Matrices: PYQ Walkthrough (2)

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Question

(CBSE 2024 style.) If A=(2314)A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}, find A−1A^{-1} and use it to solve the system 2x+3y=52x + 3y = 5, x+4y=6x + 4y = 6.

Solution — Step by Step

∣A∣=(2)(4)−(3)(1)=8−3=5|A| = (2)(4) - (3)(1) = 8 - 3 = 5

Since ∣A∣≠0|A| \neq 0, the inverse exists.

For a 2×2 matrix (abcd)\begin{pmatrix}a & b \\ c & d\end{pmatrix}, the adjugate is (d−b−ca)\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}.

So adj(A)=(4−3−12)\text{adj}(A) = \begin{pmatrix}4 & -3 \\ -1 & 2\end{pmatrix}.

A−1=1∣A∣adj(A)=15(4−3−12)=(4/5−3/5−1/52/5)A^{-1} = \frac{1}{|A|}\text{adj}(A) = \frac{1}{5}\begin{pmatrix}4 & -3 \\ -1 & 2\end{pmatrix} = \begin{pmatrix}4/5 & -3/5 \\ -1/5 & 2/5\end{pmatrix}

Write A(xy)=(56)A\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}5\\6\end{pmatrix}. So X=A−1BX = A^{-1} B:

(xy)=15(4−3−12)(56)=15(20−18−5+12)=15(27)\begin{pmatrix}x\\y\end{pmatrix} = \frac{1}{5}\begin{pmatrix}4 & -3 \\ -1 & 2\end{pmatrix}\begin{pmatrix}5\\6\end{pmatrix} = \frac{1}{5}\begin{pmatrix}20 - 18 \\ -5 + 12\end{pmatrix} = \frac{1}{5}\begin{pmatrix}2\\7\end{pmatrix}

So x=2/5x = 2/5 and y=7/5y = 7/5.

Why This Works

The matrix equation AX=BAX = B has a unique solution X=A−1BX = A^{-1}B provided AA is invertible (i.e., ∣A∣≠0|A| \neq 0). For a 2×2 matrix, the inverse formula is fast and worth memorizing — flip diagonal entries, negate off-diagonal, divide by determinant.

This method is the matrix analog of solving a single equation ax=bax = b by x=b/ax = b/a. The matrix inverse plays the role of “1/a” in the matrix world.

For 2×2 matrices, the entire inverse calculation takes 30 seconds. For 3×3, you have to compute the cofactor matrix and adjugate — 3-4 minutes. Always check if the system can be solved by simpler methods (substitution, elimination) for small cases.

Alternative Method

Use Cramer’s rule directly:

x=∣Ax∣∣A∣=∣5364∣5=20−185=25x = \frac{|A_x|}{|A|} = \frac{\begin{vmatrix}5 & 3\\6 & 4\end{vmatrix}}{5} = \frac{20-18}{5} = \frac{2}{5}

y=∣Ay∣∣A∣=∣2516∣5=12−55=75y = \frac{|A_y|}{|A|} = \frac{\begin{vmatrix}2 & 5\\1 & 6\end{vmatrix}}{5} = \frac{12-5}{5} = \frac{7}{5}

Same answer in one step.

Students forget to divide by ∣A∣|A| at the end. Always remember: A−1=1∣A∣adj(A)A^{-1} = \dfrac{1}{|A|} \text{adj}(A) — the determinant is in the denominator. If you forget, you compute the adjugate instead of the inverse.

Final answer: A−1=15(4−3−12)A^{-1} = \dfrac{1}{5}\begin{pmatrix}4 & -3 \\ -1 & 2\end{pmatrix}, x=2/5x = 2/5, y=7/5y = 7/5.

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