Problem 1
Question
Show that \(B\) is the inverse of \(A\). \(A=\left[\begin{array}{ll}7 & 4 \\ 5 & 3\end{array}\right], B=\left[\begin{array}{rr}3 & -4 \\ -5 & 7\end{array}\right]\)
Step-by-Step Solution
Verified Answer
Yes, matrix \(B\) is the inverse of matrix \(A\) since both the product \(A*B\) and \(B*A\) result in the identity matrix.
1Step 1: Understand the concept of inverse matrix
A matrix \(B\) is said to be the inverse of a matrix \(A\) if and only if the product of \(A\) and \(B\) and the product of \(B\) and \(A\) both give the identity matrix. The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere.
2Step 2: Perform matrix multiplication A*B
Multiply matrix \(A\) and \(B\) together. First, multiply the elements of the rows of the first matrix by the elements of the columns of the second matrix and sum them up. The resulting matrix would be a 2x2 matrix given by: \ \[ A*B= \left[\begin{array}{ll} 7*3+4*(-5) & 7*(-4)+4*7 \ 5*3+3*(-5) & 5*(-4)+3*7 \end{array}\right] = \left[\begin{array}{rr} 1 & 0 \ 0 & 1 \end{array}\right] \]
3Step 3: Perform matrix multiplication B*A
Repeat this process but change the order of multiplication: multiply matrix \(B\) by matrix \(A\). Matrix multiplication is not commutative, but for inverse matrices, the result should be the identity matrix irrespective of the order of multiplication. The resulting matrix would be a 2x2 matrix given by: \ \[ B*A= \left[\begin{array}{ll} 3*7+(-4)*5 & 3*4+(-4)*3 \ (-5)*7+7*5 & (-5)*4+7*3 \end{array}\right] = \left[\begin{array}{rr} 1 & 0 \ 0 & 1 \end{array}\right] \]
4Step 4: Conclusion
Since the matrix product when \(A\) is multiplied by \(B\) and \(B\) is multiplied by \(A\) both give the identity matrix, we can conclude that matrix \(B\) is indeed the inverse of matrix \(A\).
Key Concepts
Matrix MultiplicationIdentity MatrixCommutative Property
Matrix Multiplication
Matrix multiplication might seem tricky at first, but once you understand the process, it becomes quite manageable. It involves taking the rows of the first matrix and multiplying them by the columns of the second matrix. Let's break this down:
When multiplying two matrices, say matrix \( A \) with dimensions \( m \times n \) and matrix \( B \) with dimensions \( n \times p \), the resulting matrix will have dimensions \( m \times p \). Here’s how to perform this operation:
Remember, matrix multiplication is not commutative, which means \( AB \) does not always equal \( BA \); thus, the order of multiplication matters.
When multiplying two matrices, say matrix \( A \) with dimensions \( m \times n \) and matrix \( B \) with dimensions \( n \times p \), the resulting matrix will have dimensions \( m \times p \). Here’s how to perform this operation:
- For each element in the resulting matrix, calculate the sum of products of corresponding elements from the row of the first matrix and the column of the second matrix.
- Repeat this for each row of the first matrix and each column of the second matrix.
Remember, matrix multiplication is not commutative, which means \( AB \) does not always equal \( BA \); thus, the order of multiplication matters.
Identity Matrix
The identity matrix is a fundamental concept in linear algebra, acting much like the number 1 in multiplication. It's a square matrix with ones on the diagonal and zeros elsewhere, which doesn't change a matrix when it is multiplied.
For example, consider a 2x2 identity matrix:
\[ I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]
For example, consider a 2x2 identity matrix:
\[ I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]
- When you multiply any 2x2 matrix \( A \) with this identity matrix, the result is the matrix \( A \) itself.
- This property holds true for any size of square matrices.
Commutative Property
The commutative property is a well-known property in arithmetic, but unlike numbers, matrix multiplication does not follow the commutative property. This may initially surprise students who are new to matrices.
For matrices, the order of multiplication is critical:
Understanding why matrix multiplication isn’t commutative is essential for anyone diving deeper into linear algebra, as neglecting this rule could lead to incorrect solutions in assignments or mathematical proofs.
For matrices, the order of multiplication is critical:
- \( A \times B \) may yield a different result than \( B \times A \).
- Simply put, swapping the order of matrices will generally produce different products.
Understanding why matrix multiplication isn’t commutative is essential for anyone diving deeper into linear algebra, as neglecting this rule could lead to incorrect solutions in assignments or mathematical proofs.
Other exercises in this chapter
Problem 1
Find the determinant of the matrix. $$ [-5] $$
View solution Problem 1
Find \(x\) and \(y\). $$ \left[\begin{array}{rr} 4 & x \\ -1 & y \end{array}\right]=\left[\begin{array}{rr} 4 & -3 \\ -1 & 2 \end{array}\right] $$
View solution Problem 1
Determine the order of the matrix. $$ \left[\begin{array}{rrr} 0 & -3 & 0 \\ 9 & 2 & -7 \end{array}\right] $$
View solution Problem 2
Find the determinant of the matrix. $$ [6] $$
View solution