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Find matrix B, if matrix A= [(1,5),(1,2)...

Find matrix B, if matrix `A= [(1,5),(1,2)]`, matrix `C= [(2),(1)] and AB= 3C`

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To find matrix B given that \( AB = 3C \), where \( A = \begin{pmatrix} 1 & 5 \\ 1 & 2 \end{pmatrix} \) and \( C = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \), we can follow these steps: ### Step 1: Determine the order of matrix B Matrix A is a \( 2 \times 2 \) matrix, and matrix C is a \( 2 \times 1 \) matrix. Since \( AB = 3C \), we need to find the order of matrix B. - The product \( AB \) will have the same number of rows as A and the same number of columns as B. - Given that C is \( 2 \times 1 \), \( 3C \) will also be \( 2 \times 1 \). - Therefore, the number of rows in B must be 2, and the number of columns must be 1. Thus, matrix B is \( 2 \times 1 \). ### Step 2: Define matrix B Let matrix B be represented as: \[ B = \begin{pmatrix} x \\ y \end{pmatrix} \] ### Step 3: Set up the equation \( AB = 3C \) Now we can set up the equation: \[ AB = \begin{pmatrix} 1 & 5 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \cdot x + 5 \cdot y \\ 1 \cdot x + 2 \cdot y \end{pmatrix} = \begin{pmatrix} x + 5y \\ x + 2y \end{pmatrix} \] And we know that: \[ 3C = 3 \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix} \] ### Step 4: Set up the system of equations From the equation \( AB = 3C \), we have: \[ \begin{pmatrix} x + 5y \\ x + 2y \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix} \] This gives us two equations: 1. \( x + 5y = 6 \) (Equation 1) 2. \( x + 2y = 3 \) (Equation 2) ### Step 5: Solve the system of equations We can solve these equations using the elimination method. From Equation 2: \[ x + 2y = 3 \implies x = 3 - 2y \] Substituting \( x \) in Equation 1: \[ (3 - 2y) + 5y = 6 \] This simplifies to: \[ 3 + 3y = 6 \] Subtracting 3 from both sides: \[ 3y = 3 \implies y = 1 \] Now substitute \( y = 1 \) back into Equation 2 to find \( x \): \[ x + 2(1) = 3 \implies x + 2 = 3 \implies x = 1 \] ### Step 6: Write the final matrix B Thus, we have: \[ B = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \] ### Final Answer The matrix B is: \[ B = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \]
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