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An invertible matrix need not have an invertible chosen diagonal block. For example,has singular matrices as both one-by-one diagonal blocks. Neither diagonal-block formula applies in that partition. Interchanging the two rows produces the identity matrix and a usable pivot element. This is why a failed block pivot is not proof that the full matrix is a singular matrix; Pivoted LU decomposition changes the order before eliminating.

Conversely, invertible diagonal blocks do not suffice: $M=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ has $A=D=1$ but Schur complement $S=0$, so it is a singular matrix. Both hypotheses in the block matrix inverse formula have work to do.

For $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$, the homogeneous quadratic coefficient matrix isThe conic is degenerate when $Q$ is a singular matrix. Examples over the real numbers include the point $x^2+y^2=0$, the repeated line $y^2=0$, and the two straight lines in $xy=0$. The real solution set alone need not reveal the algebraic distinction: $x^2+y^2=-1$ has no real points, but its matrix is nonsingular. This determinant test is different from the conic type discriminant, which only uses the quadratic part in $x,y$.

The determinant assigns a scalar to a square matrix. For a linear map on the real plane, it measures signed area scaling: its absolute value tells how much area grows, and its sign records whether the orientation of a vector space is preserved or reversed. In three dimensions, the same idea measures signed volume. Over a field, a square matrix is an invertible matrix exactly when its determinant is nonzero. These are two views of one phenomenon: losing an independent direction collapses full-dimensional volume and makes the original input impossible to recover uniquely.^[ref]

Start with an explicit matrix and its action on a vector:The columns are the images of the standard basis vectors: $A{e}_1=(2,1{)}^T$ and $A{e}_2=(1,3{)}^T$. Consequently the unit square with corners $(0,0),(1,0),(1,1),(0,1)$ becomes the parallelogram with corners $(0,0),(2,1),(3,4),(1,3)$. Its area is five. Every small patch of this square undergoes the same linear map, so the factor describes the whole deformation, rather than just one specially chosen boundary. The matrix mixes horizontal and vertical components, but its determinant compresses that combined effect into one area factor.^[ref]

Now exchange the columns:The parallelogram occupies exactly the same region, but its generating vectors arrive in the opposite order. Moving from the first column to the second now turns clockwise rather than counterclockwise. Ordinary area remains five; the oriented volume in two dimensions changes sign. A negative determinant therefore signals reversed orientation of a vector space, not a negative amount of material. Keeping the sign makes successive reversals calculable: two reversals restore the original orientation of a vector space.

Compare a singular matrix:Its columns are linearly dependent: the second is twice the first. The unit square collapses onto the segment from $(0,0)$ to $(3,6)$, which has zero area despite nonzero length. Since $C(-2,1{)}^T=0$, the kernel of a linear map contains a nonzero vector; inputs differing by that vector have identical outputs. The collapsed picture explains why an inverse matrix cannot exist.

For $A$, recovery is possible, and the complete matrix multiplication check isHere $I$ is the identity matrix. The inverse matrix restores the original square, shrinking its area by $1/5$. Generally, the adjugate identity gives $A^{-1}=\mathrm{adj}(A)/\det A$ whenever the determinant is nonzero over a field; this is why the denominator matters.^[ref]

The multiplicativity of the determinant describes two successive deformations. Matrix multiplication means that $AB$ applies $B$ first, then $A$. TakeThe first linear map doubles area; the second multiplies the resulting area by five. Thus the unit square passes through areas $1,2,10$, with the last parallelogram bounded by $(0,0),(2,1),(4,7),(2,6)$. Direct calculation agrees:This explains physically why area factors multiply instead of adding. The general identity holds for square matrices of the same size, including singular matrices. If either stage loses a direction, later processing cannot recover the discarded information.^[ref]

The same interpretation works in three dimensions without drawing a new picture. The diagonal matrixsends the unit cube to a rectangular box with side lengths two, three, and one. Its ordinary volume is six. The minus sign comes from reversing the third direction, so the oriented volume is minus six. Replacing the last diagonal entry by zero flattens the cube into a planar region: its area may be positive, but its three-dimensional volume is zero. This distinction matters whenever a linear map changes shape: the determinant measures size in the full input dimension, rather than the size of whichever lower-dimensional object remains.

The determinant measures neither trace nor perimeter. For $A$, the trace happens also to be five, but the identity matrix in two dimensions has trace two and determinant one. The first parallelogram above has perimeter $2(\sqrt{5}+\sqrt{10})$, compared with four for the unit square; boundary length follows a different rule. Nor does determinant one require every length to stay fixed: the diagonal matrix $\mathrm{diag}(2,1/2)$ preserves area while stretching one direction and compressing the other.

These properties make the determinant useful when coordinates obscure what a deformation does. A nonzero determinant certifies that linearly independent input directions remain distinguishable; its magnitude counts the area or volume represented by a transformed cell. That second role reappears in the change of variables formula, where the Jacobian determinant corrects the sizes of small cells. Signed volume also separates a deformation that reverses the orientation of a vector space from one that merely stretches it, even when both give identical ordinary volume factors.^[ref]

For general $n$, regard the columns $v_1,\dots ,v_n$ as ordered arguments. The determinant is the unique alternating multilinear map to the scalar field normalized by $\det I=1$. The multilinearity of the determinant means linearity in each column separately, with all others held fixed:Alternation means equal columns give zero; exchanging columns negates the result. Normalization sets the oriented volume of the standard basis to one. These properties also hold in rows. For real three-dimensional columns, their determinant is the scalar triple product, and its absolute value is the enclosed volume. Over other fields, the algebraic definition still works even without a geometric notion of positive volume.^[ref]

Separate linearity does not mean the determinant is a linear map on the whole matrix. Scaling every column of an $n$-by-$n$ matrix by $t$ gives $\det (tA)=t^n\det A$. For our two-dimensional example, $\det (2A)=4\det A=20$: doubling both directions quadruples area. The power records how many independent directions are being scaled together.

The properties suggest practical computations. Adding a multiple of one column to another preserves the determinant: the extra term has repeated columns and vanishes. This is the algebra behind an area-preserving shear mapping. Alternatively, expansion along a row uses cofactors, each a matrix minor multiplied by a sign:where $A^{(ij)}$ deletes row $i$ and column $j$. The Leibniz formula for determinants supplies the general entry formula. Both calculations encode the same geometric invariant; the defining properties explain their structure without requiring memorization of a separate recipe for every size.^[ref]

On real symmetric matrices of size $n\ge 2$, the displayed pairing is symmetric because $\det (AC)=\det A\det C$. It is not bilinear: $B(I,2I)=2^n$, whereas linearity would require $2B(I,I)=2$. Its self-pairing is always nonnegative, but a nonzero singular matrix has $B(A,A)=0$, so positive definiteness fails as well. For $n=1$ it reduces to the ordinary real inner product. The Frobenius inner product $\mathrm{tr}(A^{T}C)$ supplies the appropriate bilinear, positive-definite geometry in every size.

Consider this original linear system and its augmented matrix:The first diagonal entry is zero, so it cannot serve as a pivot element. Interchange $R_1$ and $R_2$ to obtain a pivot element of one:This chooses a convenient nonzero entry for exact arithmetic; partial pivoting would instead interchange $R_1$ and $R_3$, selecting the largest absolute value, two. For our chosen ordering, the first-column elimination multipliers are $m_2=0$ and $m_3=2$. The elementary row operation $R_3\leftarrow R_3-2R_1$ givesThe next pivot element is two and the next elimination multiplier is $7/2$. Applying $R_3\leftarrow R_3-\frac{7}{2}{R}_2$ gives the upper triangular matrix systemEvery operation also changed the right-hand side. Back substitution now givesChecking the original equations, rather than only the reduced ones, verifies the answer by matrix multiplication:Thus a zero first diagonal entry was a row-order obstruction, not a singular matrix. Indeed, the diagonal product of the final upper triangular matrix is $7$; the single row interchange makes $\det A=-7$, since the two row additions preserve the determinant.

In a linear Gaussian state-space model with process noise covariance $Q$ and exact scalar observation $X_t=F{S}_t$, the predictive state covariance includes $Q$ even though there is no separate measurement noise. Here $R_t=G{P}_{t-1}{G}^{\top }+Q$ denotes predicted state covariance; it differs from the measurement-noise notation in the general Kalman filter equations. The observation innovation variance is $V_t=FG{P}_{t-1}{G}^{\top }{F}^{\top }+FQ{F}^{\top }$. Omitting the second term incorrectly removes the current process noise. If $V_t>0$, Gaussian conditional expectation gives the posterior covariance matrix $P_t=R_t-R_t{F}^{\top }F{R}_t/V_t$. This can be a singular matrix because the observed linear combination is then known exactly.

For a concrete example with two state components that are dimensionless quantities, take predicted mean $(10,0{)}^{\top }$, predicted covariance matrixand exact observation of the first component, $F=(1,0)$, $X=12$. The innovation is $2$, its variance is $4$, and the Kalman gain is $(1,1/2{)}^{\top }$. The posterior mean and posterior covariance matrix are thereforeThe observed component now has zero variance. The second component remains uncertain, but its posterior mean changes and its variance falls from $9$ to $8$ because its prediction error had nonzero correlation with that of the first. This is the mechanism by which measurements of one component can inform another.

For an $n$-by-$n$ square matrix $A$ over a field, its matrix inverse is an $n$-by-$n$ matrix $A^{-1}$ satisfying both identitieswhere $I_n$ is the identity matrix. It exists exactly when $A$ is an invertible matrix, equivalently when its matrix rank is $n$ or its determinant is nonzero. It reverses the linear map: applying $A$ and then $A^{-1}$ recovers the input, while applying $A^{-1}$ and then $A$ recovers the output.^[ref]

The matrix inverse is unique. If $B$ and $C$ both satisfy the identities, then $B=B(AC)=(BA)C=C$. Although the definition is two-sided, for finite square matrices either identity alone suffices: $BA=I$ forces a trivial kernel, whereas $AB=I$ forces surjectivity; either makes $A$ an invertible matrix by the rank-nullity theorem. Rectangular matrices may have only one of these properties.

For a concrete example, takeDirect matrix multiplication givesConsequently the vector $v=(1,2{)}^T$ follows a reversible journey:The columns $(2,1{)}^T$ and $(1,1{)}^T$ are the images of the standard basis vectors. They turn the unit square into a parallelogram; the matrix inverse restores both the square and its surrounding grid. Since $\det A=1$, this example changes shape while preserving area, illustrating that invertibility does not require preserving lengths or angles.

For comparison, the singular matrixcannot have a matrix inverse: the output $(1,0{)}^T$ cannot specify which of the two inputs occurred. Their difference $(1,-1{)}^T$ belongs to its kernel. Indeed $S(x,y{)}^T=(x+y,0{)}^T$ collapses the plane to a line, its matrix rank is one, and its determinant is zero. No subsequent function of the output alone can reconstruct every input.

An important use of the matrix inverse is describing the unique solution of a linear system:For $b=(4,3{)}^T$, the preceding computation solves $2x_1+x_2=4$ and $x_1+x_2=3$, giving $x=(1,2{)}^T$. In computation, usually solve the linear system using Gaussian elimination and a factorization such as an LU decomposition, rather than forming every entry of the matrix inverse. The inverse formula specifies the answer; a factorization provides a reusable way to compute it for each right-hand side.^[ref]

Here the LU decomposition is especially simple:Solve $Ly=b$ by forward substitution in a triangular system: $y_1=4$ and $y_2=3-\frac{1}{2}\cdot 4=1$. Then solve $Ux=y$ by backward substitution in a triangular system: $x_2=2$ and $x_1=(4-2)/2=1$. The same factors work when $b$ changes. If the whole matrix inverse is actually needed, its $j$th column is the solution of $Ax=e_j$, where $e_j$ is the $j$th standard basis vector.

Reversing several operations means undoing the last one first. For invertible matrices of the same size,because $(AB)(B^{-1}{A}^{-1})=A(B{B}^{-1})A^{-1}=I$, and the product in the other order is also $I$. This is why a change-of-basis matrix uses its matrix inverse to reverse the direction of coordinate conversion.

A multivariable Newton iteration correction must solve the linear system $D\mathbf{f}(\mathbf{x})\mathbf{s}=-\mathbf{f}(\mathbf{x})$. It exists exactly when the right-hand side is in the image of the Jacobian matrix, and is unique when that Jacobian matrix is invertible. At a singular matrix, there may be either no corrections or several; singularity alone does not decide which. For finding a critical point of a scalar function $F$, use $\mathbf{f}=\nabla F$, so the Jacobian matrix is the Hessian matrix. Equal rows of the Hessian matrix with unequal corresponding gradient entries immediately make the correction equations inconsistent.

An odd integer number of rows and columns forces a real skew-symmetric matrix to be a singular matrix. The determinant satisfies $\det A=\det A^{\top }=\det (-A)=(-1{)}^n\det A=-\det A$, so it is zero. The argument also works over any field of characteristic different from two.

Multiplicativity of the determinant gives $⟨A,B⟩=\det A\det B$. Thus C1 holds. For $n\ge 2$, C2 fails: choosing $A=B=I$ and $\lambda =2$ gives $\det (2I)=2^n\ne 2$. The same choice $A=B=C=I$ disproves C3, since $\det (B+C)=2^n\ne 2$. The nonnegativity portion of C4 holds, because $\det (A^2)=(\det A{)}^2\ge 0$, but its positive-definiteness clause fails: a nonzero singular matrix that is symmetric, such as $\mathrm{diag}(1,0,\dots ,0)$, has zero self-pairing.

Therefore this is not an inner product for $n\ge 2$. For the exceptional case $n=1$, the determinant is the scalar entry and the pairing is ordinary multiplication of real numbers, so all four conditions hold. The determinant pairing is not an inner product in higher dimension because it measures volume scaling rather than a bilinear length geometry.

The determinant is $a^2-i(-i)=a^2-1$. The adjugate matrix formula givesWhen $a=1$ or $a=-1$ we have a singular matrix, so no inverse matrix exists. This criterion holds for real or complex $a$.

For $n\ge 1$, an eigenvalue is a number $\lambda \in \mathbb{C}$ for which there is a nonzero vector $v$ satisfying $Av=\lambda v$. Equivalently, $\ker (A-\lambda I)\ne \{0\}$, or $\det (A-\lambda I)=0$. The characteristic polynomial $\det (tI-A)$ has degree $n$ and leading coefficient one. By the fundamental theorem of algebra it has a complex root $\lambda$. At that root $A-\lambda I$ is a singular matrix, so it has a nonzero kernel vector, proving that $A$ has an eigenvalue.

For all but finitely many $s\in \mathbb{C}$, the matrix $A-sI$ is invertible. Part (a), applied to $A-sI$ and $B$, then givesBoth sides are polynomials in the two variables $s,t$. Since they agree for infinitely many values of $s$, their coefficients as polynomials in $s$ agree. Setting $s=0$ yieldseven when $A$ is singular.

For Newton's method for nonlinear systems, solve $J({\mathbf{z}}_0)\mathbf{\delta }=-\mathbf{F}({\mathbf{z}}_0)$ and set ${\mathbf{z}}_1={\mathbf{z}}_0+\mathbf{\delta }$. The Jacobian matrix here isAt $(0,0)$,The correction linear system is $3{\delta }_x-2{\delta }_y=1$, $-{\delta }_y=1$. Hence ${\delta }_y=-1$, ${\delta }_x=-1/3$, and one iteration givesAt the alternative starting value $(1,0)$,This Jacobian matrix is a singular matrix: its first row is twice its second. In fact the correction equations would bewhich are inconsistent, since twice the second left side equals the first while twice its right side is $0$, rather than $-3$. Thus Newton's iteration cannot take its first step from $(1,0)$. This is failure of the correction linear system, not merely nonuniqueness of a possible correction.

To find a critical point, apply multivariable Newton iteration to the gradient, rather than to the scalar value $F$. The gradient and Hessian matrix areWrite the correction as ${\mathbf{s}}_0={\mathbf{x}}_1-{\mathbf{x}}_0$. At ${\mathbf{x}}_0=(-1,-1)$, Newton's method in optimization requiresSubtracting the second equation from the first gives $10s_y=7$, so $s_y=7/10$ and $s_x=13/10$. HenceThe gradient there isA critical point has not been found after one iteration.

At the alternative starting point $(0,2/3)$, the Hessian matrix and gradient areThe correction would need to satisfy $2s_x+2s_y=-4/3$ and $2s_x+2s_y=0$ simultaneously. This linear system is inconsistent. The standard iteration cannot take even its first step: the Hessian matrix is a singular matrix, and the required correction does not exist. This is a Newton correction compatibility condition failure, not merely an ambiguity between several corrections.

The coefficient matrix has determinantA homogeneous linear system has a nonzero solution exactly when this is a singular matrix. ThereforeAt this value all solutions are $(x,y,z)=t(1,-2,1)$, with $t\ne 0$ for a nonzero solution.

The inverse appearing in the assumption requires $\det C\ne 0$. Taking determinants of $(BC{)}^{\mathsf{T}}=-C^{-1}B$ giveswhere the minus sign comes from the odd matrix order. Multiplying by $\det C$ gives $\det B((\det C{)}^2+1)=0$. Since $C$ is real, the second factor is positive. Thus $\boxed{\det B=0}$ and $B$ is a singular matrix. This is the singularity forced by an odd-dimensional transpose-inverse identity, with the roles of the two matrix letters exchanged.

Take determinants of $(CB{)}^{\mathsf{T}}=-B^{-1}C$. In order three, multiplication of a matrix by minus one changes its determinant's sign. HenceThe assumption $\det (B^{\mathsf{T}})\ne 0$ is the same as $\det B\ne 0$, so multiplication by $\det B$ givesSince the matrices are real, the first factor is strictly positive. Thus $\boxed{\det C=0}$, proving that $C$ is a singular matrix. This is singularity forced by an odd-dimensional transpose-inverse identity; no orthogonality of $B$ was used.

Collecting the coefficients gives the coefficient matrixA square homogeneous linear system has a nonzero solution precisely when its matrix is a singular matrix. Expanding the determinant down the first column,Hence the necessary and sufficient condition is $a=0$ or $b=c$ or $b=-c$. A zero determinant gives a nontrivial kernel of a linear map; otherwise multiplication by the matrix inverse would force the zero solution.

If $I+A$ were a singular matrix, a nonzero vector $v$ would satisfy $(I+A)v=0$, so $Av=-v$. This would make $-1$ an eigenvalue of the anti-Hermitian matrix, contrary to part (a). Thus $\boxed{I+A\text{ is invertible}}$. The same argument shows $I-A$ is an invertible matrix, which will also be used below.