For two invertible matrices A and B of suitable orders, the value of $(AB)^{-1}$ is:

Question

The integer value printed by the ANSI-C program given below is ________.


int funcp(){
    static int x = 1;
    x++;
    return x;
}

int main(){
    int x,y;
    x = funcp();
    y = funcp() + x;
    printf("%d\n", (x+y));
    return 0;
}

Answer 1

Step 1: Instead of relying purely on the abstract rule, verify it using a concrete pair of small invertible matrices, since a correct general formula must also hold for any specific numeric example.
Step 2: Let $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$, both invertible since their determinants are 1 and 2.
Step 3: Compute the product first: $AB = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}$. Using the standard 2x2 inverse formula, $(AB)^{-1} = \begin{pmatrix} 0.5 & -0.5 \\ 0 & 1 \end{pmatrix}$.
Step 4: Now compute $B^{-1}A^{-1}$ separately. $B^{-1} = \begin{pmatrix} 0.5 & 0 \\ 0 & 1 \end{pmatrix}$ and $A^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$, so $B^{-1}A^{-1} = \begin{pmatrix} 0.5 & -0.5 \\ 0 & 1 \end{pmatrix}$.
Step 5: This matches $(AB)^{-1}$ exactly, while $A^{-1}B^{-1} = \begin{pmatrix} 0.5 & -1 \\ 0 & 1 \end{pmatrix}$ is a different matrix, confirming that only the reversed order gives the true inverse.\[\boxed{(AB)^{-1} = B^{-1}A^{-1}}\]

For two invertible matrices A and B of suitable orders, the value of \((AB)^{-1}\) is:

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The Correct Option is B

Solution and Explanation

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