The matrix A has x rows and x + 5 columns. The matrix B has y rows and 11 - y columns. Both AB and BA exist. What are the values of x and y respectively?
3 and 8
Understanding Matrix Dimensions for Multiplication
This problem asks us to find the dimensions of two matrices, A and B, given that both matrix products AB and BA exist. The existence of matrix products depends directly on the dimensions of the matrices being multiplied.
Matrix Dimensions and Multiplication Rules
Let's first define the dimensions of the given matrices:
- Matrix A has x rows and x + 5 columns. Its dimension is \(A_{x \times (x+5)}\).
- Matrix B has y rows and 11 - y columns. Its dimension is \(B_{y \times (11-y)}\).
For two matrices to be multiplied, the number of columns in the first matrix must be equal to the number of rows in the second matrix.
Condition for AB to Exist
For the matrix product AB to exist, the number of columns in matrix A must be equal to the number of rows in matrix B.
- Number of columns in A = x + 5
- Number of rows in B = y
So, the condition is:
\(x + 5 = y\) (Equation 1)
Condition for BA to Exist
For the matrix product BA to exist, the number of columns in matrix B must be equal to the number of rows in matrix A.
- Number of columns in B = 11 - y
- Number of rows in A = x
So, the condition is:
\(11 - y = x\) (Equation 2)
Solving for x and y using Matrix Multiplication Conditions
We now have a system of two linear equations with two variables:
- \(x + 5 = y\)
- \(11 - y = x\)
We can solve this system using substitution. From Equation 1, we already have y expressed in terms of x: \(y = x + 5\).
Substitute this expression for y into Equation 2:
\(11 - (x + 5) = x\)
Now, simplify and solve for x:
\(11 - x - 5 = x\)
\(6 - x = x\)
Add x to both sides:
\(6 = x + x\)
\(6 = 2x\)
Divide both sides by 2:
\(x = \frac{6}{2}\)
\(x = 3\)
Now that we have the value of x, substitute it back into Equation 1 (or Equation 2) to find y. Using Equation 1:
\(y = x + 5\)
\(y = 3 + 5\)
\(y = 8\)
Verifying the Matrix Dimensions
We found \(x = 3\) and \(y = 8\).
- Matrix A has \(x = 3\) rows and \(x + 5 = 3 + 5 = 8\) columns. So, A is \(3 \times 8\).
- Matrix B has \(y = 8\) rows and \(11 - y = 11 - 8 = 3\) columns. So, B is \(8 \times 3\).
Let's check the matrix multiplication conditions with these dimensions:
- For AB: A is \(3 \times 8\), B is \(8 \times 3\). The number of columns in A (8) equals the number of rows in B (8). AB exists and will be a \(3 \times 3\) matrix.
- For BA: B is \(8 \times 3\), A is \(3 \times 8\). The number of columns in B (3) equals the number of rows in A (3). BA exists and will be an \(8 \times 8\) matrix.
Since both AB and BA exist with \(x=3\) and \(y=8\), these are the correct values.
Conclusion on Matrix Dimensions
The values of x and y that satisfy the conditions for both matrix products AB and BA to exist are \(x=3\) and \(y=8\).
Revision Table: Key Matrix Concepts
| Concept | Explanation | Condition for Multiplication |
|---|---|---|
| Matrix Dimension | Represented as rows \(\times\) columns (e.g., \(m \times n\)). | Not directly involved in the condition, but defines the number of rows and columns. |
| Matrix Multiplication AB | Product of matrix A and matrix B. | Number of columns in A must equal the number of rows in B. If A is \(m \times n\) and B is \(p \times q\), then \(n = p\). The resulting matrix AB will be \(m \times q\). |
| Matrix Multiplication BA | Product of matrix B and matrix A. | Number of columns in B must equal the number of rows in A. If B is \(p \times q\) and A is \(m \times n\), then \(q = m\). The resulting matrix BA will be \(p \times n\). |
Additional Information: Solving Simultaneous Equations for Matrix Problems
Problems involving matrix dimensions often lead to systems of linear equations. Solving these systems accurately is crucial.
- Substitution Method: Solve one equation for one variable, then substitute that expression into the other equation. This reduces the system to a single equation with one variable.
- Elimination Method: Multiply equations by constants so that when the equations are added or subtracted, one variable is eliminated, leaving a single equation with one variable.
- In this problem, the substitution method was straightforward because Equation 1 already gave us an expression for y in terms of x.
Understanding how matrix dimensions affect operations like multiplication is fundamental in linear algebra.
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