Q.Examine the consistency of the following system of equations: $5x - y + 4z = 5$ $2x + 3y + 5z = 2$ $5x - 2y + 6z = -1$

Write the system as $AX=B$ and solve with $X=A^{-1}B$.

Determinant. Expanding along row 1,

so a unique solution exists.

Adjoint and inverse.

Solve. …

The idea

The matrix method packs the whole system into one equation $AX=B$, where $A$ holds the coefficients, $X$ the unknowns and $B$ the constants. If $\det A\neq0$ then $A^{-1}$ exists and multiplying on the left gives $X=A^{-1}B$ — the three values at once.

Set up

Step 1 — Determinant

Expanding along the first row,

Since $\det A=51\neq0$, $A$ is invertible and the solution is unique.

Step 2 — Cofactors and adjoint

Computing the nine cofactors gives the cofactor matrix

and the adjoint is its transpose:

$\operatorname{adj}A=\begin{bmatrix}28&-2&-17\\13&10&-17\\-19&5&17\end{bmatrix}.$ …

Method: Solving a 3×3 System via the Inverse (Adjoint) when the Determinant is Non-zero

When the coefficient determinant of a 3-variable system is non-zero, the system is automatically consistent with a unique solution, and that solution can be obtained in one formula using the adjoint.

Steps

Step 1: Write $AX=B$ and compute $\det(A)$

Expand along the first row using the checkerboard sign pattern. A non-zero result confirms $A^{-1}$ exists and the system is consistent with exactly one solution.

Step 2: Compute all nine cofactors of $A$

For entry $a_{ij}$, the cofactor $C_{ij}=(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the $2\times2$ minor left after deleting row $i$ and column $j$. Keep the sign pattern $\begin{pmatrix}+&-&+\\-&+&-\\+&-&+\end{pmatrix}$ in front of you while doing this.

Step 3: Assemble the adjoint by transposing the cofactor matrix

$\operatorname{adj}(A) = \left(C_{ij}\right)^{T}$ …

Common Mistakes

Mistake 1: Using the cofactor matrix directly as the adjoint, forgetting to transpose it

Why it's wrong: the adjoint is the TRANSPOSE of the cofactor matrix, not the cofactor matrix itself. Here the cofactor matrix is $\begin{bmatrix}28&13&-19\\-2&10&5\\-17&-17&17\end{bmatrix}$, and only after transposing does it become $\operatorname{adj}(A)=\begin{bmatrix}28&-2&-17\\13&10&-17\\-19&5&17\end{bmatrix}$. Skipping the transpose swaps off-diagonal entries and produces a completely wrong $A^{-1}$, hence wrong $x,y,z$. Correct approach: always write out the cofactor matrix first, then explicitly transpose it (flip across the main diagonal) to get the adjoint.

Mistake 2: A checkerboard sign error on one of the nine cofactors …