Questions: Find a 2 × 2 matrix A and a 2 × 1 column matrix B for which AX=B has no solution Which of the following, if substituted into AX=B, produces no solution? A. A=[1 2; 2 4], B=[3; 14] B. A=[1 2; 2 3], B=[3; 5] C. A=[1 2; 5 3], B=[4; 5] D. A=[3 4; 8 12], B=[7; 3]

Find a 2 × 2 matrix A and a 2 × 1 column matrix B for which AX=B has no solution

Which of the following, if substituted into AX=B, produces no solution?
A. A=[1 2; 2 4], B=[3; 14]
B. A=[1 2; 2 3], B=[3; 5]
C. A=[1 2; 5 3], B=[4; 5]
D. A=[3 4; 8 12], B=[7; 3]

Transcript text: Find a $2 \times 2$ matrix $A$ and a $2 \times 1$ column matrix $B$ for which $A X=B$ has no solution Which of the following, it substituted into $\mathrm{AX}=\mathrm{B}$, produces no solution? A. $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 4\end{array}\right], B=\left[\begin{array}{r}3 \\ 14\end{array}\right]$ B. $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right], B=\left[\begin{array}{l}3 \\ 5\end{array}\right]$ C. $A=\left[\begin{array}{ll}1 & 2 \\ 5 & 3\end{array}\right], B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$ D. $A=\left[\begin{array}{rr}3 & 4 \\ 8 & 12\end{array}\right], B=\left[\begin{array}{l}7 \\ 3\end{array}\right]$

Solution

Solution Steps

To determine which combination of matrices $ A $ and $ B $ results in the system $ AX = B $ having no solution, we need to check if the matrix $ A $ is singular (i.e., its determinant is zero) and if the augmented matrix $[A|B]$ is inconsistent. This can be done by calculating the determinant of $ A $ and checking the rank of $ A $ and the augmented matrix $[A|B]$.

Step 1: Define the Matrices

We are given four options for matrices $ A $ and $ B $:

\[ A_1 = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}, \quad B_1 = \begin{bmatrix} 3 \\ 14 \end{bmatrix} \]

\[ A_2 = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, \quad B_2 = \begin{bmatrix} 3 \\ 5 \end{bmatrix} \]

\[ A_3 = \begin{bmatrix} 1 & 2 \\ 5 & 3 \end{bmatrix}, \quad B_3 = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \]

\[ A_4 = \begin{bmatrix} 3 & 4 \\ 8 & 12 \end{bmatrix}, \quad B_4 = \begin{bmatrix} 7 \\ 3 \end{bmatrix} \]

Step 2: Check Determinant of $ A $

To determine if the system $ AX = B $ has no solution, we first check if the determinant of $ A $ is zero (i.e., $ A $ is singular).

\[ \text{det}(A_1) = \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} = 1 \cdot 4 - 2 \cdot 2 = 0 \]

\[ \text{det}(A_2) = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = 1 \cdot 3 - 2 \cdot 2 = -1 \]

\[ \text{det}(A_3) = \begin{vmatrix} 1 & 2 \\ 5 & 3 \end{vmatrix} = 1 \cdot 3 - 2 \cdot 5 = -7 \]

\[ \text{det}(A_4) = \begin{vmatrix} 3 & 4 \\ 8 & 12 \end{vmatrix} = 3 \cdot 12 - 4 \cdot 8 = 36 - 32 = 4 \]

Step 3: Check Rank of $ A $ and Augmented Matrix $[A|B]$

Since $ \text{det}(A_1) = 0 $, $ A_1 $ is singular. We need to check the rank of $ A_1 $ and the augmented matrix $[A_1|B_1]$.

\[ \text{rank}(A_1) = 1, \quad \text{rank}([A_1|B_1]) = 2 \]

Since the ranks are different, the system $ A_1 X = B_1 $ has no solution.

For the other matrices, since their determinants are non-zero, they are non-singular, and thus the systems $ A_2 X = B_2 $, $ A_3 X = B_3 $, and $ A_4 X = B_4 $ have solutions.

Final Answer

The option that produces no solution is:

\[ \boxed{\text{A}} \]

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