Let A=[3−6−12] Construct a 2×2 matrix B such that AB is the zero matrix. Use two different nonzero columns for B.

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liannemdh · Accepted Answer

Let [3−6−12] We need to construct a 2×2 matrix B such that AB is the zero matrix that is AB=0 Let B=[abcd] Find the product of AB AB=[3−6−12][abcd] =[3a+(−6)c3b+(−6)d(−1)a+2c(−1)b+2d] Since AB=0, then =[3a−6c3b−6d−a+2c−b+2d]=[0000] Two matrices are equal if their corresponding inputs are equal, so we get a system of equations 3a−6c=0 3b−6d=0 −a+2c=0 −b+2d=0 From the first equation we get a=2c Replace this in the third, we get −2c+2c=0⇒0=0 Thus, this is true. From the second we get b=2d, replace this in the fourth equation to get −2d+2d=0⇒0=0 Therefore, we get a=2c and b=2c We choose some arbitrary values for a and b and get in the first case a=2⇒c=1 b=4⇒d=2 Thus, we get B=[1224] Find the product of AB AB=[3−6−12][2412] =[3⋅2+(−6)⋅13⋅4+(−6)⋅2(−1)⋅2+2⋅1(−1)⋅4+2⋅2] =[0000] In the second case we get a=4⇒c=2 b=8⇒d=4 Find the product of AB =[3−6−12][4824] =[3⋅4+(−6)⋅23⋅8+(−6)⋅4(−1)⋅4+2⋅2(−1)⋅8+2⋅4] =[0000] Hence, [2412][4824]

fudzisako · Accepted Answer

To find a matrix B such that AB is the zero matrix, we can set up the equation AB = 0 and solve for B.Let A=[3−6−12].We want to find a matrix B such that AB = 0. Let&#039;s represent B as [xyzw], where x, y, z, and w are the entries of B.Now, we can write the matrix equation AB = 0 as:A[xyzw]=[3−6−12][xyzw]=[3x−6z3y−6w−x+2z−y+2w]=[0000].We need to find nonzero values for x, y, z, and w that satisfy this equation. Let&#039;s choose two different nonzero columns for B:Column 1: [11],Column 2: [23].Substituting these values into the equation AB = 0, we have:A[1213]=[3−6−12][1213]=[3(1)−6(1)3(2)−6(3)−1(1)+2(1)−1(2)+2(3)]=[0000].Therefore, the matrix B that satisfies AB = 0 with the chosen columns is:B=[1213].

Jazz Frenia · Accepted Answer

Result:B=[2110]Solution:First, let&#039;s find 𝐛1 such that A𝐛1=0.We have the equation A𝐛1=0. Substituting the given matrix A and the column vector 𝐛1, we get:[3−6−12][b11b21]=[00]Simplifying the matrix multiplication:[3b11−6b21−b11+2b21]=[00]This gives us the following system of equations:3b11−6b21=0−b11+2b21=0To solve this system, we can express b11 and b21 in terms of a parameter. Let&#039;s choose b21=1 as a parameter.From the first equation, we have 3b11−6=0, which gives b11=2. Therefore, 𝐛1=[21].Now, let&#039;s find 𝐛2 such that A𝐛2=0, and 𝐛2 is linearly independent from 𝐛1.We can choose 𝐛2=[10].Thus, the matrix B is:B=[2110]Therefore, AB is the zero matrix.

Andre BalkonE · Accepted Answer

Step 1:Given matrix A: A=[3−6−12]To construct matrix B, we need to find two different nonzero columns for B. Let&#039;s denote the columns of B as B1 and B2.We know that for AB to be the zero matrix, each column of A must be multiplied by the corresponding column of B resulting in a column of zeros.Let&#039;s start by finding B1:Multiplying matrix A with B1 should result in the zero vector:AB1=[3−6−12][b11b21]=[00]Solving this equation, we get:3b11−6b21=0 (equation 1)−b11+2b21=0 (equation 2)From equation 1, we can express b21 in terms of b11:b21=12b11Step 2:Now, let&#039;s find B2:Multiplying matrix A with B2 should result in the zero vector:AB2=[3−6−12][b12b22]=[00]Solving this equation, we get:3b12−6b22=0 (equation 3)−b12+2b22=0 (equation 4)From equation 3, we can express b22 in terms of b12:b22=12b12Step 3:Now, we can construct matrix B using the obtained values:B=[b11b12b21b22]=[b11b1212b1112b12]We have successfully constructed matrix B such that AB is the zero matrix.

Let\begin{bmatrix}3&-6\\-1&2\end{bmatrix}Construct a 2\times2

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