Question

Find \(A B\) and \(B A\) given $$A=\left(\begin{array}{ll}1 & 2 \\\3 & 6\end{array}\right), \quad B=\left(\begin{array}{cr}10 & 4 \\\\-5 & -2\end{array}\right)$$ Observe that \(\mathrm{AB}\) is the null matrix; if we call it \(0,\) then \(\mathrm{AB}=0,\) but neither \(\mathrm{A}\) nor B is 0. Show that A is singular.

Short answer

AB = 0, BA = \begin{pmatrix} 22 & 44 \ -11 & -22 \end{pmatrix}, A is singular.

Step-by-step solution

Define Matrices

Identify matrices A and B. Here we have: \[ A = \begin{pmatrix} 1 & 2 \ 3 & 6 \end{pmatrix}, \ B = \begin{pmatrix} 10 & 4 \ -5 & -2 \end{pmatrix} \]

Calculate AB

Multiply matrix A by matrix B:\[ AB = \begin{pmatrix} 1 & 2 \ 3 & 6 \end{pmatrix} \begin{pmatrix} 10 & 4 \ -5 & -2 \end{pmatrix} \]

Determine Each Entry of AB

Compute each entry of the AB matrix: \[ AB_{11} = (1 \times 10) + (2 \times -5) = 10 - 10 = 0 \] \[ AB_{12} = (1 \times 4) + (2 \times -2) = 4 - 4 = 0 \] \[ AB_{21} = (3 \times 10) + (6 \times -5) = 30 - 30 = 0 \] \[ AB_{22} = (3 \times 4) + (6 \times -2) = 12 - 12 = 0 \] Therefore, \[ AB = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \]

Calculate BA

Multiply matrix B by matrix A:\[ BA = \begin{pmatrix} 10 & 4 \ -5 & -2 \end{pmatrix} \begin{pmatrix} 1 & 2 \ 3 & 6 \end{pmatrix} \]

Determine Each Entry of BA

Compute each entry of the BA matrix: \[ BA_{11} = (10 \times 1) + (4 \times 3) = 10 + 12 = 22 \] \[ BA_{12} = (10 \times 2) + (4 \times 6) = 20 + 24 = 44 \] \[ BA_{21} = (-5 \times 1) + (-2 \times 3) = -5 - 6 = -11 \] \[ BA_{22} = (-5 \times 2) + (-2 \times 6) = -10 - 12 = -22 \] Therefore, \[ BA = \begin{pmatrix} 22 & 44 \ -11 & -22 \end{pmatrix} \]

Show A is Singular

A matrix is singular if its determinant is zero. Calculate the determinant of A: \[ \text{det}(A) = (1 \times 6) - (2 \times 3) = 6 - 6 = 0 \] Since determinant of A is zero, matrix A is singular.

Key concepts

Singular Matrix

In linear algebra, a matrix is called singular if it does not have an inverse. This characteristic is determined by the matrix's determinant. If the determinant is zero, the matrix is singular.
For example, consider matrix A: \[ A = \begin{pmatrix} 1 & 2 \ 3 & 6 \end{pmatrix} \]
To check if it's singular, we compute its determinant: \[ \text{det}(A) = (1 \times 6) - (2 \times 3) = 6 - 6 = 0 \]
Since the determinant is zero, matrix A is indeed singular.
Singular matrices are important because they signify a system of linear equations that either has no solutions or has infinitely many solutions.
Always remember: if you get a determinant of zero, you're dealing with a singular matrix!

Null Matrix

A null matrix, or zero matrix, is a matrix in which all elements are zero.
For example, a 2x2 null matrix looks like this: \[ 0 = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \]
In the given exercise, when we multiplied matrix A by matrix B, we obtained the null matrix: \[ AB = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \]
This is an interesting case because neither matrix A nor B is a null matrix, but their product is. This typically suggests that at least one of the matrices is singular because it transforms another matrix into a null matrix through multiplication.
Understanding null matrices is crucial for solving systems of equations and related linear algebra concepts.

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from its elements.
For a 2x2 matrix, the determinant is computed as follows: \[ \text{det}(A) = a_{11}a_{22} - a_{12}a_{21} \]
For the matrix: \[ A = \begin{pmatrix} 1 & 2 \ 3 & 6 \end{pmatrix} \]
We calculate the determinant: \[ \text{det}(A) = (1 \times 6) - (2 \times 3) = 6 - 6 = 0 \]
Because the determinant is zero, matrix A is singular.
Determinants are crucial because they provide information about the matrix. They tell us whether a matrix is invertible, and by extension, whether a system of linear equations has a unique solution.