Given the matrices A = \begin{pmatrix} x & 1 \\ 4 & 2 \end{pmatrix} and B = \begin{pmatrix… — JAMB UTME

Given the matrices $A = \begin{pmatrix} x & 1 \\ 4 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 \\ 1 & x \end{pmatrix}$ , what is the sum of all real values of $x$ for which the matrix $C = AB - B^T$ is singular?

$\frac{5}{2}$ Answer
B
$\frac{13}{2}$
C
$1$
D
$-\frac{5}{2}$

Answer

The sum of all real values of

x

\frac{5}{2}

Evaluating

AB

yields

\begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix}

and subtracting

B^T = \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix}

gives matrix

C = \begin{pmatrix} 2x - 1 & 4x - 1 \\ 7 & x + 12 \end{pmatrix}

. Setting

\det(C) = (2x - 1)(x + 12) - 7(4x - 1) = 0

results in the quadratic equation

2x^2 - 5x - 5 = 0

. Since the discriminant

\Delta = 65 > 0

, both roots are real, and their sum is given by

-\frac{b}{a} = \frac{5}{2}

Step-by-Step Solution

Compute the matrix product

AB

AB = \begin{pmatrix} x & 1 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 1 & x \end{pmatrix} = \begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix}

Matrix multiplication requires taking the dot product of rows of the first matrix with columns of the second matrix.

Find the transpose of matrix

B

, denoted

B^T

B^T = \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix}

The transpose of a matrix is formed by swapping its rows and columns.

Compute matrix

C = AB - B^T

C = \begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix} = \begin{pmatrix} 2x - 1 & 4x - 1 \\ 7 & x + 12 \end{pmatrix}

Perform element-wise subtraction of matrix

B^T

from matrix

AB

Set the determinant of matrix

C

to zero for singularity

\det(C) = (2x - 1)(x + 12) - 7(4x - 1) = 2x^2 + 23x - 12 - (28x - 7) = 2x^2 - 5x - 5 = 0

A matrix is singular if and only if its determinant is zero.

Find the sum of the real roots of the quadratic equation

2x^2 - 5x - 5 = 0

Discriminant

\Delta = (-5)^2 - 4(2)(-5) = 65 > 0

. By Vieta's formula,

x_1 + x_2 = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}

Since the discriminant is positive, two distinct real roots exist, and their sum is given by

-\frac{b}{a}

Key Concept

Singular matrix condition, matrix multiplication, transpose operations, and Vieta's formulas.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic