Question

Which of the following matrices are invertible?
$$\text{A} = \left[\begin{array}{cc} 2 & 3 \\ 10 & 15 \end{array}\right], \quad \text{B} = \left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & -1 & 3 \\ 1 & 2 & 3 \end{array}\right], \quad \text{C} = \left[\begin{array}{ccc} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 4 & 6 & 8 \end{array}\right], \quad \text{D} = \left[\begin{array}{ccc} 2 & 4 & 2 \\ 1 & 1 & 0 \\ 1 & 4 & 5 \end{array}\right]$$

• A. both A and B
• B. only C
• C. only A
• D. only D

Hint:

Always scan the rows and columns for easy dependencies before doing calculation math! In matrix B, $R_1 = R_3$ instantly tells you $|\text{B}|=0$. In matrix C, noticing $R_1 + R_2 = R_3$ eliminates it in a glance, saving you from expanding large determinants!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given four matrices (A, B, C, D) and need to determine which of them are invertible.

Step 2: Key Formula or Approach:
A square matrix is invertible if and only if it is non-singular, meaning its determinant is non-zero ($|\text{M}| \neq 0$). If the determinant equals zero ($|\text{M}| = 0$), the matrix is singular and cannot be inverted.

Step 3: Detailed Explanation:
Let's evaluate the determinants of each matrix systematically:
1. For Matrix A:
$$|\text{A}| = \begin{vmatrix} 2 & 3 \\ 10 & 15 \end{vmatrix} = (2 \times 15) - (3 \times 10) = 30 - 30 = 0$$ Since $|\text{A}| = 0$, A is singular (not invertible).
2. For Matrix B:
Observe that the first row $R_1 = [1, 2, 3]$ and the third row $R_3 = [1, 2, 3]$ are completely identical. By properties of determinants, if any two rows or columns are identical, the determinant is automatically zero:
$$|\text{B}| = 0 \quad (\text{not invertible})$$ 3. For Matrix C:
Observe the row operations: adding the first row to the second row gives $R_1 + R_2 = [1+3, 2+4, 3+5] = [4, 6, 8]$, which is exactly equal to the third row $R_3$. Since the rows are linearly dependent ($R_3 = R_1 + R_2$), the determinant must be zero:
$$|\text{C}| = 0 \quad (\text{not invertible})$$ 4. For Matrix D:
Let's expand the determinant along the first row:
$$|\text{D}| = \begin{vmatrix} 2 & 4 & 2 \\ 1 & 1 & 0 \\ 1 & 4 & 5 \end{vmatrix} = 2\begin{vmatrix} 1 & 0 \\ 4 & 5 \end{vmatrix} - 4\begin{vmatrix} 1 & 0 \\ 1 & 5 \end{vmatrix} + 2\begin{vmatrix} 1 & 1 \\ 1 & 4 \end{vmatrix}$$ $$|\text{D}| = 2(5 - 0) - 4(5 - 0) + 2(4 - 1)$$ $$|\text{D}| = 2(5) - 4(5) + 2(3) = 10 - 20 + 6 = -4$$ Since $|\text{D}| = -4 \neq 0$, matrix D is non-singular and therefore invertible.

Step 4: Final Answer:
Only matrix D is invertible, which corresponds to option (D).