Question

10. (a) Consider the following matrices: A = ( 2 ) B = (3) and C = (-3) w = Find the det(A). [1] (ii) Is the matrix A singular? Justify your answer. [1] (iii) Suppose the AB = C. Determine the values of x and y. [3] (iv) Is it possible to find the matrix product BA? State why or why not. [2]

Answer 1

i) We have to find the determinant of A.

We can do this as:

$\det(A)=|\begin{bmatrix}{3} & {6} \\ {1} & {-2}\end{bmatrix}|=3*(-2)-6*1=-6-6=-12$

ii) We have to find if the matrix A is singular.

Singular matrix have determinant equal to 0.

This is not the case for A, as its determinant is -12. Then, A is not a singular matrix.

iii) We have to find the values of x and y so that AB = C.

We have to write the matrix multiplication and we will obtain a system of linear equations:

We can now solve the system of equations by adding 3 times the second equation to the first equation:

$\begin{gathered} 3(x-2y)+(3x+6y)=3(-3)+(-3) \\ 3x-6y+3x+6y=-9-3 \\ 6x+0y=-12 \\ x=\frac{-12}{6} \\ x=-2 \end{gathered}$

We can now use the second equation to find the value of y:

$\begin{gathered} x-2y=-3 \\ x+3=2y \\ y=\frac{x+3}{2} \\ y=\frac{-2+3}{2} \\ y=\frac{1}{2} \end{gathered}$

The values are x = -2 and y = 1/2.

iv) When we want to multiply two matrices, the required condition is that the number of columns of the matrix on the left is equal the number of rows of the matrix on the right.

In the case of A(2x2) and B(2x1), when we do A*B this condition is satisfied.

But when we try to multiply BA, the number of columns of B is not equal to the number of rows of A, so the matrix mulitplication is not possible.

Answer:

i) det(A) = -12

ii) A is not singular because singular matrices have determinant equal to zero.

iii) x = -2 and y = 1/2.

iv) Is not possible because the number of columns of the first matrix has to be equal to the number of rows of the second matrix.