | 0.2 | 0.1 | 0.3 | 0.2 | 0.2
The variance of X will be:
The angle between two lines whose direction ratios are proportional to \( (\sqrt{3} - 1) \), \( (-\sqrt{3} - 1) \), and -4 is:
