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: State the invertibility test.
A square matrix is invertible exactly when its determinant is non-zero; a zero determinant means it is singular.
Step 2: Test A.
$|A|=\begin{vmatrix}2&3\\10&15\end{vmatrix}=2\cdot15-3\cdot10=30-30=0$, so A is not invertible.
Step 3: Test B.
In B the first row $(1,2,3)$ equals the third row $(1,2,3)$. Two identical rows force the determinant to be 0, so B is not invertible.
Step 4: Test C.
In C, row1 + row2 $=(1+3,2+4,3+5)=(4,6,8)$, which is row3. The rows are dependent, so $|C|=0$ and C is not invertible.
Step 5: Test D.
Expand along the first row: $|D|=2\begin{vmatrix}1&0\\4&5\end{vmatrix}-4\begin{vmatrix}1&0\\1&5\end{vmatrix}+2\begin{vmatrix}1&1\\1&4\end{vmatrix}=2(5)-4(5)+2(3)=10-20+6=-4$.
Step 6: Conclude.
Only D has a non-zero determinant ($-4$), so only D is invertible, matching option (4).
\[ \boxed{\text{only } D} \]