Linear Algebra

### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2023 Q.58)** **1. Governing Concept & Definition**: - **Topic**: Engineering Mathematics & Matrix Determinants - For the $2 \times 2$ matrix $A = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix}$, we evaluate the determinant of the matrix polynomial: $\det(M) = |A^4 + 3A^2 - 5A + 6I|$ **2. Detailed Step-by-Step Calculation**: - **Step 1: Matrix Powers**: $A = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix}$ $A^2 = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix} = \begin{pmatrix} 1(1)+2(3) & 1(2)+2(5) \\ 3(1)+5(3) & 3(2)+5(5) \end{pmatrix} = \begin{pmatrix} 7 & 12 \\ 18 & 31 \end{pmatrix}$ $A^4 = A^2 \cdot A^2 = \begin{pmatrix} 7 & 12 \\ 18 & 31 \end{pmatrix} \begin{pmatrix} 7 & 12 \\ 18 & 31 \end{pmatrix} = \begin{pmatrix} 7(7)+12(18) & 7(12)+12(31) \\ 18(7)+31(18) & 18(12)+31(31) \end{pmatrix} = \begin{pmatrix} 265 & 456 \\ 684 & 1177 \end{pmatrix}$ - **Step 2: Linear Combination $M = A^4 + 3A^2 - 5A + 6I$**: $M = \begin{pmatrix} 265 & 456 \\ 684 & 1177 \end{pmatrix} + 3\begin{pmatrix} 7 & 12 \\ 18 & 31 \end{pmatrix} - 5\begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix} + 6\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ $M_{11} = 265 + 3(7) - 5(1) + 6 = 265 + 21 - 5 + 6 = 287$ $M_{12} = 456 + 3(12) - 5(2) + 0 = 456 + 36 - 10 = 482$ $M_{21} = 684 + 3(18) - 5(3) + 0 = 684 + 54 - 15 = 723$ $M_{22} = 1177 + 3(31) - 5(5) + 6 = 1177 + 93 - 25 + 6 = 1251$ $M = \begin{pmatrix} 287 & 482 \\ 723 & 1251 \end{pmatrix}$ - **Step 3: Determinant Calculation**: $\det(M) = (287)(1251) - (482)(723) = 359037 - 348486 = 10551$ **3. Final Answer**: - **Calculated Value**: **10551** - **Official GATE Accepted Range**: **10551 to 10551**