We are given two matrices, $A$ and $B$, and we need to find the matrix product $BA$. The matrix $A$ is a $3 \times 3$ matrix: $A = \begin{bmatrix} 2 & 1 & 0 \\ 7 & 2 & 8 \\ 1 & 0 & 4 \end{bmatrix}$ The matrix $B$ is a $3 \times 3$ matrix: $B = \begin{bmatrix} 1 & 1 & 4 \\ 8 & 3 & 1 \\ 3 & 2 & 0 \end{bmatrix}$

AlgebraMatrix MultiplicationLinear Algebra
2025/3/8

1. Problem Description

We are given two matrices, AA and BB, and we need to find the matrix product BABA.
The matrix AA is a 3×33 \times 3 matrix:
A=[210728104]A = \begin{bmatrix} 2 & 1 & 0 \\ 7 & 2 & 8 \\ 1 & 0 & 4 \end{bmatrix}
The matrix BB is a 3×33 \times 3 matrix:
B=[114831320]B = \begin{bmatrix} 1 & 1 & 4 \\ 8 & 3 & 1 \\ 3 & 2 & 0 \end{bmatrix}

2. Solution Steps

To find the matrix product BABA, we multiply the rows of BB by the columns of AA.
The resulting matrix will be a 3×33 \times 3 matrix.
BA=[114831320][210728104]BA = \begin{bmatrix} 1 & 1 & 4 \\ 8 & 3 & 1 \\ 3 & 2 & 0 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ 7 & 2 & 8 \\ 1 & 0 & 4 \end{bmatrix}
BA11=(1)(2)+(1)(7)+(4)(1)=2+7+4=13BA_{11} = (1)(2) + (1)(7) + (4)(1) = 2 + 7 + 4 = 13
BA12=(1)(1)+(1)(2)+(4)(0)=1+2+0=3BA_{12} = (1)(1) + (1)(2) + (4)(0) = 1 + 2 + 0 = 3
BA13=(1)(0)+(1)(8)+(4)(4)=0+8+16=24BA_{13} = (1)(0) + (1)(8) + (4)(4) = 0 + 8 + 16 = 24
BA21=(8)(2)+(3)(7)+(1)(1)=16+21+1=38BA_{21} = (8)(2) + (3)(7) + (1)(1) = 16 + 21 + 1 = 38
BA22=(8)(1)+(3)(2)+(1)(0)=8+6+0=14BA_{22} = (8)(1) + (3)(2) + (1)(0) = 8 + 6 + 0 = 14
BA23=(8)(0)+(3)(8)+(1)(4)=0+24+4=28BA_{23} = (8)(0) + (3)(8) + (1)(4) = 0 + 24 + 4 = 28
BA31=(3)(2)+(2)(7)+(0)(1)=6+14+0=20BA_{31} = (3)(2) + (2)(7) + (0)(1) = 6 + 14 + 0 = 20
BA32=(3)(1)+(2)(2)+(0)(0)=3+4+0=7BA_{32} = (3)(1) + (2)(2) + (0)(0) = 3 + 4 + 0 = 7
BA33=(3)(0)+(2)(8)+(0)(4)=0+16+0=16BA_{33} = (3)(0) + (2)(8) + (0)(4) = 0 + 16 + 0 = 16
BA=[1332438142820716]BA = \begin{bmatrix} 13 & 3 & 24 \\ 38 & 14 & 28 \\ 20 & 7 & 16 \end{bmatrix}

3. Final Answer

[1332438142820716]\begin{bmatrix} 13 & 3 & 24 \\ 38 & 14 & 28 \\ 20 & 7 & 16 \end{bmatrix}

Related problems in "Algebra"

We are asked to solve four problems: (a) Expand and simplify the expression $6(2y-3) - 5(y+1)$. (b) ...

Algebraic SimplificationExponentsDifference of SquaresEquationsFactorization
2025/7/22

We are asked to simplify the expression $(a^{-2}b^3)^{-2}$, writing the answer with positive powers.

ExponentsSimplificationPower Rules
2025/7/22

A group of children bought a certain number of apples. If each apple is cut into 4 equal pieces and ...

System of EquationsWord Problem
2025/7/21

The problem asks to simplify the expression $\frac{x+1}{y} \div \frac{2(x+1)}{x}$.

Algebraic simplificationFractionsVariable expressions
2025/7/21

A group of children bought some apples. If each apple is divided into 4 equal pieces and 1 piece is ...

Linear EquationsSystems of EquationsWord Problem
2025/7/21

We need to find the value of the expression $6 + \log_b(\frac{1}{b^3}) + \log_b(\sqrt{b})$.

LogarithmsExponentsSimplification
2025/7/20

We need to solve the following equation for $y$: $\frac{y-1}{3} = \frac{2y+1}{5}$

Linear EquationsSolving EquationsAlgebraic Manipulation
2025/7/20

We are given the equation $\frac{3x}{2} = \frac{x-1}{7}$ and need to solve for $x$.

Linear EquationsSolving EquationsAlgebraic Manipulation
2025/7/20

The problem is to solve the equation $\frac{m}{5} = \frac{m+5}{10}$ for $m$.

Linear EquationsSolving EquationsVariables
2025/7/20

The problem is to solve the linear equation $\frac{2x}{5} = 3x + 1$ for $x$.

Linear EquationsEquation SolvingVariable Isolation
2025/7/20