Understanding Singular Matrices and Determinants

A matrix is considered singular if its determinant is equal to zero. The determinant is a scalar value that can be calculated from the elements of a square matrix. For a 2x2 matrix $\begin{bmatrix} p & q \\ r & s \end{bmatrix}$, the determinant is calculated as $ps - qr$.

In this problem, we are given three singular matrices, A, B, and C. This means the determinant of each matrix is zero. We can use this property to find the unknown values $a$, $b$, and $c$, and then calculate the product $abc$.

Step-by-Step Solution to Find a, b, and c

Finding the Value of 'a' using Matrix A

Matrix A is given by:

\[ A = \begin{bmatrix} 1 & 4 \\ 3 & 2a \end{bmatrix} \]

Since matrix A is singular, its determinant $|A|$ must be 0.

The determinant of A is calculated as:

\[ |A| = (1)(2a) - (4)(3) = 2a - 12 \]

Setting the determinant to zero:

\[ 2a - 12 = 0 \] \[ 2a = 12 \] \[ a = \frac{12}{2} \] \[ a = 6 \]

So, the value of $a$ is 6.

Finding the Value of 'b' using Matrix B

Matrix B is given by:

\[ B = \begin{bmatrix} 3b & 5 \\ a & 2 \end{bmatrix} \]

We have found that $a=6$. Substituting this value into matrix B:

\[ B = \begin{bmatrix} 3b & 5 \\ 6 & 2 \end{bmatrix} \]

Since matrix B is singular, its determinant $|B|$ must be 0.

The determinant of B is calculated as:

\[ |B| = (3b)(2) - (5)(6) = 6b - 30 \]

Setting the determinant to zero:

\[ 6b - 30 = 0 \] \[ 6b = 30 \] \[ b = \frac{30}{6} \] \[ b = 5 \]

So, the value of $b$ is 5.

Finding the Value of 'c' using Matrix C

Matrix C is given by:

\[ C = \begin{bmatrix} a + b + c & c + 1 \\ a + c & c \end{bmatrix} \]

We have found that $a=6$ and $b=5$. Substituting these values into matrix C:

The element in the top-left corner is $a+b+c = 6+5+c = 11+c$.

The element in the bottom-left corner is $a+c = 6+c$.

So, matrix C becomes:

\[ C = \begin{bmatrix} 11 + c & c + 1 \\ 6 + c & c \end{bmatrix} \]

Since matrix C is singular, its determinant $|C|$ must be 0.

The determinant of C is calculated as:

\[ |C| = (11 + c)(c) - (c + 1)(6 + c) \]

Setting the determinant to zero:

\[ (11 + c)(c) - (c + 1)(6 + c) = 0 \]

Expand the terms:

\[ (11c + c^2) - (6c + c^2 + 6 + c) = 0 \] \[ 11c + c^2 - (c^2 + 7c + 6) = 0 \]

Remove the parenthesis, remembering to distribute the negative sign:

\[ 11c + c^2 - c^2 - 7c - 6 = 0 \]

Combine like terms:

\[ (11c - 7c) + (c^2 - c^2) - 6 = 0 \] \[ 4c + 0 - 6 = 0 \] \[ 4c - 6 = 0 \] \[ 4c = 6 \] \[ c = \frac{6}{4} \] \[ c = \frac{3}{2} \]

So, the value of $c$ is $\frac{3}{2}$.

Calculating the Value of abc

We have found the values of $a$, $b$, and $c$:

• $a = 6$
• $b = 5$
• $c = \frac{3}{2}$

Now, we calculate the product $abc$:

\[ abc = (6)(5)\left(\frac{3}{2}\right) \] \[ abc = (30)\left(\frac{3}{2}\right) \] \[ abc = \frac{30 \times 3}{2} \] \[ abc = \frac{90}{2} \] \[ abc = 45 \]

The value of $abc$ is 45.

Summary of Calculations

Here is a summary of the values found for $a$, $b$, and $c$, and the final product:

This step-by-step process using the singular matrix property helps us find the required values and solve the problem.

Revision Table: Key Concepts

Additional Information: Properties of Determinants

Determinants have several useful properties in linear algebra:

• A square matrix is singular if and only if its determinant is zero. This was the core concept used here.
• The determinant of a product of matrices is the product of their determinants: $|AB| = |A||B|$.
• The determinant of the identity matrix is 1.
• The determinant of a matrix is the same as the determinant of its transpose: $|A| = |A^T|$.
• If a matrix has a row or column of zeros, its determinant is zero.
• If a matrix has two identical rows or columns, its determinant is zero.

These properties are fundamental in solving various matrix problems and understanding the characteristics of linear transformations.