Finding Matrix A Using Matrix Equation AX = B

The problem asks us to find the matrix A, given the matrix equation AX = B and the specific matrices X and B. We are provided with:

• Matrix X: \( {\rm{X}} = \left[ {\begin{array}{c} 3&{ - 4}\\ 1&{ - 1} \end{array}} \right] \)
• Matrix B: \( {\rm{B}} = \left[ {\begin{array}{c} 5&2\\ { - 2}&1 \end{array}} \right] \)
• The equation: AX = B

To find matrix A, we can use the concept of matrix inverses. If matrix X has an inverse (\(X^{-1}\)), we can multiply both sides of the equation AX = B by \(X^{-1}\) on the right:

\(AX \cdot X^{-1} = B \cdot X^{-1}\)

Since \(X \cdot X^{-1} = I\) (the identity matrix), and \(AI = A\), the equation simplifies to:

\(A = BX^{-1}\)

So, the first step is to find the inverse of matrix X, denoted as \(X^{-1}\).

Calculating the Inverse of Matrix X

For a 2x2 matrix \( M = \left[ {\begin{array}{c} a&b\\ c&d \end{array}} \right] \), the inverse \(M^{-1}\) is given by the formula:

\( M^{-1} = \frac{1}{\det(M)} \cdot \text{adj}(M) \)

where \(\det(M)\) is the determinant of M, and \(\text{adj}(M)\) is the adjugate (or adjoint) of M.

Step 1: Find the Determinant of X

Matrix X is \( {\rm{X}} = \left[ {\begin{array}{c} 3&{ - 4}\\ 1&{ - 1} \end{array}} \right] \). The determinant of a 2x2 matrix \( \left[ {\begin{array}{c} a&b\\ c&d \end{array}} \right] \) is \(ad - bc\).

\(\det(X) = (3)(-1) - (-4)(1)\)

\(\det(X) = -3 - (-4)\)

\(\det(X) = -3 + 4\)

\(\det(X) = 1\)

Since the determinant is 1 (non-zero), the inverse of matrix X exists.

Step 2: Find the Adjugate of X

For a 2x2 matrix \( \left[ {\begin{array}{c} a&b\\ c&d \end{array}} \right] \), the adjugate is found by swapping the diagonal elements (a and d) and changing the sign of the off-diagonal elements (b and c).

Matrix X is \( \left[ {\begin{array}{c} 3&{ - 4}\\ 1&{ - 1} \end{array}} \right] \). Here, a=3, b=-4, c=1, d=-1.

\(\text{adj}(X) = \left[ {\begin{array}{c} d&{-b}\\ {-c}&a \end{array}} \right] = \left[ {\begin{array}{c} {-1}&{-(-4)}\\ {-1}&3 \end{array}} \right]\)

\(\text{adj}(X) = \left[ {\begin{array}{c} {-1}&4\\ {-1}&3 \end{array}} \right]\)

Step 3: Calculate the Inverse of X

Now, we use the formula \( X^{-1} = \frac{1}{\det(X)} \cdot \text{adj}(X) \).

\( X^{-1} = \frac{1}{1} \cdot \left[ {\begin{array}{c} {-1}&4\\ {-1}&3 \end{array}} \right] \)

\( X^{-1} = \left[ {\begin{array}{c} {-1}&4\\ {-1}&3 \end{array}} \right] \)

Calculating Matrix A by Multiplying B and X Inverse

Now that we have \(X^{-1}\), we can calculate A using \(A = BX^{-1}\).

Matrix B is \( {\rm{B}} = \left[ {\begin{array}{c} 5&2\\ { - 2}&1 \end{array}} \right] \) and \(X^{-1} = \left[ {\begin{array}{c} {-1}&4\\ {-1}&3 \end{array}} \right] \).

\( A = \left[ {\begin{array}{c} 5&2\\ { - 2}&1 \end{array}} \right] \left[ {\begin{array}{c} {-1}&4\\ {-1}&3 \end{array}} \right] \)

To multiply two 2x2 matrices, we perform the following calculations:

Let \( B = \left[ {\begin{array}{c} b_{11}&b_{12}\\ b_{21}&b_{22} \end{array}} \right] \) and \( X^{-1} = \left[ {\begin{array}{c} x_{11}^{-1}&x_{12}^{-1}\\ x_{21}^{-1}&x_{22}^{-1} \end{array}} \right] \). The resulting matrix \( A = \left[ {\begin{array}{c} a_{11}&a_{12}\\ a_{21}&a_{22} \end{array}} \right] \) has elements:

• \(a_{11} = b_{11}x_{11}^{-1} + b_{12}x_{21}^{-1}\)
• \(a_{12} = b_{11}x_{12}^{-1} + b_{12}x_{22}^{-1}\)
• \(a_{21} = b_{21}x_{11}^{-1} + b_{22}x_{21}^{-1}\)
• \(a_{22} = b_{21}x_{12}^{-1} + b_{22}x_{22}^{-1}\)

Applying this to our matrices B and \(X^{-1}\):

• \(a_{11} = (5)(-1) + (2)(-1) = -5 - 2 = -7\)
• \(a_{12} = (5)(4) + (2)(3) = 20 + 6 = 26\)
• \(a_{21} = (-2)(-1) + (1)(-1) = 2 - 1 = 1\)
• \(a_{22} = (-2)(4) + (1)(3) = -8 + 3 = -5\)

So, matrix A is:

\( A = \left[ {\begin{array}{c} { - 7}&{26}\\ 1&{ - 5} \end{array}} \right] \)

Comparing with the Options

Let's compare our calculated matrix A with the given options:

Our calculated matrix A, \( \left[ {\begin{array}{c} { - 7}&{26}\\ 1&{ - 5} \end{array}} \right] \), matches the matrix in Option 1.

Revision Table: Matrix Calculation Steps

Additional Information: Properties of Matrix Operations

Working with matrix equations like AX = B involves fundamental matrix operations and properties:

• Matrix Multiplication: This operation is not commutative, meaning \(AB \neq BA\) in general. In our problem, the order matters: \(A = BX^{-1}\) is correct, but \(A = X^{-1}B\) would be incorrect.
• Matrix Inverse: The inverse of a square matrix M, denoted \(M^{-1}\), exists if and only if its determinant \(\det(M)\) is non-zero. If the determinant is zero, the matrix is singular and has no inverse.
• Identity Matrix (I): The identity matrix acts like the number 1 in scalar multiplication. For any matrix M, \(MI = IM = M\). Also, \(M M^{-1} = M^{-1} M = I\).
• Solving Matrix Equations: Equations like AX=B or XA=B can often be solved using matrix inverses, provided the inverse exists. If AX=B, then \(A = BX^{-1}\). If XA=B, then \(X = BA^{-1}\) (assuming A has an inverse). Note the order of multiplication.

Understanding these properties is crucial for solving matrix problems efficiently and accurately.